help-19

[ember shell]

Positive infinity

[signal oar]

Well think about what you get if you plug in e.g. 1

[ember shell]

Uhhh ok

[signal oar]

$f(x) = 3 \sqrt{2x - 7} - 8$.

$f(0) = 3 \sqrt{2(0) - 7} - 8 = 3 \sqrt{-7} - 8$.

[ember shell]

Fx equals to 3sqrt-5 minus

Or smth

[ember shell]

I thought we were doing 1

😭

[signal oar]

Yeah, we can try x = 1 too

So for x = 0, what can you say?

Is it fine or not

[ember shell]

No

I don't think so

[signal oar]

(Why?)

[ember shell]

Uhhh it's not like a full answer ig? It's not a single number like 1 for example

Idk

[signal oar]

The problem is the square root

[ember shell]

Ye

[signal oar]

The square root can only have a positive argument

[ember shell]

Oh

[signal oar]

But in this case, it is $\sqrt{-7}$.

[ember shell]

So it has to be 8 or above?

The domain?

[signal oar]

Don't jump to that conclusion too fast.
What do we need for $\sqrt x$ for it to be defined?

Like what restriction can you put on x

[ember shell]

For z to be positive

X

Uhhh

[signal oar]

Yes

and 0 too, right?

sqrt(0) = 0, that's fine, defined.

[ember shell]

The sqrt of 0 is just 0

Tho

[signal oar]

Yeah, so it's defined too

[ember shell]

Oh

Ok

[signal oar]

We only worry about things like $\sqrt{-1}$, $\sqrt{-2}$ or $\sqrt{-4}$. \ Can you think of their values? Because there actually is no value in $\mathbb R$.

[ember shell]

Uh...

[signal oar]

For a solution of sqrt(-1), we would need something squared to be equal to -1

And we can't have that

[ember shell]

Ye

[signal oar]

Anything squared is greater or equal to 0

[ember shell]

Ic

[signal oar]

So for $\sqrt x$, we can put the restriction $x \geq 0$.

[ember shell]

Oh

Ok I think I get ut

It

[signal oar]

Now here, you don't just have x, but 2x - 7

But the same holds, for $\sqrt{2x - 7}$, we need $2x - 7\geq 0$.

[ember shell]

So would that be like x>= 8

Wait no 7

[signal oar]

Not really, be careful, $2x - 7 \geq 0$ is what we need.

[ember shell]

Oh yeah 2x

I forgor

Uhh

[signal oar]

You can treat this like an equation

Add 7 to both sides

[ember shell]

Like then it would be

4.4

[signal oar]

2x >= 7

[ember shell]

4.t

4.5

😭😭

[signal oar]

x >= 3.5

Yeah, almost, I think that was just a wrong calculation on your part, but the idea should be correct

[ember shell]

Ye

Ok

[signal oar]

Yeah, so we need x >= 3.5

[ember shell]

Is thst how you write it?

Or is it something else

[signal oar]

There is nothing else in the function that could give you something undefined, only the square root is what you have to worry about, because a square root of a negative number is not defined

[signal oar]

You can write it in different ways

[ember shell]

Oh

My teacher taught is with the parenthesis

Thingies

[signal oar]

Interval notation: $[3.5, \infty)$. \ Set-builder notation: ${x \in \mathbb R \mid x \geq 3.5}$ are two examples.

[ember shell]

Ye

Ok so how do I find the range then?

Ehm.....

@signal oar 😭

[signal oar]

It's all the values that your function can take on. [f(x) = 3 \sqrt{2x - 7} - 8.] What values can a square root take on?

[ember shell]

Oh wait

So would the range be uhhh

[3.5,infinity)?

Hello?

...ok

[signal oar]

Nope

[ember shell]

Oh

Uhhhh

[signal oar]

What values can a square root take on?

Like you can imagine its graph

[ember shell]

Positive ones

[signal oar]

,w plot sqrt(x)

Yeah, including 0, as you can see

[ember shell]

Ye

[signal oar]

Now, $3 \sqrt{2x - 7}$ will go from $0$ to $\infty$ (since $3$ is just a factor on it, and the $2x - 7$ is picked to be $\geq 0$ either way, so it's just like $\sqrt x$) \But we have a $-8$ on it too. That will shift the range

So it will not go from just 0 to infinity, but from ... (hopefully you can fill this in) to infinity

[ember shell]

Oh since it's 2x?

So it could be -3.5 technically

Right

?

[signal oar]

(don't consider the 2x - 7, it's inside of the square root and not relevant here)

[ember shell]

Ok

So it would be 7 to infinity right?

[signal oar]

We already know that of our function [f(x) = 3 \sqrt{2x - 7} - 8,] the first part, $3 \sqrt{2x - 7}$ is from $0$ to $\infty$. \ It'd not be $7$ to $\infty$, no. \ So we take that $0$ to $\infty$ and shift it down by $8$ because the second part is a $-8$.

So now it's not from 0 to infinity, but from -8 to infinity

[ember shell]

Oh

So it could be

[-8,infinity)?

[signal oar]

Yeah

[ember shell]

Ohh

Ok ic

So then for 16.28

It would be like

Uhh

Infinity? As the domain or smth

Idk I'm kinda stupid

[signal oar]

[f(x) = \frac{x + 2}{x - 2}.] What $x$ do you think can you plug in? \ (Or asked in reverse: Are there any $x$ that make this impossible to determine (undefined))?

[ember shell]

Uhh like you can plug in 0 right?

[signal oar]

Yeah, sure

2/-2

That's -1

[ember shell]

Ye

Is thst not allowed

?

[signal oar]

It is allowed

[ember shell]

Oh ok

[signal oar]

If you can determine a value for it, then it's allowed

I could determine it's -1

So it's allowed

[ember shell]

Oh wait so would the domain be

Like

Infinity

Negative infinity

Two open brackets?

[signal oar]

(-infty, infty)?
Almost, but there is one value you forgot to consider

[ember shell]

Ihh what

Uhh

[signal oar]

Consider $\frac 1 x$. For what value would that be impossible to determine (undefined)?

[ember shell]

Oh is it 2

[ember shell]

😭

[signal oar]

Yes

[ember shell]

Ah ic

[signal oar]

Because then, you have 4/0

We don't know what that is

[ember shell]

Ye

But every other value works?

[signal oar]

So everything except 2

[ember shell]

How do you even write thst down

😭

[signal oar]

Yeah, only the denominator can cause problems

Because you can't divide through 0

[ember shell]

Ye

[signal oar]

[(-\infty, 2) \cup (2, \infty)] or [{x \in \mathbb R \mid x \neq 2}.]

[ember shell]

No

[signal oar]

Then maybe you are supposed to use the second

[signal oar]

How did you in your class write this? With curly braces {} (set builder notation) or in interval notation (so for example [5, 10])?

[ember shell]

I have no clue...

The second one doesn't look any better tbh

[signal oar]

Well, the $\cup$ symbol means "union". \ So for example, with $[1, 2] \cup [4, 5]$, you are saying "1 to 2 OR 4 to 5"

[ember shell]

Oh

[ember shell]

Oh ic

[signal oar]

And we are saying here that it's from - infinity to 2), so not including 2, or from (2 to infinity, again not including 2

So in total, we did not include 2

But every other value

[ember shell]

Ohh

[signal oar]

You could also say $\mathbb R \setminus {2}$.

All of R without 2

That's probably the shortest

[ember shell]

Oh

Ok

So for example loke

Like

20.1.1 for the domain would this be correct?

[signal oar]

Yeah

[ember shell]

Ok and then the range would be like

Uhh

Give me a second

I'm stupid

[signal oar]

Saying that is useless and just unnecessary, don't...

[ember shell]

Uh ok

[signal oar]

So you have $\sqrt{2 + x}$, that's just the same as $\sqrt x$ when looking at the range.

And the range of $\sqrt x$ is $[0, \infty)$, right?

[ember shell]

Uhh ok

[ember shell]

Yee

Yes

[signal oar]

So now you have a -5

[ember shell]

Yeah

[signal oar]

That shifts everything down by 5, so your range will be \$[0 - 5, \infty - 5] = [-5, \infty]$.

[ember shell]

Yeah

So ur saying

That's what the range would be?

[signal oar]

Yes

[ember shell]

Why are they both closed tho?

[signal oar]

Oops, the one at infinity shouldn't be closed ofc

The one at -5 should be closed

[ember shell]

Oh lol

😭

[signal oar]

infinity is not a number, so we never close that one

[ember shell]

Yeah okk

[signal oar]

I made a typo

[ember shell]

No it's ok

[signal oar]

You can also imagine this graphically

,w plot sqrt x

[clever fjord]

Results provided by WolframAlpha

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Requested by Kepe#5235

[ember shell]

So the problem to the rights would be [7,infinity)?

[signal oar]

And now we are shifting that down by -5

,w plot sqrt x - 5

[clever fjord]

Results provided by WolframAlpha

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

Requested by Kepe#5235

[ember shell]

Oh

Thst makes a lot of sense now ye

Btw

Ehst does the @ symbol mean in math

😭

[signal oar]

Which one do you mean?

[ember shell]

I have no clue uhh ok wait let me show you ig

[signal oar]

Sure

[ember shell]

this should be like the second to last thiing

uhhh

idk it was given to me

[static totem]

they say what it means

[ember shell]

uhhh

wut

i just dont understand the format then

or what its trying to ask me

ig

[signal oar]

It's not a common symbol, they define it as that expression

[ember shell]

oh ic

[signal oar]

We can define any symbol to mean anything

[ember shell]

oh so its just there to make me pissed or smth

good to know

[static totem]

in a sense

[ember shell]

lol

so if we're not doing a square root problem

[signal oar]

They want to say $\frac{a^3 - b^3}{a - b}$ shorter than having to write all of that out all the time for all variables $a$ and $b$, probably.

[ember shell]

oh ye

ok

[ember shell]

uhhhhh

elaborate

pls

[signal oar]

$a@b$ is defined to mean $\frac{a^3 - b^3}{a - b}$. So now you can say $1@3$ and it means the same thing as [\frac{1^3 - 3^3}{1 - 3}.]

[ember shell]

ohhh

ic ic

[ember shell]

ok last thing

uhh

i think i already asked too meth bleh

much

i hate spelling

uhh

so if the problem isnt a uhh

square root problem thing

uhh lemme show you

[signal oar]

You can ask as much as you want, this is your help channel, lol

[ember shell]

oh yeah

its because this is like

stuff i needa catch up on

😭

so i haev to figure out most of it out by muself

uhh

so for example

on this problem

how would i find the range

since its not the same format

as the other ones

did i confuse you?

[signal oar]

Think about what this takes on for large x

[ember shell]

wdym large x

[signal oar]

It'll be a huge number divided by a huge number

And the +2 and -2 won't matter

Like imagine we pick $x = 100000$.

[ember shell]

a

ah ic

uhh that huge number would still work though technically

[signal oar]

Then $f(100000) = \frac{100000 + 2}{100000 - 2}$. That is $\frac{100002}{99998}$. This will be a bit larger than $1$ but very near to it.

[ember shell]

yeah

[signal oar]

Now take an even larger number. f(1000000000000).
That will be even nearer to 1, only a tiny bit larger than it.

[ember shell]

ok

[signal oar]

Well, it will never reach 1 if we take larger and larger values.

[ember shell]

so would that just be even closer to 1?

[signal oar]

It will always be a tiny bit larger than it

[ember shell]

so the range would be uhh

[signal oar]

Wait, now look at x to -infinity

[ember shell]

oh

ok so for example like

-4?

that would be like

1/3

right?

after being simplified

[signal oar]

$f(-100000) = \frac{-100000 + 2}{-100000 - 2} = \frac{-99998}{-100002} = \frac{99998}{100002}$, just a bit smaller than $1$. For even smaller (more negative $x$), it'll also just always be a bit smaller than $1$, but never reaching $1$.

So for x to +infinity, it doesn't reach 1, for x to -infinity, it doesn't reach 1

So now let's look at when x gets near to 2, since that's where there is a hole in the domain

[ember shell]

Uhhh

Oh

It's a little bit less than one

So would it just be [-1,1]

[signal oar]

From the left (so $x < 2$), if we go very near $2$ (but we're not allowed to reach it, since it's not in the domain), if we pick for example $x = 1.99$, then we'll have [f(1.99) = \frac{1.99 + 2}{1.99 - 2} = \frac{3.99}{-0.01} = -399.] For values even nearer to $2$, left from $2$, the range will go to $-\infty$. \[10pt] Now from the right (so $x > 2)$, if we go very near $2$, if we pick for example $x = 2.01$, then we have [f(2.01) = \frac{2.01 + 2}{2.01 - 2} = \frac{4.01}{0.01} = 401.] Now for values even nearer to $2$, right from $2$, it'll got to $\infty$.

[ember shell]

Ok uhh

So thst would be the rsnge?

Or am I just wilding

[signal oar]

Ok so now imagine the graph, from the left, it approaches 1 from the bottom (since it's less than 1). Then, it suddenly becomes very negative.

Now, starting at the very right, it approaches 1 from the top (since it's greater than 1). Then, it suddenly becomes very big.

The graph will probably help you follow all my arguments

[ember shell]

....huh

[signal oar]

,w plot (x+2)/(x-2)

[ember shell]

Ah ic

[ember shell]

Ok

[signal oar]

So the range is $\mathbb R \setminus {1}$.

[signal oar]

After this argumentation

[ember shell]

....

Ok 😭

Is there no other way to write it?

[signal oar]

The range?

Or the argument

[ember shell]

Uh the range

😭

[signal oar]

$(-\infty, 1) \cup (1, \infty)$ or ${x \in \mathbb R \mid x \neq 1}$ are some.

[signal oar]

I'd personally prefer the R \ {1} one

[ember shell]

Oh ic

U r very helpful

[signal oar]

I'd suggest really take your time with these, imagine how it looks like, plotting/sketching these quickly can help

[ember shell]

Oh okk

I don't wanna sound like a creep but can I friend you if I need help later

Because this is very confusing

And there's like a lot more things I needs do today

😅

[signal oar]

I'm not that much on Discord, but if you open a help channel, there'll pretty much always be someone available

[ember shell]

Oh ic

Ty anyways

Lol

That's about it

Bye :D

[signal oar]

Bye

[ember shell]

.close

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[tidal barn]

So

[tidal barn]

How many arrangements of the letters A, B, C, ..., O, P exist such that by removing some letters, one cannot obtain any of the words PONK, DOBA, or COP?

A-P is 16 numbers, therefore 16!

How to continue

[odd edge]

@tidal barn Has your question been resolved?

[leaden quiver]

I wonder if its easier to remove all cases that include the words ponk doba and cop

[tidal barn]

You must remove some sequences

Because for example POABENK

is actually what cant be in it

Cuz u remove ABE

and u have PONK

idk how to do ir

[open onyx]

the only interesting part is to prevent overcounting

find the smallest set of removed letters that satisfies the property the question asked for

[tidal barn]

Well

The theorem

Principal of inclusion and exclusion

Needs to be used

[open onyx]

we’ll get to it

[tidal barn]

U think we can do it quickly in 15mins? Ive been sitting on this for 2h, i need to submit it to teacher

Rn i have there 16! - (16 4) - (16 4) - (16 3) + 3

16 4 thing is combinatirict number

[open onyx]

cutting down from 16! is fine but i don’t know what your idea with 16c4 and 16c3s is

[tidal barn]

Its the number of how to choose 4 positions among 16 total

For PONK

then for DOBA

and finally for COP

[open onyx]

so if you remove ABCD for PONK (?) then you have 12 letters left
how does the second 16c4 work

[tidal barn]

I dont know how to explain tho

I choose 4 positions for word PONK among 16

POABCENFGHK

Example

[odd edge]

@tidal barn Has your question been resolved?

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[harsh prairie]

i am so lost solving this question... can anyone help me explain how to find the values of a and b given I found the equation to cn+2?

[harsh prairie]

The correct answers I've gotten was by plugging in a value for n then testing the result in all the inputs so, altough I got most of them correct, idk wth is going on

[valid spoke]

that looks hard. what grade you in?

[harsh prairie]

uni

[valid spoke]

university?

[harsh prairie]

<@&286206848099549185>

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@harsh prairie Has your question been resolved?

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@harsh prairie Has your question been resolved?

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[glass shadow]

i need some more help

[glass shadow]

how would i do these

[forest sky]

,rotate

[forest sky]

take the two points on the lines given. pick one point as your start point and one point as your end point.

• find the "rise": how many grid lines above or below is the second point compared to the first point? this is positive if it's above, negative if it's below.
• find the "run": how many grid lines left or right is the second point compared to the first point? this is positive if it's to the right, negative if it's to the left
• find the slope: the slope is equal to rise/run. make sure you keep the proper sign

[glass shadow]

how would i do this?

help

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[celest grotto]

can someone help me with the first question

[celest grotto]

the answer is no

i guess it’s because quadratic doesn’t have an inverse function?

[half fulcrum]

ans is yes i believe

oh wait idk actually

[serene terrace]

f is a straight line not quadratic

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[fossil moth]

Question of relative velocity

I find it difficult to understand, can someone please help

The answer is a

[static totem]

doesn't make much sense

oh the units are different

[fossil moth]

Likewise, that's why I asked the question lol

Oh

Just noticed as well

[bronze epoch]

First, convert all units to m/s. Then, the velocity of the bullet relative to the thief is: velocity(police) + muzzle velocity - velocity(thief)

[fossil moth]

Ok

Why are we adding the velocity of muzzle though?

I thought it was already relative to the ground, seems like it's relative to police

[bronze epoch]

The muzzle velocity is relative to the gun, which is traveling with the police at 45 km/h

[static totem]

that's what it means, how much speed the bullet gains relative to the gun

[fossil moth]

Ohh, that much was Intuitive I think because when it said muzzle velocity, it didn't mention relative to what

[fossil moth]

Ohh

[fossil moth]

I'm getting 72*5/18

20m/s

i found the velocity in km/hr first then I'm converting into m/s..the answer is not matching..

[noble forge]

did you convert muzzle velocity to km/h

[fossil moth]

Ohh no

I didn't see that one was also in m/s mb

Oh now I'm getting 150m/sec

Thanks

.close

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[fossil moth]

This one is more confusing

If the driver finds other train to be moving in same direction on same track with velocity v, the other train should be move ahead so no collision would happen

And if it's from behind, the driver should be accelerating, not retarding

[signal yacht]

what about none

[fossil moth]

Lemme see answer

[fossil moth]

Looks like the answer is A instead

[signal yacht]

oh

it says v1 v2

ig your question has a typo

[fossil moth]

I can't form a descriptive image of the sutuation

Ohh

[signal yacht]

yess

[fossil moth]

.close

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[rose hare]

hello

[desert marlin]

.close

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[mystic saffron]

Correct Right?

[desert marlin]

looks fine

[mystic saffron]

Alright

[mystic saffron]

i think this is good aswell

Just double checking

.close

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[desert marlin]

!occupied

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[fast dirge]

help i dont know how to do this

<@&286206848099549185>

[desert marlin]

!15mins

[odd edge]

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.

[desert marlin]

!status

[odd edge]

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

[fast dirge]

1

[desert marlin]

do algebra

[fast dirge]

wha

[desert marlin]

use algebra

to solve for y2

you know how you can like

multiply both sides of an equation by the same thing

add or subtract the same things from both sides

right?

[fast dirge]

yeah

[desert marlin]

okay

so do that here

try some things

you probably don't want that denominator

try to get rid of it

undo the division

what undos division?

multiplication

right

so that's the first thing you should try

[fast dirge]

can u jus show me step by step on how u would solve it

.close

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[cunning bronze]

10th sum

[lyric thicket]

what do I have to look at?

sorry

[odd edge]

@cunning bronze Has your question been resolved?

[cunning bronze]

Question no 10th

[odd edge]

@cunning bronze Has your question been resolved?

[odd edge]

@cunning bronze Has your question been resolved?

[odd edge]

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[civic storm]

Problem: For any given married couple, P(husband retired) = 0.7
P(wife retired) = 0.4
P(husband retired given wife retired) = 0.8
Choose 2 couples, P(one wife retired and one husband retired) = ?

[civic storm]

Where's my mistake:
P(1 wife retired) = 1 - 0.4² - 0.6² = 0.48
P(1 husband retired | one wife retired)
= 1 - P(2 husbands retired | 1 wife retired) - P(0 husbands retired | 1 wife retired)
= 1 - 0.8 • 0.7 - (1 - 0.8) • (1 - 0.7)
= 1 - 0.8 • 0.7 - 0.2 • 0.3 = 0.38
Answer = 0.38 • 0.48 = 0.1824
(correct answer = 0.2016)

[boreal crag]

Probably in the P(one husband retired given one wife retired part)

I'm not sure how you calculated that

[civic storm]

Adding a footnote

[boreal crag]

The logic doesn't seem right

[civic storm]

Edited

[boreal crag]

Hmm

Yes recheck that step

What's P(0 husbands retired | 1 wife retired) and why

[civic storm]

Edited again, but it should be 0.2 • 0.3

[boreal crag]

no, I don't think it should be

you need P(husband retired | wife didn't retire)

Which is definitely not 0.7

[civic storm]

Sorry I'm not used to conditional prob notations

P(0 husbands retired | 1 wife retired) is (1 - 0.8) • (1 - 0.7)
Former for husband with retired wife, latter for husband with unretired wife

[boreal crag]

ok, P(2 husbands retired | 1 wife retired)

The .8 is for the husband who's wife retired, right

What about the .7

[civic storm]

Husband whose wife not retired

Wait

!!!

[boreal crag]

But .7 isn't the probability of a husband retiring, given his wife didn't retire

[civic storm]

P(husband retired | wife not retired) is not just 0.7

[boreal crag]

Right

[civic storm]

Let me work this out

Yea thanks a lot

I forgot that having a condition changes everything

I'd go with the Venn Diagram method

.close

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[mortal stump]

is taking lim of ln(1+a)^1/a the same as ln [lim of (1+a)^1/a] ?

[boreal crag]

Yeah

Because ln is a continuous function, you can move it inside/outside limits

[mortal stump]

but when u have ln8 and ln2 ^3, they're not the same

[boreal crag]

oh yeah they wrote it down poorly

It should be $\ln((1+a)^\f1a)$

[clever fjord]

thewizardofOU

[boreal crag]

As written down

[mortal stump]

oh i see

I misremembered how ln works

I thought the 1/a goes on the outside

thxs for clearing that up

.close

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[boreal crag]

yw

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[viscid flint]

ye

[odd edge]

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[mystic saffron]

how to solve for a?

[mystic saffron]

ik my hand writting is bad but you guys get the point

[errant jay]

Whats the rest of the question

[mystic saffron]

thats it

owhw

wait

now

CP = PH = OH = CO

and CP = 5

[mystic saffron]

THIS IS NOT 3D BTW

WAIT isnt a = 2.5?

.close

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[fervent hound]

$\int \sqrt{1-x^3} \dd x$

[clever fjord]

Adam Chebil

[fervent hound]

How do I evaluate this??

[formal ermine]

difference of cubes?

[fervent hound]

Then?

[formal ermine]

maybe a u sub

have a go and send your working

[fervent hound]

I alreasy tried that, but I didn't know what to do after difference of cubes

[fervent hound]

${ \int \sqrt{ \left( 1-x \right) \left( 1+x+x^2 \right)} \dd x$

[formal ermine]

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

[cold urchin]

U sub doesn’t work here

[formal ermine]

i realised

[cold urchin]

This integral doesn’t have elementary solution

You can use binomial to get the first few terms but that’s it

,w integral of sqrt(1-x^3)

[clever fjord]

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

Results provided by WolframAlpha

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

Requested by pure1144

[fervent hound]

How did u know that?

[cold urchin]

Because I’ve seen it before

[fervent hound]

So there's no function such that its derivative is sqrt(1-x³) ??

[cold urchin]

Not elementary ones

[fervent hound]

What does elementary mean here ?

[cold urchin]

see https://en.m.wikipedia.org/wiki/Elementary_function

[fervent hound]

Ok ty

.close

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[wintry viper]

For 8a?

[wintry viper]

Can some tell me is this is right?

[worn wren]

jsut to confirm, what is on thy 3rd row from the bottom, you wrote (1/2)e raised to the power of -20(1/2), or to the power of -2(1/2)?

either way, -20(1/2) nor -2(1/2) arent equal to 1

[wintry viper]

My bad my hand writing pretty bade it’s supposed to be -2x

It’s supposed to be -2xe^-2x + e^-2x

=0

[worn wren]

(1/2)e^-2(1/2) is not equal to (1/2)e^1

so the second line from the bottom is wrong

[wintry viper]

Oh

[worn wren]

do you want to know what exactly you missed or should i let you figure it out yourself?

[wintry viper]

Excatly what I missed I have been stuck for a while

[worn wren]

-2(1/2) is not equal to 1, but -1

[wintry viper]

🤦

I feel so dumb

[worn wren]

other than that, everything is perfect (i think)

[wintry viper]

Ok thank you

[worn wren]

dw i once told my teacher that 2+2 = 2

[wintry viper]

😭sometimes maths does that to us

Thanks again

.close

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[quasi sparrow]

what are x_i and how do they relate to your matrix A

you still need to define the x_i

yes them being orthogonal is important

you need to use that fact in your proof

also you should do part a before doing part b

none of your calculation proves anything

struggling with what exactly?

part a) tells you to show $||x_1 + x_2||^2 = ||x_1||^2 + ||x_2||^2$

[clever fjord]

riemann

[quasi sparrow]

assuming $x_1, x_2$ are orthogonal.

[clever fjord]

riemann

[quasi sparrow]

use that and expand the left side

you've expanded them both correctly so far

but you haven't proven equality yet

apply the first or second one

yes

that's not how proofs work

you start with one side ONLY and do algebra on it until you get the other side

you can basically delete everything to the right of the equal signs

no

delete doesn't mean replace with original

like if you type
abc = def

and delete def

what do you get

yes delete and remove are synonyms

still haven't removed those

after your last line here, then you use orthogonality to get the result

look up the definition of orthogonal

should be in your book/notes

[clever fjord]

QuasiStar 超新星

[quasi sparrow]

this only makes sense in 2 dimension

you have an n dimensional vector

the definition of orthogonal in n dimensions is their dot product is zero

so this is correct

[odd edge]

@hazy gazelle Has your question been resolved?

[quasi sparrow]

You keep putting things you need to delete

Everything to the right of the equal sign on the first two lines should be deleted

[odd edge]

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[lament walrus]

!show

[odd edge]

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

[south plume]

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{bc+ac+ab}{abc}$

[clever fjord]

artemetra

[south plume]

that's clear right?

i made it into one fraction

the lcm of a, b, and c is a*b*c

[south plume]

i'm just asking, did you understand what i did here

not yet bro

[lament walrus]

haha be patient

[south plume]

i'm not giving away solutions like that

$$\frac{bc+ac+ab}{abc}=1$$
$$\implies bc+ac+ab=abc$$

[clever fjord]

artemetra

[lament walrus]

if the sum of three number are equal to 1 and the multiply of the three numbers also equal 1 what thats mean?

[south plume]

a=1-b-c
->
bc+(1-b-c)c+(1-b-c)b=(1-b-c)bc

bc+c-bc-c^2+b-b^2-bc=bc- b^2c - bc^2

[slate radish]

-> (b-1)(c-1)(b+c) = 0

[lament walrus]

u still didnt solve or ur done?

its not too hard just think a lil bit

i think we dont solve here?

🙂

[mystic saffron]

bro what is that 😭 😭
god i know i will hate maths even more in the coming years

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[gentle abyss]

help

[gentle abyss]

i need help

[upbeat hearth]

with

[gentle abyss]

the square root of 4

[upbeat hearth]

ur trolling?

[gentle abyss]

no i am new to this

i am homeschooled and my mom isn’t home right now

[upbeat hearth]

u can legit type it into google and big 2 will show up

its 2

[gentle abyss]

🔥

ok

[keen mason]

when a*a = b, then sqrt(b) = a, for a >= 0

if ur actually not trolling

[upbeat hearth]

?

not true xd

now true

[keen mason]

mb dude

[upbeat hearth]

[keen mason]

https://media.discordapp.net/attachments/954963057514258452/1162769717472284723/Untitled-1.gif?ex=653d24af&is=652aafaf&hm=d84c9648469e29cb9d215e95cf638d17e76dbbdd9af7c314128674e54bcf08bd&

[gentle abyss]

huh

[keen mason]

" Square root of a number is a value, which on multiplication by itself, gives the original number"

[odd edge]

@gentle abyss Has your question been resolved?

[odd edge]

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[desert marlin]

I've been working on this proof. After trying a few values, it's like 1.6... so I'm thinking it is the golden ratio. Anyways, the sequence isn't monotonic, it alternates like up and down.
So I tried some subsequences
like the even subsequence I have proven is increasing, and the odd is decreasing
if I have that they are both bounded in the proper direction respectively both by phi, then I'm done
but I don't know how to get started on showing this bound. Any help?

[summer cradle]

do you have any facts that will help you prove (a_n) converges?

not find the limit but just that it converges

[desert marlin]

If it was monotone and bounded

which it is not

if it was cauchy

which is hard

if both the even and odd subsequences converge to the same value (which is what I'm trying)

since the even and odd subsequences are monotonic I just have to show they're bounded

[summer cradle]

i'll just add that if it does converge, say to $X$, then we have $$\lim_{n\to\infty} a_n = 1 + \frac{1}{\lim_{n\to\infty} a_{n-1}}$$ or equivalently $$X = 1 + \frac{1}{X}$$

[clever fjord]

layla

[summer cradle]

which does narrow it down to the golden ratio

[desert marlin]

right

[summer cradle]

(it's either the golden ratio or that other solution but that one is negative)

[desert marlin]

right

so maybe it'd be easier to show it is cauchy

but I got stuck trying that aswell

[summer cradle]

i'm not sure yet but just wanted to point out it will be enough to show (a_n) converges

[iron bear]

very fibby

[summer cradle]

we could like, do some linear algebra

but not sure that's the intended idea

[iron bear]

i will note that 1.5 <= a_n <= 2 for n>1

[desert marlin]

Bounded yes but not monotonic

[iron bear]

(so you can prove that the sequence is contractive)

[desert marlin]

I'm not sure what contractive means

[iron bear]

|a_(n+1) - a_n| <= r|a_n - a_(n-1)| for some constant r<1

[desert marlin]

hm I don't think that's the expected way, never seen it before

[summer cradle]

snow you're so smart

i'm sorry i called you bad at math i take it back

[desert marlin]

does that work?

[iron bear]

you can also try this

the sequence is always rational

and the numerator/denominator are the fibonacci numbers

[desert marlin]

So are you both indicating that it would not be possible to show these are odd/even subsequences are bounded ?

that's what I wanted to try

but if I can't do that I can try other stuff I guess

[iron bear]

well they're certainly bounded

[desert marlin]

Right but bounded by phi

like

I've shown the even sequence is monotone decreasing

if I can show it has GLB=phi

[iron bear]

i think you can just prove that they converge to the same limit

[desert marlin]

right

[iron bear]

without bounding by phi

[desert marlin]

I was trying to use monotone sequence theorem

[iron bear]

yes

[desert marlin]

to prove they converge to the same limit

how would you do it without that?

[iron bear]

well we know the sequence is bounded

and the odd/even terms are monotone

so that guarantees their convergence

then you know what they converge to

[desert marlin]

well I know

but I haven't shown it

without knowing their sup and inf

right?

[iron bear]

the recurrence relation will tell you exactly what they converge to

[desert marlin]

OHHH

I CAN DO ALGEBRA ON THAT

[iron bear]

because of what layla wrote before

yeah

[desert marlin]

I can do that algebra

not on the whple sequence

but on the subsequences

too

okay

dang

[iron bear]

ye

[desert marlin]

very nice

ty snow

[iron bear]

[desert marlin]

I noticed some mistakes in my algebra for one of the directions but other than that I'm good now

.close

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[iron bear]

great

[odd edge]

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[rain tundra]

How's the sum of roots equal to zero here?

[low locust]

vieta

multiply out (x-x1)(x-x2).. and compare the coefficient of x^3 with the given polynomial

[rain tundra]

hmm okay I get it now, thanks

.close

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[lethal cipher]

Ola

[lethal cipher]

I need help rq

These different parameters

ax+bx+cx+d

d is on y-achse

c on x-achse

a is amplitude

what does b do

[odd edge]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[lethal cipher]

all good

[boreal crag]

what

that makes zero sense

[odd edge]

@lethal cipher Has your question been resolved?

[lethal cipher]

i was talking about sinus function

and yh ur right b period

pi pi half

i found a very good video so im good

[odd edge]

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[mystic saffron]

how did it get 7 as the maximum value

[tropic copper]

it has a restricted domain

[mystic saffron]

i still don't know how the answer is 7

ohh

but how did it get 7

[boreal crag]

from the graph

the bolded section is the section that's actually part of the domain

where is the highest point on that section located

[hardy panther]

None of what you said sounds right

(I sent that like 8 minutes ago)

[lethal cipher]

more than half of it was german

_<

[hardy panther]

Lol, and it was more than hour old

I'm on my phone using data, there was a connection issue

[lethal cipher]

Yea happens, hope ur wifi gets better

my wifi broke once and after that i got a new one delivered

2 days like without good wifi

🪿📐

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@mystic saffron Has your question been resolved?

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[winter rapids]

Am I getting this right?

Find the probability of each outcome when a biased die is rolled, if rolling a 2 or rolling a 4 is three times as likely as rolling each of the other four numbers on the die and it is equally likely to roll a 2 or a 4.

P(1) = x

P(2 or 4) = 3 (P(1) + P(3) + P (5) +P(6))
P(2 or 4) = 3(x + x + x + x)
P(2 or 4) = 12x
P(2) = 6x
P(4) = 6x

P(1) + P(2) + ... + P(6) = 1
6x + 6x + 4x = 1
16x = 1
x = 1/16

P(2) = P(4) = 6/16 or 3/8
P(1) = P(3) = P(5) = P(6) = 1/16

[orchid torrent]

I think the question meant three times the probability of rolling a 1 (which is equal to that of a 3, 5, or 6)

[odd edge]

@winter rapids Has your question been resolved?

[odd edge]

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[junior ether]

why is the solution to $\frac{x}{2x-1}-1\geq 0$ not just $x\leq 1$

[clever fjord]

Jash

[junior ether]

x/(2x-1)>=1

x>=2x-1

0>=x-1

1>=x

[remote prairie]

because 0 doesnt work

so its less than or equal to 1 just not 0

[open onyx]

multiplying by a negative number reverses the direction of inequality
e.g. 1<2 but -1>-2

[junior ether]

so how do u solve it

[open onyx]

you have to split the problem into the cases where the value you want to multiply is strictly positive or negative

[junior ether]

oh

.close

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• Show your work, and if possible, explain where you are stuck.
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[glossy drift]

Pls help

Not for #5 tho

[grizzled tide]

!status

[odd edge]

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

[glossy drift]

Um

I dont know where to begin

FOR QUESTION 6

A and B

[velvet totem]

do you know how to find the maximum of a function

[glossy drift]

No

Im dumb man

[velvet totem]

you've done calc right

[glossy drift]

No

[velvet totem]

okay

is this for algebra

[glossy drift]

Is this like a university math server….

[glossy drift]

Um im canadian so its just called math

So idk

[velvet totem]

okay

are you familiar with completing the square

[glossy drift]

Yes

[velvet totem]

so you can write any quadratic in this form. a scales (stretches/squishes) it vertically, b translates it horizontally, and c translates it vertically

[velvet totem]

an untranslated parabola (b=0, c=0) has its vertex at 0,0

[glossy drift]

Ok but on a test i cant just use my phone to do all that (its tmrw so)

[velvet totem]

yeah

just

[velvet totem]

when the parabola is translated the vertex is translated with it so the vertex will be at (b,c)

[velvet totem]

so you'd complete the square

[blissful musk]

Can also use -b/2a

[velvet totem]

idk how well im explaining this

[blissful musk]

Faster

[velvet totem]

yeah

[glossy drift]

Ok i did that n got -4.9 (t + 5)^2 -125.14

Wrong right

[velvet totem]

yeah

[glossy drift]

So then what is it

[blissful musk]

Try using -b/2a it's easier

Gives you the x value of vertex then sub in for y value

[small kettle]

just solve the quadratic equation for it for A

it will give two x values

replace those two x values with T and test them out

you may even get a negative

take the highest x value you got from the 2

and then replace it with T, thats the max heigh

now to find how long the arrow would be in the air

you got the answer for A with what i said above, so replace h with the answer you got

for example, lets say it was 20 or something, now the question would become

20 = -4.9t^2 +50t + 2.3

i assume you know how to solve from here

switch 20 to the other side and it becomes

-4.9t^2+50t-17.7

do the quadratic equation again

[glossy drift]

OHHH

[small kettle]

and boom'

thats the time

[glossy drift]

okay also how do u do C tho..

-b/2a then quadratic formula?

[small kettle]

its literally just the same thing lol

they just changed the equation a little

do the same steps

[glossy drift]

What abt the 40% gravity…

[small kettle]

that was already accounted when they changed the equation

[glossy drift]

OH

[small kettle]

it becomes this with 40% less

[glossy drift]

Okay

[small kettle]

all they did was change the x^2 variable

[glossy drift]

Ok and do u know how to do #7

That 2 confusing

1

[small kettle]

ill tell in a few minutes

[glossy drift]

Alr

[small kettle]

a

ok so i had to go somewhere but i figured it out while walking

so they lose 400 people per 0.10 increase

that means 4000 people per 1 increase

how many times does 4000 go in 12000

4000*3 = 12000

because obviously 4*3 = 12

that means they need to increase it to the price of 3 dollars to lose all their cusotmers

from some testing, it seems that at 2 dollars, they make the max revuene.

the formula is R = P*(12000-400 multipled by (p-1/0.10))

R = revunene
p = price

[small kettle]

at either 8400 customers, which is 1.9 dollars, they make 15960 dollars. at 7600 customers and 2.1 price, they make 15,960. and it only goes downhill from there

but at 2 dollars, at 8000 customers, they get 16000, whcih is the highest

@glossy drift

[glossy drift]

oh alr

[small kettle]

Type .close

[odd edge]

@glossy drift Has your question been resolved?

[odd edge]

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