help-0

[ocean seal]

RedstonePlayz09

[tawny condor]

In general:

[warped horizon]

Oehh

It works like that too huh

[tawny condor]

$\parens{\frac{x}{y}}^2 = \frac{x}{y} \cdot \frac{x}{y} = \frac{x \cdot x}{y \cdot y} = \frac{x^2}{y^2}$

[ocean seal]

RedstonePlayz09

[warped horizon]

Yeah im reading this algebra book, the examples given are easy but when it comes to the exercises part

It gives like different types of questions which is good but it would be better if there were some examples on how to solve em

[warped horizon]

Hm i see, so what do i do next ?

[tawny condor]

From where

From this, probably multiply both the numerator and denominator by a

to get rid of the 1/a

[warped horizon]

Oeh alr

And i think i should multiply the sq root with root a² ?

[tawny condor]

,rotate

[tawny condor]

Just write it out with * a on the outside first

Then write a as sqrt(a^2)

and then combine the square roots

Do it one step at a time, take it slow

[warped horizon]

Kay

[tawny condor]

Once you get this fully and move on to more advanced topics, you can omit all of the steps

[warped horizon]

Oeh i see

Can u suggest me some books

Like the advanced one's

[tawny condor]

Sorry I don't really have anything to recommend

Just keep up with the one you're using right now

Maybe look in #book-recommendations

[warped horizon]

Oeh alr

this right ?

[tawny condor]

Let me check

You forgot to multiply the denominator by a aswell

Should be 4a

[warped horizon]

oh ye

4a

[tawny condor]

If you only multiply it on the numerator, the value changes

But other than that, very good

[warped horizon]

tq lol but u just been leading it

[tawny condor]

You're learning and that's good

[warped horizon]

hm so now i can cancel a^2 here right ?

[tawny condor]

Yeah

In the fraction

[warped horizon]

hm alr

O i think

i see smth here

all the numbers here

are divisible by 2 so

i cam take 2 common

wait no

4

@tawny condor i think i got it, but what about the 4 in the sq root

[tawny condor]

delete ur last pic

This

,rotate

[tawny condor]

alright

How did you cancel a 2 out

[warped horizon]

here

[tawny condor]

That's wrong

[warped horizon]

oh it is ?

[tawny condor]

Your square root doesn't have a 2 in front of it

It's like saying

$\frac{2x + y}{2} = x + y$

[ocean seal]

RedstonePlayz09

[tawny condor]

Which is completely wrong

[warped horizon]

oeh

the square root doesn't have a 2 in front of it

[tawny condor]

The 2 is dividing the 2x AND the y

Or in your case

-2(1 - 3a) AND sqrt(...)

[warped horizon]

oh lemme see

[tawny condor]

$\frac{-2(1 -3a) \pm \sqrt{36a^2 - 24a + 4 + 168a^2}}{4a}$

[ocean seal]

RedstonePlayz09

[tawny condor]

Simplify more

You get:

$\frac{-2(1 -3a) \pm \sqrt{204a^2 - 24a + 4}}{4a}$

[ocean seal]

RedstonePlayz09

[tawny condor]

$\frac{-2(1-3a)}{4a} \pm \frac{\sqrt{204a^2 - 24a + 4}}{4a}$

[ocean seal]

RedstonePlayz09

[warped horizon]

oehhh

i see

[tawny condor]

(Splitting the fraction)

Because

$\frac{x+y}{z} = \frac{x}{z} + \frac{y}{z}$

[ocean seal]

RedstonePlayz09

[warped horizon]

yeahh

[tawny condor]

Now try to continue

[warped horizon]

got it got it

$\frac{-(1-3a)}{2a} \pm \sqrt{51a^2 - 6a+1}}{2a}$

how do i type whole thing as 2a

[ocean seal]

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

[tawny condor]

It's fine

as long as you got it

[warped horizon]

hm yeah

thank u so much for the help

[tawny condor]

no problem!

[warped horizon]

can i add u ?

[tawny condor]

Sure

[warped horizon]

.close

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[tacit flint]

save me please!

what i did was square both sides

so i got

x^2+2xy+y^2 >= 4xy

idk if im doing this correctly but idk where to go from there ;-;

[minor needle]

it's almost done

your steps are good

[tacit flint]

!

where to next?

[minor needle]

subtract 4xy from both sides (simplify the inequality)

[tacit flint]

oh so i get
x^2 - 2xy + y^2 >= 0

oh thats a difference of two squares?

so do i simplify it to that?

[minor needle]

rather square of difference

but yes

[tacit flint]

ohhh i see

[minor needle]

aka

(x - y)^2 >= 0, right?

[tacit flint]

yeah

AH

i see!

and that will always be positive

ok gotcha thank you!

[minor needle]

(nonnegative, because it's 0 when x = y)

[tacit flint]

yeah gotcha thanks!

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[nova axle]

I need help with d

[lone heart]

Please don't occupy multiple help channels.

[nova axle]

I closed the other one

.close

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[tacit flint]

it says i cant use a calculator here so how would i tackle this?

[tall shuttle]

$\tan^2 x = \sec^2 x - 1$

[ocean seal]

tales

[tacit flint]

OH of course

thank you haha nvm ik what to do thanks

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[wheat isle]

did i draw this diagram correctly

[pulsar aspen]

Looks correct, but make sure that the 7.8 km line is longer than the 5.8 one, for the comfort of the reader.

[wheat isle]

ah yeah just ignore that

im not sure what angle im supposed to find

this?

[pulsar aspen]

I don't know either. English is not my first language.

[wheat isle]

okay

i think im supposed to find that angle anyway

[barren shuttle]

yeah i'd say so too

[vale wigeon]

no

you want the bearing of the depot from the station

that angle currently marked is the other way around

[wheat isle]

whattt

oh

?

[vale wigeon]

no

bearing is the CLOCKWISE angle between NORTH and the direction

[barren shuttle]

start on the blue line and keep going clockwise until you hit the red one

no?

[vale wigeon]

no

[barren shuttle]

damn

[vale wigeon]

the blue line points west not north

you need to draw a north line emanating from the STATION and start from that

[barren shuttle]

ahh

[wheat isle]

i dont follow

[vale wigeon]

we have been over this

this is the definition of bearing

[wheat isle]

but i am starting clockwise

[vale wigeon]

but you're starting from the wrong direction for one

[wheat isle]

so im confused. my diagram is correct but none of the angles i labelled is right

i dont see any other angle it could be

[vale wigeon]

it is the clockwise angle between north and the direction from Station to Depot

[wheat isle]

oh so my diagram is wrong

[vale wigeon]

your diagram had the north line in the wrong place

[wheat isle]

why is the north there?

[vale wigeon]

because we're looking at a bearing from the station

therefore we want a north line emanating from the station

and not from the depot, because we don't want the bearing of anything from the depot

[wheat isle]

would i be using 180 degrees to find that angle?

[vale wigeon]

i don't know what you mean by that but the phrase "use 180 degrees" sounds a little silly to me

but here is what i would take note of

bearing + green angle + right angle = 360°
green angle can be computed via basic trig

i guess by your standards this would be considered as "using 360 degrees"?

[waxen flame]

Here's a prettier picture. 😉

[wheat isle]

now what is my standards supposed to mean

[wheat isle]

wonderful

[vale wigeon]

the odd turn of phrase that you used

the meaning of which is still lost on me

[wheat isle]

well i thought we could use the straight line thing

which is 180 deg

[vale wigeon]

"the straight line thing"?

[wheat isle]

so um

straight lines are 180 degrees

so i thought i could use that

[vale wigeon]

i dont have the energy to scrutinize your wording more but let the record state it still stinks.

[wheat isle]

why do we use 360 degrees instead of 180 degrees?

[waxen flame]

Yes, you can add 180 degrees to that to get the bearing.

[vale wigeon]

because the back side of my left ankle itched at 05:22:19 UTC today

obviously

(sarcasm)

[wheat isle]

damn

i thought it was legit

[wheat isle]

so i found the angle of the triangle to be 53 degrees so, 180+53 then?

[waxen flame]

Yes.

edit : No

[vale wigeon]

no

i can tell you for sure that the angle marked as θ in kookiemon's diagram is less than 45°

[waxen flame]

90 - 53.

[vale wigeon]

^

like lets not conflate an angle with its complement ok

[wheat isle]

Why is it 90-53

[vale wigeon]

how did you get your 53° angle?

tell us how you calculated it and we'll tell you whether that actually gives you anything from the diagram.

[wheat isle]

this is legible

tan^-1(7.8/5.8)

[vale wigeon]

that's where your angle is

it is not the θ that @waxen flame marked.

[vale wigeon]

chickenscratch

[wheat isle]

itsss fineee

why wouldnt we add 90 degrees to the 53 to get theta instead of subtracting it

[vale wigeon]

do you think that this diagram could conceivably imply θ = 90+53 and not θ+53 = 90? Y/N

[wheat isle]

Possibly

[vale wigeon]

you see two angles labeled 53 and theta

you see them make a right angle when put together

you wonder whether this could be interpreted any other way than θ+53=90

[wheat isle]

ohhh

i get it now

so after getting 37 degrees from that do we add that to 180?

[vale wigeon]

idk do we

no we need to do a song and dance before proceeding with such an operation

[wheat isle]

https://tenor.com/view/possibly123-star-wars-possibly-gif-19412181

Ohhhh thyyy thirty seven will you add to my one eightyyyyyyyyy clap clap

180+37=217 deg

[vale wigeon]

we need to strike a specially made drum exactly 17 times on the first full moon after an even-numbered Monday at exactly 01:33:55 UTC, having signed and notarized in triplicate the necessary paperwork to perform such a ritual, which has to be prepared at least one (1) month but not more than two (2) months in advance.

[wheat isle]

Im glad that my question prompted you to write such professional advice

oddly specific too

anywasy im gonna sleep

ill keep that procedure in mind when dealing with trig

[vale wigeon]

it's anything but

[wheat isle]

must be the itch on the ankle

.close

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[glossy wharf]

to get equivalent capacitance, I just need to sum all capacitance value?

[upbeat coral]

no

there are rules just like for resistors

@glossy wharf

capacitors in parallel (like C2, C3, C4) you sum

capacitors in series, you sum the reciprocals

its like the opposite of resistance

[lone heart]

@glossy wharf Has your question been resolved?

[glossy wharf]

so all the value would jst be reciprocal and I add them all up?

[upbeat coral]

so the rule for capacitance in series is

$\frac{1}{C_t} = \frac{1}{C_1} + \frac{1}{C_2} + ....$

[ocean seal]

Mind Trickx

[glossy wharf]

so what should I do then to get the overall capacitance of the circuit

[upbeat coral]

so

start with the general structure (which is a parallel circuit with three lines)

lets do the first line

the first line has C1 in series with a parallel grouping of C2, C3, C4

the first step is to figure out equivalent capacitance of the parallel grouping

and then to use the series formula to figure out total capacitance in line with C1

does this make sense so far?

@glossy wharf

[glossy wharf]

wait so we're going to use 2 types of getting the value?

get the capacitance of c2,3,4 then add to c1 is thay right?

[upbeat coral]

yes

now, whats the capacitance of c2,3,4

[glossy wharf]

10f

[upbeat coral]

exactly

now use the formula to get that capacitance combined with c1

[glossy wharf]

3/10?

then 10/3 since it is reciprocal right?

would be 3.3 sumthing?

[upbeat coral]

yes

now lets do the second line

u need a calculator for this

get the total capacitance of C5,6,7 using the series formula

[glossy wharf]

alright hold on

180/101

then 2nd line wiuld be sum like 1.8

right?

[upbeat coral]

yes

its like 1.782 smthg

now lets do the third line

[glossy wharf]

alr

15/2

7.5

right?

[upbeat coral]

yep

now

we have the values for each of the three lines, and we know each of the lines is connected in parallel

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[upbeat coral]

.reopen

[lone heart]

✅

[upbeat coral]

so we just add each value from each line

so 10/3 + 180/101 + 15/2

gives us

12.6155

which is the equivalent capacitance

hope that helped!!

[glossy wharf]

205/106

[upbeat coral]

?

[glossy wharf]

oh righttt I added them fractions wrong

yea I get it, but now how can I get the charge and volts of each capacitance?

[upbeat coral]

use the fomrula charge = capacitance * voltage

you now know C and V for the whole circuit

so you can get charge for the whole circuit as well

now in parallel, each line has the same voltage

(which is 50)

and in series, each capacitor, or group of capacitors has the same charge

use those two facts to get the charge and voltage for every capacitor

[glossy wharf]

so c1 would be 25?

[upbeat coral]

wdym?

[glossy wharf]

since the each line have 50 voltage, then c1 would have 25 and c234 would also have 25

or no?

[upbeat coral]

no

the voltages don't have to be equal

just the charges

so we know that TOTAL VOLTAGE across line 1 is 50 V

[glossy wharf]

yea

[upbeat coral]

so Q1 / C1 = V1

[glossy wharf]

sooo c1*v1 ?

[upbeat coral]

gives u q1

but u don't know v1

which is why you need to use the c2,3,4 combo

we know that q2/c2 = q3/c3 = q4/c4

cause they are all in parallel

we also know that q2 + q3 + q4 = q1

[glossy wharf]

what next?

[upbeat coral]

ok

so lets actually change track a little bit to make it more simple

cause im getting too complex

lets combine c2,3,4 into one

know we know that q1/c1 + q(2,3,4 combined)/c(2,3,4 combined) = 50

and we know that q1 = q(2,3,4 combined)

now solve for both values of q

[glossy wharf]

how can I solve it?

[upbeat coral]

well we know c1 = 5

and c(2,3,4 combined is 10)

so $\frac{q1}{5} + \frac{q1}{10} = 50$

[ocean seal]

Mind Trickx

[upbeat coral]

just solve this equation for q1

which is also equal to q(2.3.4)

solving this

q1 = q(2,3,4 combined)

which is equal to 500/3

so q1 = 500.3

does that make sense? @glossy wharf

[glossy wharf]

yea so now we can get the V right?

[upbeat coral]

yes

now we can get V1

becasue v1 = q1/c1

and know we know both

we can also get

v(2,3,4,) by doing q(2,3,4,)/c(2,3,4 combined)

[glossy wharf]

I see but how we gon get each charge for 2,3, and 4?

[upbeat coral]

well, q2/c2 = q3/c3 = q4/c4

right

and all equal v(2,3,4,)

[glossy wharf]

ohhh

[upbeat coral]

now just do these steps for the other two lines and you should get the q and v values

for all the capacitors

[glossy wharf]

it would work even for 2nd line?

it's a series line

[upbeat coral]

well a series has all the charages equal to each other

so q5 = q6 = q7

and you know all the values for c5.c6.c7

[glossy wharf]

ohhhh I got it

thanks I will try to solve all on my own now

[upbeat coral]

np np

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@glossy wharf Has your question been resolved?

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[warped horizon]

Hlo, so i almost got the answer

[warped horizon]

But i have to eliminate that a+b there

Am i missing somwthing M

?

[dusty heron]

Hello

[warped horizon]

Hlo

[dusty heron]

I wish you good luck mate

See ya

[warped horizon]

Um

Alright

Nvm i got the answer

.close

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[round valley]

I am having trouble coming up with an equation to represent when the two cards would have the same money left on them.

[pulsar aspen]

Just equate them

[soft pulsar]

Yes, don't overthink it. They have the same amount when they are equal to each other by definition!

[round valley]

Ah ok, Thanks for the help!

.close

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[wispy salmon]

hey guys, about the green where he takes the mr^2 and turns in into inertia, shouldn't he take the 1/2 along with it if its a disc?

[wispy salmon]

or rather, how do i know when to take 1/2 and when not to when i turn in into i

[soft pulsar]

Can you explain why you think he should take the 1/2 along if it's a disk?

The 1/2 is part of the kinetic energy equation. So why would you use it when finding just the inertia in a?

[wispy salmon]

because disc inertia is mr^2/2 @soft pulsar

he says that the general inertia formula is just i=mr^2

so i kind of get why he didnt take the 1/2 along with it, but it leads me to question when do we treat the inertia as a disc and when not

[soft pulsar]

Oh, I understand what you mean now. What he probably meant was to just show how he derived the expression
$KE = \frac{1}{2} I \omega ^2$

[ocean seal]

Cosack

[soft pulsar]

And then from there it's up to you to implement the correct inertia equation based on your shape

[wispy salmon]

$KE = \frac{1}{4} mr^2 \omega ^2$

[ocean seal]

ImFlower

[wispy salmon]

for disc

?

[soft pulsar]

Yep. Looks right

I think he was just deriving the expression more than anything

And just means for you to plug in

$I = \frac{1}{2}mr^2$ in the bottom blue box

[ocean seal]

Cosack

[wispy salmon]

alright thanks, it seems like i got a mistake in my homework then

disc/wheel falls from 25m , even if we treat the wheel as a ring and not a disc it still wont be what the answer say in the book v=(9.36)

Anyway thanks @soft pulsar

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[warped horizon]

the x^2's here don't and can't cancel eachother right ?

[tacit arch]

They do not

[warped horizon]

kay

thank u

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[warped horizon]

hm how can i turn this into a quadractic equation

[fleet vale]

square both sides (but first you have to make sure that what's under the root is positive)

or in other words, to be able to square the equation, x has to be bigger or equal than -1

[warped horizon]

oeh

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[warped horizon]

so the x + 1 under the root is positive right ?

[fleet vale]

not always, since you can't take the root of a negative number

[warped horizon]

yeah

[fleet vale]

and for example for x=-2, you would have square root of -3

so you first have to point out that x must be bigger or equal than -1

[warped horizon]

i see, so how do we know if its like bigger than -1

is it like, the value is given ? or smth else

or is it just a condition that it has to be greater than -1 cuz no neg number's under root

[fleet vale]

,graph sqrt(x+1)

,w graph sqrt(x+1)

[ocean seal]

Results provided by WolframAlpha

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Requested by ! Ilya#2861

[fleet vale]

you see how for anything less than x=-1 there is nothing

[warped horizon]

oeh

yeah

[fleet vale]

well then it doesn't exist, don't worry about it

[warped horizon]

kay

[fleet vale]

but you have to write down that x must be bigger or equal than -1

because

if you square, and find something less than -1 for x at the end, that would not be an answer

you get why?

[warped horizon]

ahh i see

yeah

so we know that x can't be -1 n it must be greater

so we can square on both sides

[fleet vale]

yes

and solve for x

[warped horizon]

hmm i tried squaring before

i keep getting x^2 = a^2 (x + 1)

and i square root on both sides ?

to remove the square on x^2 ?

[fleet vale]

no

[warped horizon]

hm what do i do

[fleet vale]

have you learnt what the discriminant is?

[warped horizon]

hmm nope i didn't see it in this book

i know its b^2 - 4 ac doe

and that it tells us how many solutions exist ?

[fleet vale]

yeah, you need that to be able to solve

yep

[warped horizon]

thas all ik about it

[warped horizon]

oeh

[fleet vale]

you can't solve it without it I think, as far as I know

[warped horizon]

oh i see

hmm can u tell me how ?

[fleet vale]

how the discriminant works?

[warped horizon]

how can we use it here for solving this

[fleet vale]

well is the question just solve this equation or no?

[warped horizon]

this is the answer, so ig i have to get it into a quad. eq and idk how

[fleet vale]

okay

so

[warped horizon]

find its roots

[fleet vale]

so you have x^2=a^2(x+1)

[warped horizon]

yeah

[fleet vale]

you have to collect all terms on one side

so you have =0 on one side

[warped horizon]

oeh

so x^2/a^2(x+1) = 0

[fleet vale]

no, you can subtract a^2(x+1) on both sides

you don't divide by a^2(x+1)

[warped horizon]

oo i see

[fleet vale]

so you would get?

[warped horizon]

x^2 - a^2 (x+1) = 0

[fleet vale]

yep

then distribute a^2

[warped horizon]

x^2 - a^2 x - a^2 = 0

[fleet vale]

yep, now you may know a quadratic equation is in the form of: ax^2+bx+c=0

what would a, b and c be?

here

[warped horizon]

oeh

i think a = 1 , b = -a^2 and c = -a^2 ?

[fleet vale]

yes, well done

[warped horizon]

ah i see

now i just use the quad. formula

[fleet vale]

yes

exactly

I always use the discriminant first, but here it is not needed since a is not one value but could be anything

[warped horizon]

hm how did u know to like subtract a^2(x+1) on both sides though ?

[fleet vale]

you do everything to get =0 on the right

i will give an example

[warped horizon]

oeh

[fleet vale]

x^2= ( x+1 ) / 2

how would I get everything on the left?

it's the same as before

[warped horizon]

i think i will get 2 x^2 - (x+1) = 0 ?

[fleet vale]

yes, but you couldve just subtracted both sides by ( x + 1 ) / 2 , but what you've got is the same, since you first multiplied everything by two and the subtracted both sides by (x+1)

[warped horizon]

ah i see

[fleet vale]

it is the same, since you could (if you wanted) divide both sides by 2 but 0 wouldnt change

it is the same equation

[warped horizon]

oehh i see

thank u so much for the help and ur time

[fleet vale]

np good luck 🙂

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[blazing quest]

for the second one i got the answers x=e^3 and x=-4/3 but in the marking scheme it says -4/3 is not possible. can anyone explain why?

[hushed locust]

you can't take the logarithm of a negative number

[blazing quest]

thank you. i forgot about that

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[river swan]

Would I be right in saying that:
the domain of a function is the range of its inverse
The range of a function is the domain of its inverse

then what would the domain and range of the function gf(x) be?

[river swan]

Just cause I dont fully understand this graph/ diagram

[mortal trellis]

you labelled the arrows wrong

the one starting at "domain of g" should be labelled g

assuming an inverse exists, yes domain and range switch places

[river swan]

so what is the actual final range and domain of gf(x)

[mortal trellis]

gf or fg?

[river swan]

both I guess

[mortal trellis]

well for fg the arrow labelled with fg is correct

you only need to switch the labels of the other two arrows

for gf you would have to draw the same picture but switch f and g everywhere

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[keen glade]

"if the amplitudes of the angles of a triangle are proportional to the numbers 1, 2 and 3, then the triangle has one angle whose amplitude is greater than 120°" is this correct?

[keen glade]

or is it right angled?

[keen glade]

thought about it, it can't be right angled

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[brisk cedar]

Can someone help me with part b of question 3

I got to

x = 1 + sinx - cosx

The x being theta cus idk how to write

<@&286206848099549185>

[minor needle]

so now

try to rewrite sin(x) - cos(x) in the form they want

[brisk cedar]

How would u tho because wouldn’t you need sinxcosx

Like in that form

[minor needle]

sin(x) - cos(x) into Asin(x - a)

[brisk cedar]

So if u expanded sin(x - a) u would get sinxcosa-cosxsina

Unless I’m missing smthing

Which is why I’m confused on that formula that relates to sinx - cosx

How that* I ment to say

[minor needle]

use

[ocean seal]

Modus

[brisk cedar]

Yeah sorry my bad

[brisk cedar]

Ahh ok

[brisk cedar]

Thanks I get it now

Appreciate it

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[ashen basin]

Is there a fast way to do this via ti84?

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[ashen basin]

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[vapid steppe]

why is it that if ur trying to solve sin(theta)=.7325429 in [0,360 deg) that when u take sin inverse and get theta=47.1 degrees, to find the other angle you dont do 90+theta but rather do 180-theta

[nimble latch]

the unit circle might help

you can see that sin45 and sin(180-45) is the same

[vapid steppe]

so if u were trying to solve cos(theta)=.97540415 with theta in [0,360) without unit circle, how would u do it

[nimble latch]

,w cos(x) = .97560415

[ocean seal]

Results provided by WolframAlpha

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

Requested by shik#9660

[nimble latch]

are u given a calculator

[vapid steppe]

yea

[vapid steppe]

ya

[nimble latch]

just type cos inverse of that angle

[vapid steppe]

yea that gives one of them

how do u find the other one cuz its not always 180-theta

[nimble latch]

oh

for cos, it's 360 - base angle

you can see that in the unit circle

for tan it's 180 + base angle

[vapid steppe]

wait is it because cos is positive in Q1 and Q4

so the Q1 angle would just be cos^-1(theta)

and the Q4 angle would be 360-cos^-1(theta)

[nimble latch]

yeah

[vapid steppe]

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[noble onyx]

1. Prove that there exists a unique prime number of the form n^2 + 2^n - 3, where n is a positive integer.

= n^2 + 2n - 3

= (n - 1) (n + 3)

One of the factors must equal one for the expression to be prime. Find an that makes one of the factors equal to one.

Let n = 2

= (2 - 1) (2 + 3)

= 1 * 5

= 5

n^2 + 2n - 3 represents the prime number 5.

Now let's prove that this expression is unique to 5 and no other prime numbers.

We can check the only possible values of n in our positive integer domain with n = 1 and n > 2 in the factored form of our expression.

Let n = 1
= (n - 1) (n + 3)

(1 - 1) (1 + 3) = 0
= 0 * 4 = 0

n cannot equal one as it leads to 0 which is a non-prime number.

Now let’s look at the other possible values of n.

Let n > 2

= (n - 1) (n + 3)

Since n > 2, each factor will always result in being greater than one - meaning that no value of when n > 2 will give us factors for a prime number.

Therefore, we have proven that there exists a unique prime number, 5, of the form n^2 + 2n - 3, where n is a positive integer.

[noble onyx]

Would this be a valid proof and is the logic acceptable.

[vapid shuttle]

this is chatGPT <@&268886789983436800>

[noble onyx]

?

I have the writing history on my google docs

[next brook]

Doesn't quite look like ChatGPT's writing style to me.

[vapid shuttle]

a spaced line in between each sentence for no reason? detectors online say it is but oh well

[next brook]

On the other hand, it doesn't look right to me -- how did you get from n^2 + 2^n - 3 to (n - 1) (n + 3)? Somehow 2^n became 2n ...

[noble onyx]

the copy and paste from google docs didnt include the ^

i'll edit the message sorry

[noble onyx]

easier to read?

[vapid shuttle]

listen, it is fine. agree to disagree. I got a different result posting your original message into that.

[nimble latch]

unrelated but what's wrong with chatgpt tho

[next brook]

We don't allow using it to answer questions, because it is often subtly wrong, but very convincingly wrong.

[nimble latch]

well, then even if op here used chatgpt, isnt it still fine since they arent answering a question

[next brook]

Yeah. There's a bit of internal discussion how much we want to entertain questions that start with "ChatGPT said this, is it right / what's wrong with it", but as of right now it's not a the ban-on-sight level that ChatGPT answers are.

Anyway, as the responding moderator I hereby declare that Aight seems to be good. Carry on, citizens!

[next brook]

Is it a typo that your first line still says n^2 + 2^n - 3 rather than n^2 + 2n - 3?

If it is, then the proof looks sound.

[noble onyx]

thanks. would this be considered an exhaustive proof? im unsure because we looked at n < 2 generally or is it a mix of an exhaustive proof

[next brook]

I cannot imagine what more one could want.

[lone heart]

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[tacit flint]

i dont get part b ;-;

doesnt the two solutions mean the two x intercepts? im confused

[hasty hill]

nope it means at what values of p will cos 2x = p have exatly 2 solutions

[tacit flint]

i dont fully understand

[hasty hill]

ok

tell me graphically how will you solve cos2x = 0

[tacit flint]

well the 2 shrinks the x values by 2

so at 0 it would be values like 45 etaac

etc*

?

[hasty hill]

cool

now think about this you have to curves y = f(x) and y = g(x)..how will you solve for f(x) = g(x) graphically

you can draw y = f(x) and y = g(x) and find the points where these 2 curves intersect to get the required points

similarly you have to do the same for cos2x = 1

y = cos2x curve is already given..y = 1 is very simple

now plot both of them roughly and you will understand how many intersection points are there

similarly do for other values like y = -1, y = 5, y = 10

[tacit flint]

hmm alright i think i see what u mean

[tacit flint]

OH

i realised where ive gone wrong now

ive been treating them as the same equation

i think ik what to do now

thanks!!!

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[opal yacht]

Can you explain fx

[placid zinc]

Product rule

[opal yacht]

oh

[placid zinc]

It's very similar to the regular derivative of xe^x

[opal yacht]

thanks

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[brave junco]

can anyone tell me what the name of the method/identity that gets from the first one to the second is? thanks.

[placid zinc]

Sum identity for tan

[brave junco]

big thanks

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[split lantern]

Using the Schwarz-Christoffel formula, I got $$f^{\prime} (z) = \frac{A}{(z+1)^{\frac12} (z-1)^{\frac12}}$$. How do I proceed?

[ocean seal]

Orange905

[split lantern]

I am not sure how to deal with the square root. More specifically, the branch cut

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@split lantern Has your question been resolved?

[split lantern]

<@&286206848099549185>

[lone heart]

@split lantern Has your question been resolved?

[split lantern]

ok so i apparently have to ignore the branch cut and make it so that i pick the vertices and x_n's in a fashion that is the same 'orientation' as moving across the real axis to the right

i got f(z) = 2/pi * arccosh(z) - i so i assume i'm right

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[nova axle]

I have forgotten how to do this. Can someone help?

.close

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[quick sinew]

hi

[quick sinew]

ChatGPT gave me this:

for 8.a)
How did we get 0 0 0 0 0 for the first row?

[wary stream]

First off, you shouldn't use chatgpt to do math as it's wrong a lot

[median oar]

(especially logic, it's a predictive language algorithm, it can't think, understand or do logical reasoning tasks)

[quick sinew]

this is all wrong?

[wary stream]

And two, the first three columns are the propositions, you're finding all the possible combinations of 1s and 0s

[quick sinew]

yea

[wary stream]

It could be, havn't check it personally but I've seen chatgpt wrong a lot

[quick sinew]

So wait

P Q P exclusive or Q
1 1 0
1 0 1
0 1 1
0 0 0

we must use this right?

[wary stream]

Yes

[quick sinew]

alright

So

not sure what R is

how do i get the values for r?

[wary stream]

R is just another proposition

[quick sinew]

HUH?

so how would that look like..

[wary stream]

https://www.youtube.com/watch?v=IDTNWDHKQDk&ab_channel=GVSUmath

[quick sinew]

i think i understand everything else

[quick sinew]

Why are we setting R = true for the first 4 rows?

ohhh are we just trying to find all possiblities

[wary stream]

That's what I said

And two, the first three columns are the propositions, you're finding all the possible combinations of 1s and 0s

[quick sinew]

alr thanks

[quick sinew]

i got the same answer as chatgpt

is it right?

ima guess it is bc i got the same answer

anyways for question 8.b)

[wary stream]

It just wants you to describe that expression in words

[quick sinew]

alr

so that would be

The expression pOqOr evaluates to true exactly when a statement is either true or false on both sides. Here we have 3 characters, thus we must see if its true or false by grouping the two groups first (for example pOq or qOr) and then evaluate the other side to get one overall answer of true or false. For example, if pOq was true, r must be false in order for the expression to evaluate to true. If we have both false or true on both sides, we will generate false.

@wary stream like that?

/close

!close

[median oar]

.close

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[ivory lantern]

Becky has the following assignment marks in her French class: 80, 92, 90, 75, 90, and 85. Find the
mean, median, and mode grade of the six assignment marks.

[median oar]

what have you tried

[ivory lantern]

nothing

[median oar]

do you understand what the question is asking?

[ivory lantern]

yeah

[median oar]

so what is stopping you

[ivory lantern]

I just want to see what the answer is to compare with mine

since theres no answer key

[median oar]

show your answer first

we are not an answer key

[ivory lantern]

Mean grade: Approximately 85.33
Median grade: 87.5
Mode grade: 90

u think thats correct?

[median oar]

yes

that is correct

[ivory lantern]

all of them?

[median oar]

yes

[ivory lantern]

ight thanks

[median oar]

don't just post a quesiton and say nothing

[median oar]

you also lied here

dont do that

[ivory lantern]

Sorry i just wanted to see if I was on the right track

if I say I know what im doing no one will help

[wary stream]

If you wanted to see if you were on the right track, then you should post all necessary info, ie the question and the work so people can check to see if your are on the right track

[median oar]

this is not true

[ivory lantern]

ight i will for other questions.

[median oar]

if that was a previous experience, im sorry, but that usually odesn't happen

[ivory lantern]

.close

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[neon rock]

how do i know if i should add or minus the equations in solving by elimination

[sour dove]

do you have a specific example?

[neon rock]

oh yea sure give me a min

(2x+3y=6)
(15x-21y=-44)

like how do i know whether to add the two equations up or minus it

[sour dove]

so when you're trying to eliminate a variable, you want to have them cancel. It doesn't matter whether you choose to eliminate x or y in your example, just as long as one of them remains that's all that matters.

So for example, if we wanted to eliminate y, notice how you have a positive y and a negative y? Multiply the first equation by 7 and then add the two equations together. That'll leave you with 29x = -2.

Similarly, say if we want to eliminate x, then we would need to multiply either the 1st equation by -15/2 or the 2nd equation by -2/15.

so there's not a "right" way to do it per se. There are certainly easier ways. For example I would eliminate y in this case since you don't need to multiply by a fraction

[neon rock]

ohh ok thank

thanks

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[red vault]

How do i find this?

[wary stream]

You can use point slope or slope intercept form

[red vault]

Been a while, I have to relearn all of this for my finals

Ill try slope intercept form

[wary stream]

Google/youtube is very useful

[sour dove]

it's not much to it. Find the slope by doing m = (y2 - y1)/(x2 - x1), then do y - y* = m(x - x*), where x*, y* are either pair of points (just don't mix them)

[red vault]

Ok i think i remember

[lone heart]

@red vault Has your question been resolved?

[sour dove]

@red vault are you having trouble?

[red vault]

So like this?

[sour dove]

so far so good, but your math is a bit off. 13 - 7 = 6, not 7

so m = 3/6 = 1/2

[red vault]

Oops

[sour dove]

np 👍

so then plug in one of the (x, y) points into y = .5x + b and solve for b

[red vault]

Oh ok i see

Ill do that

[red vault]

What when i have this?

[sour dove]

I think you have those flipped around. You did x = .5y + b

remember the x is the left number and y is the right number: (x,y)

[red vault]

Oh yeah i did oops

[sour dove]

but after that isolate b to solve for it

[red vault]

Think i got it

[red vault]

What about this?

[lone heart]

@red vault Has your question been resolved?

[nimble fern]

I'd suggest to close the channel and open a new one for a new question.

it's hard to keep track with the progress if there are multiple seperated questions in one post.

[red vault]

Got it

.close

[lone heart]

Channel closed

Closed by @red vault

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[vapid shuttle]

I am trying to understand this definition here

I have made an example, and I am wondering if I am using it right

let's say U={1,2,3}

then we can choose e=.1 which is > 0

and this would give {.9, 1.1, 1.9, 2.1, 2.9, 3.1}

which is not a subset of U

so U is not open

is that what this basically means?

[placid zinc]

Also .95, and .99, and others are your set

[vapid shuttle]

why sadcat chartbit

[vapid shuttle]

okay I thought this was SUS, but idk why

[placid zinc]

All real numbers such that (1 - 0.1, 1 + 0.1) U (2 ...

Are a subset of your set

[vapid shuttle]

of {1, 2, 3} ??

[pseudo ice]

Notation

[placid zinc]

Yeah. This is all a part of the reals!

[vapid shuttle]

I don't know what the notation should be

[keen plinth]

very sussy notation going on here

[vapid shuttle]

that is part of my confusion peoples

[placid zinc]

Hmm? I think I understand

[tacit arch]

Austin needs to knowtation

[placid zinc]

I wouldn't consider a finite set for this, that's just confusing.

[vapid shuttle]

well I'd agree because I certainly am confused

[placid zinc]

Is [1, 2) open? Hint: consider the point 1.

[jagged raptor]

no open interval about 1 is contained in {1} so its not open

[keen plinth]

$U = \set {1, 2, 3}$ is not open because for any $\epsilon > 0$, $1 \in U$ for example would not have [(1 - \epsilon, 1 + \epsilon) \subsete U]

[vapid shuttle]

and why is me choosing a specifc epsilon > 0 to demonstrate that, not good?

[keen plinth]

because the definition is a there exists statement

so to fail it, you need to fail it for all epsilon

[vapid shuttle]

oh

for all x, and for all epsilon

ew

[jagged raptor]

no for at least one x and for all epsilon

[vapid shuttle]

it says for every x though

[jagged raptor]

when we negate we need to show there exists an x where every epsilon ball around it is not contained in U

[keen plinth]

for example, $U = (0.999, 1.001)$, you could choose $1 \in U$ but certainly choosing $\epsilon = 0.1$ would fail
[
(0.99, 1.01) \nsubsete U
]

[keen plinth]

nevertheless, U is open

[vapid shuttle]

why is it open

wouldn't just choosing any e >0 and then having .999-e would be out of the set

[keen plinth]

well (0.999, 1.001) is usually called an open interval

[vapid shuttle]

so it don't work?

[keen plinth]

so it had better be open

but you can prove this

[vapid shuttle]

I guess I am just having trouble understanding what it means for a subset of R to be open

this is the first time I am seeing the definition

[jagged raptor]

i think this might be a notation issue

[jagged raptor]

like what do you mean by this

[vapid shuttle]

I meant the set with those elements in it

[keen plinth]

a subset U of R is open if every point in U has an open interval surrounding it fully contained in U

[jagged raptor]

yeah thats not what we should have

[vapid shuttle]

how large can these open intervals be though?

does that matter?

[keen plinth]

any size you'd like

[jagged raptor]

no just one has to exist

[vapid shuttle]

there just has to be one

[jagged raptor]

yes

[vapid shuttle]

okay

[keen plinth]

as long as its fully inside U

so an open interval is open "by definition"

[placid zinc]

A subset U of R is open if every point has wiggle room

[keen plinth]

because every point in an open interal is covered by the full interval

[jagged raptor]

U is open if every point x in U has at least one epsilon ball containing x and contained in U

[vapid shuttle]

so why would something like [0,1] not be open

is this because

take the 1 element

we can't add any positive epsilon to it

and still be contained?

[keen plinth]

yes

[jagged raptor]

yes

[placid zinc]

👏

[jagged raptor]

1 \in [0,1] has no epsilon ball around 1 contained in [0,1]

[keen plinth]

[ (1 - \epsilon, 1 + \epsilon) \nsubsete [0, 1] ] for any $\epsilon > 0$

[vapid shuttle]

(0,1) though, why is this open?

[placid zinc]

Or, in the better notation, the point 1 has no wiggle room.

[jagged raptor]

well take any x \in (0,1). There is certainly an e>0 so that (x-e, x+e) is contained in (0,1). just take e = min(x,1-x)

[keen plinth]

[
(x - \epsilon, x + \epsilon) \subsete (0, 1)
]
for all $x$ by taking $\epsilon = \min\set{1 - x, x}$

[placid zinc]

I dare you to find me a point in (0,1) where you can't construct an open set around it

Double dog dare.

[keen plinth]

but in some contexts, you dont care about the open interval being centred around x

[jagged raptor]

i did it

[keen plinth]

in which case, the open interval that will cover any x in (0, 1) can be taken to be (0, 1) itself

(0, 1) is just open "by definition"

[placid zinc]

😵

[vapid shuttle]

is the definition for being open, not what I sent originally?

I thought that what we were messing around with was the definition for being open

[jagged raptor]

well its stealing it from the topological definition

[placid zinc]

So note the "quotes". You can prove that (0,1) is open, but historically, we defined openness to capture sets like (0,1).

[keen plinth]

open in R just means you are a union of open intervals

[jagged raptor]

which is equivalent to your definition (in the image)

[vapid shuttle]

I think this might be what is next on my thing

[jagged raptor]

U open = U is a union of open intervals (basis elements) = every x in U has an open interval (basis element) containing x and contained in U

[vapid shuttle]

Okay I think that I understand better now

thank you everyone

[vale wigeon]

ah so you are learning topology huh

union of opens is open

betting the next one has you show that finite intersection of opens is open, and invites you to think why the same can't be said of infinite intersections

[vapid shuttle]

you're a wizard

[keen plinth]

LOL

[jagged raptor]

💀

[keen plinth]

expectable

[vapid shuttle]

is this the part about inviting me to think?

if so, wow

[vale wigeon]

no

[vapid shuttle]

dang

I am still impressed nonetheless

XD

[vale wigeon]

but this is still worth thinking about

is Q open

how does the openness of Q or lack thereof mesh with the fact you no doubt know that both Q and its complement are dense

[vapid shuttle]

hmmm

I think Q is not open

because of denseness of irrationals

[jagged raptor]

yep

take x \in Q then any (x-e,x+e) will not be contained in Q

[vapid shuttle]

I proved that one, the denseness of irrationals

I thought it was cool

[vapid shuttle]

any? or some

[jagged raptor]

any

[vapid shuttle]

like or are you meaning that we can always choose the e to make any

but not for all e

[jagged raptor]

for every e>0 (x-e,x+e) is not a subset of teh rationals since there will be an irrational in (x-e,x+e)

[vapid shuttle]

oh

I keep not thinking of the ( , ) being an interval

I am thinking of sets this entire time

[jagged raptor]

it is an inteval

and an interval is a set

when we think of (a,b) as a subset of R, we are saying (a,b) = {x \in R : a<x<b}

[vapid shuttle]

oh that is nice

[jagged raptor]

given your definition of open, the definiton of a subset U of R being not open is there exists an x \in U such that for all e>0, (x-e,x+e) is not a subset of U

so we mean every (x-e,x+e) when showing not open

[vapid shuttle]

okay great

this was very helpful

I am going to sit on this for a bit before trying the questions

thank you everyone!

.close

[lone heart]

Channel closed

Closed by @vapid shuttle

Use .reopen if this was a mistake.

[lone heart]

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Send your question here to claim the channel.

Remember:
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• 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 sparrow]

I'm forgetting, what is the best way to go about solving something like this?

[shadow sparrow]

$12t^2-48t=0$

[ocean seal]

Huntifer

[shadow sparrow]

I feel like its wrong but my brain is thinking subtract (or divide?) 48(t) by both sides, then 12?