help-36

[timid sinew]

I know the answer but just try plugging in values first to try

[tranquil pine]

ok

i will try 1 and 2

[timid sinew]

One thing

When you plug in values

[tranquil pine]

yes?

[timid sinew]

try to look at the answer choices

[tranquil pine]

ok

[timid sinew]

1 doesnt seem like a good value to plug in, as it doesnt really match any of the answer choices

[tranquil pine]

ok

is it b?

[timid sinew]

yep

[tranquil pine]

ok

thx

.close

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[drifting matrix]

draw a tangent line and compare the slopes

[cerulean rapids]

now?

[drifting matrix]

compare the slopes

[cerulean rapids]

got it. thanks

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[drifting matrix]

np

[cerulean rapids]

.reopen

[final saddle]

✅

[cerulean rapids]

how do I find derivative of 1 here?

[astral moss]

derivative at x=1?

[cerulean rapids]

yeah

[astral moss]

is it asking you to do it algebraically or from the graph?

[cerulean rapids]

draw tangent and find the slope?

[astral moss]

ah ok

[cerulean rapids]

its multiple choice question

[astral moss]

drawing a tangent line might be a good option, as the potential answers are quite different from each other

you can immediately rule out a majority of the options though

once you determine whether the slope of the tangent line is positive or negative

[cerulean rapids]

how can i know whether the slope is positive or negative?

without using the formula

[astral moss]

by looking at the shape of the graph

if the tangent line would be sloping downwards, you immediately know the slope of it is negative

[cerulean rapids]

is the slope negative here?

[astral moss]

yes by the direction the tangent line is going in

it goes from the top-left to the bottom-right

[cerulean rapids]

yeah

there are two possible options now

-2 and -1/4

[astral moss]

yep

now you can roughly estimate it by doing rise over run

[cerulean rapids]

-2

[astral moss]

ye

[cerulean rapids]

thanks

[astral moss]

np

[cerulean rapids]

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[jagged nacelle]

find % change in de-Broglie wavelength when a potential difference of 3V is applied to deaccelerate and electron with initial KE = 4eV

[jagged nacelle]

can you solve this using differentiation?

or error analysis

fyi

,, \lambda = \frac{h}{mv}

[soft zealot]

Aetherfly

[jagged nacelle]

where h is a constant

and m is mass of electron

v is velocity

[cinder wedge]

0.2 percent

[jagged nacelle]

how'd you do that

[cinder wedge]

Let me write it out

[jagged nacelle]

did you do that?

[cinder wedge]

To find the percentage change in the de Broglie wavelength, we first need to calculate the initial and final de Broglie wavelengths.

The de Broglie wavelength (( \lambda )) is given by the formula:

[ \lambda = \frac{h}{p} ]

where ( h ) is Planck's constant and ( p ) is the momentum of the particle.

For an electron, the momentum (( p )) can be calculated using its kinetic energy (( KE )):

[ KE = \frac{1}{2} mv^2 ]

[ p = \sqrt{2mKE} ]

where ( m ) is the mass of the electron and ( v ) is its velocity.

Given that ( KE = 4 , \text{eV} ) and ( V = 3 , \text{V} ), we can calculate the final kinetic energy (( KE_f )) using the energy change due to the potential difference.

[ KE_f = KE_i - eV ]

where ( e ) is the elementary charge.

Then, we can calculate the final momentum (( p_f )) and hence the final de Broglie wavelength (( \lambda_f )).

Finally, we can calculate the percentage change using the formula:

[ % \text{ change} = \frac{\lambda_f - \lambda_i}{\lambda_i} \times 100% ]

Let's calculate it.

First, let's calculate the initial momentum (( p_i )) and the initial de Broglie wavelength (( \lambda_i )).

Given:

• Initial kinetic energy (( KE_i )) = 4 eV
• Elementary charge (( e )) = 1.6 × 10^-19 coulombs
• Mass of the electron (( m )) ≈ 9.109 × 10^-31 kg (approximately)

Using ( KE_i = \frac{1}{2}mv^2 ) and ( p_i = \sqrt{2mKE_i} ), we can find ( p_i ).

[ p_i = \sqrt{2 \times 9.109 \times 10^{-31} \times 4 \times 1.6 \times 10^{-19}} ]

[ p_i ≈ 3.36 \times 10^{-24} , \text{kg m/s} ]

Now, let's find ( \lambda_i ) using ( \lambda_i = \frac{h}{p_i} ).

[ \lambda_i = \frac{6.626 \times 10^{-34}}{3.36 \times 10^{-24}} ]

[ \lambda_i ≈ 1.97 \times 10^{-10} , \text{m} ]

Next, we'll calculate the final kinetic energy (( KE_f )) using ( KE_f = KE_i - eV ).

[ KE_f = 4 - (1.6 \times 10^{-19} \times 3) ]

[ KE_f ≈ 4 - 4.8 \times 10^{-19} ]

[ KE_f ≈ 3.9992 , \text{eV} ]

Now, we'll find the final momentum (( p_f )) using ( p_f = \sqrt{2mKE_f} ).

[ p_f = \sqrt{2 \times 9.109 \times 10^{-31} \times 3.9992 \times 1.6 \times 10^{-19}} ]

[ p_f ≈ 3.355 \times 10^{-24} , \text{kg m/s} ]

Finally, we'll calculate the final de Broglie wavelength (( \lambda_f )) using ( \lambda_f = \frac{h}{p_f} ).

[ \lambda_f = \frac{6.626 \times 10^{-34}}{3.355 \times 10^{-24}} ]

[ \lambda_f ≈ 1.974 \times 10^{-10} , \text{m} ]

Now, let's calculate the percentage change:

[ % \text{ change} = \frac{\lambda_f - \lambda_i}{\lambda_i} \times 100% ]

[ % \text{ change} = \frac{1.974 \times 10^{-10} - 1.97 \times 10^{-10}}{1.97 \times 10^{-10}} \times 100% ]

[ % \text{ change} ≈ 0.2% ]

So, the percentage change in the de Broglie wavelength when a potential difference of 3V is applied to decelerate an electron with an initial kinetic energy of 4eV is approximately 0.2%.

[soft zealot]

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

ika

Now, we'll find the final momentum (\( p_f \)) using \( p_f = \sqrt{2mKE_f} \).

\[ p_f = \sqrt{2 \times 9.109 \times 10^{-31} \times 3.9992 \times 1.6 \times 10^{-19}} \]

\[ p_f ≈ 3.355 \times 10^{-24} \, \text{kg m/s} \]

Finally, we'll calculate the final de Broglie wavelength (\( \lambda_f \)) using \( \lambda_f = \frac{h}{p_f} \).

\[ \lambda_f = \frac{6.626 \times 10^{-34}}{3.355 \times 10^{-24}} \]

\[ \lambda_f ≈ 1.974 \times 10^{-10} \, \text{m} \]

Now, let's calculate the percentage change:

\[ \% \text{ change} = \frac{\lambda_f - \lambda_i}{\lambda_i} \times 100\% \]

\[ \% \text{ change} = \frac{1.974 \times 10^{-10} - 1.97 \times 10^{-10}}{1.97 \times 10^{-10}} \times 100\% \]

\[ \% \text{ change} ≈ 0.2\% \]

So, the percentage change in the de Broglie wavelength when a potential difference of 3V is applied to decelerate an electron with an initial kinetic energy of 4eV is approximately 0.2%.
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               not set up for use with LaTeX.

See the LaTeX manual or LaTeX Companion for explanation.
Type  H <return>  for immediate help.
 ...                                              
                                                  
l.53 \[ p_f ≈
                3.355 \times 10^{-24} \, \text{kg m/s} \]
You may provide a definition with
\DeclareUnicodeCharacter```

[jagged nacelle]

uh

but that's incorrect

[cinder wedge]

I hate compile errors

[cinder wedge]

Wdym?

[jagged nacelle]

the correct answer is 100% increase

[cinder wedge]

Wait so why do you want help for

[jagged nacelle]

.

[cinder wedge]

Yea of course. Let me try to do it with that

[jagged nacelle]

<@&286206848099549185>

<@&286206848099549185>

[humble swift]

no

if it is 100% change no

only for small change u can use

[jagged nacelle]

oh

[humble swift]

yes

[jagged nacelle]

so what does the answer mean when you differentiate?

like what does it represent

[humble swift]

it will still be representing the rate

[jagged nacelle]

,, \frac{\Delta \lambda}{\lambda} = \frac{-1}{2} \frac{\Delta K}{K}

[humble swift]

but its not accurate for large stuff

[soft zealot]

Aetherfly

[jagged nacelle]

no like what if you plug in all the values?

[humble swift]

didnt understand

wdym

[jagged nacelle]

like

[jagged nacelle]

this equation

what if you put

delta K

and K

[humble swift]

yes

[jagged nacelle]

in there

[humble swift]

u will still get the same thing

error only

but the error will have error in it

[jagged nacelle]

oh

[humble swift]

yep

so u cant use that for large changes

[jagged nacelle]

but

I didnt know 100% change

[humble swift]

u can use it upto like 10% or so

[jagged nacelle]

in the beginning

what if I differentiate in the first place

my ans is wrong

how do I know it

do it using the normal method to check?

or should I not use it normally

[humble swift]

lemme tell u

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[final saddle]

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

@crisp maple Has your question been resolved?

[crisp maple]

1

[loud sundial]

Do yk the basic differentiation rules?

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@crisp maple Has your question been resolved?

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[serene smelt]

I'm struggling to work through this problem using trig identities, I'm able to solve it using l'hopital for the sake of checking my answer but every time i attempt to expand it out i get a incorrect number

[fathom walrus]

Don’t use lhopital

There’s a specific identity for cos(2x) that works for this

[fathom walrus]

Try using that

[serene smelt]

so if you have that over cos(x) - sin(x) would you be able to simplify it to just be cos(x) - sin(x)

[fathom walrus]

No

Use difference of squares

[serene smelt]

so like

(cosx+sinx) (cosx-sinx)

?

[young bridge]

✅

now cancel

[serene smelt]

i wasnt actually aware you could use difference of squares with trig identities

that's good to know

[young bridge]

it doesn't matter what x or y is. x^2 - y^2 = (x-y)(x+y)

i dont think ive ever seen an exception

[serene smelt]

Fair. I've just never taken a proper trig course so this stuff is unfamiliar for me for sure. After cancelling out things i did get the correct answer and i appreciate the help everyone

gonna close this out but it genuinely is v appreciated o7

.close

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[urban socket]

I couldn't figure out any statement, i really could use some help

[woeful light]

What grade u ib

In

[urban socket]

oh

Is the question that easy 🗿

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@urban socket Has your question been resolved?

[urban socket]

alr i solved the problem

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[feral siren]

can someone help me with the last part please

[tranquil pine]

!status

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6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[feral siren]

This is for b

What did I do wrong?

It’s supposed to be the same but negative x^2

[tranquil pine]

Oh , I didn't see the h there

Sorry,not familiar with this

[feral siren]

Dw

Nvm

I see why

This is tanh not tan

So it would be -tanh^2 + 1 = to sec^2

Is anyone able to help me with c?

I tried

But got no where

[mighty sandal]

so you've got to integrate 1/(1-x^2)

how you gonna do that

[feral siren]

By parts

I did that

But I think it did something wrong

[mighty sandal]

nope

[feral siren]

Cus it not working

Really?

[mighty sandal]

partial fractions

[feral siren]

OHHHHHHHHHH

I try that

I did it

Thank you

[mighty sandal]

no problem

gl for wednesday

[feral siren]

thank you

hopefully i do well

i guess its pretty obviously further maths

.close

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[late pasture]

Would someone please be willing to help? At the bottom of the paper I need to factor the k*2(k^2+6k+11) /6 to be equal to the right hand side of p(k). But I can't figure it out

[ornate knot]

this doesnt factor into right side

[late pasture]

Oh I did something wrong?

[ornate knot]

(k+1)(k+2) = k^2 + 3k + 2

ill check ur work wait

this is not true

[late pasture]

Does that make all my work useless since I contuine with the assumption that the base case was true?

[ornate knot]

it would be true for someother integer tbh

i dont think it would affect ur working too much ( i hope ) , kinda rusty at induction proofs

[late pasture]

Dang, would you know what I should do next then?

[ornate knot]

what is ur original question

[late pasture]

At the very top in blue

Hopefully you can read it

[ornate knot]

i can

but the series seems strange

[late pasture]

It is strange

[ornate knot]

i dont get something

why are u taking the last 3 terms in ur induction proof?

the way i done it , i only took the last term

so base case P(1)
n = n(n+1)(n+2)/6
1 = 1(1+1)(1+2)/6
1 = 1

[late pasture]

I thought we take the whole part after the +...+

Or are we only taking the last term?

[ornate knot]

i only took last term

[late pasture]

Ah okay

[ornate knot]

i have no clue about this , its better to close and make new channel

[late pasture]

Kk thanks again

. Close

.close

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[storm karma]

lim 1/x
x->inf

[storm karma]

Is it 0+ or 0

<@&286206848099549185>

[river reef]

0

[storm karma]

Ok

@river reef mc laurin expansions of e^x is valid for all x or for near 0 only?

[river reef]

sry i dont know that too

[static plinth]

for all x

the interval of convergence is all real values for the maclaurin of e^x

[storm karma]

But if we write expansion using taylor series it's different

[hallow quiver]

Need help plez

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@storm karma Has your question been resolved?

[craggy bloom]

Can you translate it?

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[fierce sun]

what is it meant when it says "the right sides must have b1+b2+b3 = 0? Thanks

[fierce sun]

So for example if the right side added to 0. there would be a solution?

i think i got it

.close

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[slow nacelle]

verwende satz des pythagors um dem Höhe zum wissen

[drowsy epoch]

alles klar

[drowsy epoch]

,rccw

[slow nacelle]

375m^2+b m^2 = 625m^2

[slow nacelle]

aha

[drowsy epoch]

ich wollte das bild umdrehen haha

[slow nacelle]

same

sonst kriege ich nacken schmerzen

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[rain ermine]

I need to calculate the double integral, but i need to use a change of variables to compute this. The solutions used let u = xy, v = y/x, but why do they do this specifically and how do i determine that?

[final saddle]

@rain ermine Has your question been resolved?

[final saddle]

@rain ermine Has your question been resolved?

[final saddle]

@rain ermine Has your question been resolved?

[lethal estuary]

they do it that way because it's a rectangle in parameter space

[rain ermine]

but how do i come up with those substitions? is there a recipe to follow to get the correct subtitution

[lethal estuary]

I don't think there's a one-size-fits-all method

but for this one, it's based on the equations you're given

if u = xy, what is the range for u

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[thorn hollow]

gui'

[thorn hollow]

does anyone know w here the digits here come from

from here

https://media.discordapp.net/attachments/1236096878693715968/1240774146829582398/IMG_2746.jpg?ex=6647c879&is=664676f9&hm=15e58c9db19c90cd50bf3cd5f1ee0f32892db738534345b0f96abf7c17722a8f&=&format=webp&width=901&height=676

I know that you should add the border lines of the histogram, but I'm doing something wrong

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@thorn hollow Has your question been resolved?

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[barren belfry]

help please

i dont even know where to begin

[lethal estuary]

calculate the areas?

[barren belfry]

so like area under it?

so id make a triangle from 0-4

then a rectangle from 4-6

then a triangle 5-8

or just a big trapezoid

[lethal estuary]

well you don't have to calculate all of them even

some of them will be negative, right

[barren belfry]

yeah

the entire areas gonna be negative

[lethal estuary]

so they'll cancel some of the positive ones

[frank shuttle]

whom needs help

[barren belfry]

i got -3

[barren belfry]

me por favor

[lethal estuary]

show your work

[barren belfry]

ok

area under the x axis is -13/2

average of the bases is 13/2 * 1

over the x axis is 7/2 *2

[lethal estuary]

area under the x axis?

[barren belfry]

yes

cause thats negative?

[lethal estuary]

that is not true

[barren belfry]

and then its -13/2+7/2

dammit

[lethal estuary]

there are only two small triangles below the x axis

[barren belfry]

i thought it was just the area

[lethal estuary]

yes, the area between the curve and the axis

[frank shuttle]

I got -3

I think mines correct

[lethal estuary]

can you shade in what area you're calculating and show me?

[frank shuttle]

@lethal estuary I think you’re wrong

[lethal estuary]

only the portions between the curve and the x-axis

[barren belfry]

so just the top part?

[lethal estuary]

no...

let me try to show

[barren belfry]

omg im helpless

ok thank

[lethal estuary]

those 3 portions

[barren belfry]

tf

why

[lethal estuary]

those are the parts that are between the curve and the axis

[barren belfry]

oh

i got 2

ok you think so?

this is the other one

im pretty sure its right

im just second doubting myself

[lethal estuary]

looks right

[barren belfry]

ok thank you

is this one {c-a (g(x)-f(x)?

.close

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[tidal pollen]

!help

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To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[marsh temple]

.close

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[fathom stream]

How could I use the well ordering principle here

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[shut wasp]

confused

i dont know how to go about answering

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@shut wasp Has your question been resolved?

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@shut wasp Has your question been resolved?

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@shut wasp Has your question been resolved?

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[tranquil pine]

Whats up guys, hope your having a good night

[tranquil pine]

They want me to find the area of the shaded area

So we know 2 cm

Would I just do that with 5 cm or 5cm - 1

[royal gust]

"do that"?

[tranquil pine]

Like what numbers to use to find area

[lunar minnow]

ur gonna be finding the area of two triangles and finding the difference of those numbers

[timid sinew]

find area of the total larger right triangle

[lunar minnow]

remember area of a triangle is 1/2 x base x height

[timid sinew]

subtract area of smaller right triangle

[tranquil pine]

Here's the whole problem they asked, Find the area of the shaded region in the figure below.

[timid sinew]

nvm

forget that

[lunar minnow]

i dont think you need hypot

[timid sinew]

hehe

[lunar minnow]

yea

[timid sinew]

im sleep deprived

[lunar minnow]

so first find the area of the big triangle

like the whole thing

[timid sinew]

yeah just (1/2)*bh

[tranquil pine]

One second let me do some black magic

large on top
small on bottom

Give me a second cause I know thats wrong

2.71?

For shaded

[timid sinew]

em

[tranquil pine]

That aint it tho

So

I imagine im dumb

[timid sinew]

wheres the sqrt(21) from

isnt it just 5?

cuz base times height

ur trying to find the area of the smaller triangle and subtract from larger

[tranquil pine]

Aint it like 2.5?

[clear moon]

do you still need help with this?

[tranquil pine]

IS it 2.5?

[clear moon]

uhh lemme work it out rq

big triangle has area 1/2 * 3 * 5 = 7.5, unshaded triangle has area 1/2 * 1 * 5 = 2.5

so the shaded part has area 7.5 - 2.5 = 5

[tranquil pine]

Ah I see now

Thank you

[clear moon]

actually you don't even need to subtract them lol, you have the base and height

[tranquil pine]

yeah I see

Im dumb

.close

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[timid sinew]

...

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[jade fable]

Wrong place

[timid sinew]

i can tell

that was quite interesting

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[solid lark]

I need help with this meteorology problem , I know I need to use something related to stadistics but idk what exactly.
Translated Text:
Complete the temperature information series for the Jauja Station using data from the Huancayo-Huayao Station. Likewise, complete the temperature information series for the Comas Station using data from the Huancayo and Jauja Stations.

[solid lark]

15 mins have passed <@&286206848099549185>

[tranquil pine]

!status

[final saddle]

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

[solid lark]

1

[final saddle]

@solid lark Has your question been resolved?

[tranquil pine]

Bro has the me city profile

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[paper crane]

Hey

[static pier]

Yo

[paper crane]

wrong one

This

for a

it says find a

[static pier]

Ya

Arc length

r*\theta

[paper crane]

is this corrcet
?

[static pier]

Probably not

For a?

Where is the pi factor coming in?

[static pier]

^^

[paper crane]

I have no clue

one sec

[paper crane]

?

[static pier]

That is the formula for arc length

Wdym?

[paper crane]

thought it was this

[faint fern]

That's the same formula. They were trying to write in latex

[paper crane]

was gunna say

i dont understand any of that

still having problems w this question

18cm

for a ?

[paper crane]

.

4.9cm for b ?

[faint fern]

What did you multiply?

[paper crane]

12.2 x 1.4

just rounded

[faint fern]

And that gives?

[paper crane]

17.08

[faint fern]

To 1 decimal place

[paper crane]

17.0

aight

i see

i genuienly dont know where 18 came from

sorry

[faint fern]

It's 17.1

[paper crane]

appreciate the help

rounded

yes

[faint fern]

Yeah no problem

For the second problem, I'm not sure what B refers to

Is it the (larger) sector area or the angle?

[paper crane]

yea

its right under neath

[faint fern]

Yeah 4.9 is right. I thought 35 was the circumference so I got confused

[paper crane]

alright perfect

and for c

4.34

or since its 1 decimal 4.3

[final saddle]

@paper crane Has your question been resolved?

[faint fern]

Try again and look at the data you have

[paper crane]

0.7 rad

im so tired i didnt even notice that thanks for catching that

[faint fern]

No problem 👍

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[misty sequoia]

y = 3x^2, y = 9x - 6

What segment of the graph do I have to calculate?

[echo torrent]

What is the question?

[misty sequoia]

Find the area between those 2

[echo torrent]

Zoom out the diagram

You would see the two intersections

Find the area between the points in which those two points intersect

[misty sequoia]

Do I have to calculate the area under the x-axes too?

[echo torrent]

No

Also

It's very ambiguous

When you don't mention what the question is asking

We could have calculated anything

Since you did not specify anything

[misty sequoia]

The question is to calculate the area restricted by them

[echo torrent]

Yeah, so, between the two points of intersection.

Not with the x-axis

[misty sequoia]

The interception points are x = 2 and x = 1

[echo torrent]

Yes.

Evaluate the integral and bound it

[echo torrent]

I know.

[misty sequoia]

But there is a small part of area under the x-axes

[echo torrent]

You said that it's between the curves

[misty sequoia]

Restricted by the curves*

[echo torrent]

So we won't really count that.

[misty sequoia]

What if there was one more restriction, y = 0

Do I have to find the area between x axes then?

[echo torrent]

Then it would be the integral from 0 to 1

Bounded by y=0 and the two curves

[misty sequoia]

I'm asking about the part under y = 0

[echo torrent]

Oops

Sorry

No

Not under the x-axis then.

[misty sequoia]

And what if there is no y = 0?

[echo torrent]

Well then it would be the integral from 1 to 2

[echo torrent]

Yes

[misty sequoia]

Only that pretty small part?

[echo torrent]

Yes

[misty sequoia]

Thank you very much

[echo torrent]

Please close the channel.

[misty sequoia]

.close

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[glad steeple]

Integers:
Multiplication of integers:
Show in general that the definition of multiplication is independent of the choice of representatives from an equivalence class.
Hint: at one point add $$b'c+a'd$$ to both sides of the equation:

This was my idea so far:
$$(a,b)\sim(a',b') \wedge (c,d)\sim(c',d')$$
Def:$$a+b'=b+a' \wedge c+d'=d+c'$$
Def: $$(a,b)\odot(c,d)=(ac+bd,ad+bc)$$
z.z.: $$(a,b)\odot(c,d)\sim(a',b')\odot(c',d')$$
$$= (ac+bd,ad+bc) \sim(a'c'+b'd',a'd'+b'c')$$
$$=(ac+bd)+(a'd'+b'c')=(ad+bc)+(a'c'+b'd')$$

[soft zealot]

Benschko

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[odd garden]

Find the value of n if nC3 + (n+1)C2 + (n+2)C1 =72

[iron mist]

Do you know what nCr is?

And if so, what have you tried?

[odd garden]

ive tried the usual method using algebra but i believe the working should be substantially shorter, so i was wonder if there was a shorter method to do this question

[iron mist]

Do you mind sending your solution?

[zealous wigeon]

yeah

[odd garden]

^

[iron mist]

Well one thing you could have done is to cancel out the factorials on the bottom with ns

That way it’s not as many variables

You could also surmise that your answer is a positive integer (because factorials are weird otherwise), so from the cubic, you could say $n(n^2+11)=420$ and then trial and error factors (clearly the second one is larger than the first, so you can minimize your guesses)

[soft zealot]

CST (please ping when replying)

[iron mist]

But yeah the basic idea is just to find the cubic, stumble onto n=7 working, and then verifying nothing else does

Long division or synthetic, you arrive at a quadratic with imaginary roots

And then yeah discriminant will do the job there

[odd garden]

are there any identities that can help shorten the working or is using algebra the only option?

[iron mist]

Honestly I don’t think so

[odd garden]

what would you cancel out exactly and where would you do it?

[iron mist]

Well for example

nC3

[soft zealot]

CST (please ping when replying)

[odd garden]

ohhhhh

[iron mist]

So when you add the fractions together

You only have to multiply constants to make the denominators equal

Instead of multiplying a lot of variables

[odd garden]

yes i understand now

thank you

[iron mist]

no problem

[odd garden]

.close

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[odd garden]

.reopen

[final saddle]

✅

[odd garden]

wait sorry i have another question im not sure about

if nCr + nC(r+1) where n and r are integers, show that n is odd

these are the hints i currently have

[tiny oar]

is it + or =?

[odd garden]

sorry its =

[rich berry]

try writing up the equation and then removing the n! from th enumerator

[tiny oar]

Use the formula of nCr and nCr+1, then simplify using n! = n (n-1)!

[distant shoal]

nCr = nC(r+1)
or , r!(n-r)! = (r+1)!(n-r-1)!
or, r+1 = n-r
n = 2r + 1

[odd garden]

are those all different methods or one method?

[iron mist]

it is sadly the complete solution, which helpers are not supposed to give outright 💀

[vital oak]

Give me solution to my problem

[odd garden]

sorry could you please explain the working between line 2 and 3

[rich berry]

cancel out r! and (n-r-1)! from both sides

if you expand

(r+1)! = (r+1) x r!

similarly, (n-r)! = (n-r) x (n-r-1)!

[iron mist]

Similar to what we did here btw

[odd garden]

ohh yep

ok

[rich berry]

what do u have so far

have you tried expanding part i) on both sides

and rearrangig

[odd garden]

ill try that

thanks

[final saddle]

@odd garden Has your question been resolved?

[odd garden]

<@&286206848099549185>

[odd garden]

<@&286206848099549185>

[distant shoal]

Oh sorry

I didn't know that coz I joined yesterday

[distant shoal]

In the first question , try to expand and take the common elements and factor it

And whatever's left inside which was uncommon , just solve it

[odd garden]

alright thanks

do you know how to do part ii)? thats the main part im not sure about

[distant shoal]

nCr + nC(r+1) = (n+1)C(r+1)

Use this

Wait nvm

[final saddle]

@odd garden Has your question been resolved?

[distant shoal]

Yea u need to do this

Use this

Write the series in expanded form

The first term is 2C2

It can be written as 3C3

As 2C2 = 3C3

[distant shoal]

Then use this formula

Add the first two terms

[distant shoal]

After adding, again the sum and the third term should come in this form

[final saddle]

@odd garden Has your question been resolved?

[odd garden]

ohh ok

thank you

.close

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[tranquil pine]

doing b

$(x,y)=(3-t,-2-4t)$

[soft zealot]

Remlis

[tranquil pine]

$r=(3,-2)+t(-1,-4)$

[soft zealot]

Remlis

[tranquil pine]

$m=\tfrac41$

[soft zealot]

Remlis

[tranquil pine]

$m_{\perp}=-\tfrac14$

[soft zealot]

Remlis

[tranquil pine]

$y=-\tfrac14 x+C$

[soft zealot]

Remlis

[tranquil pine]

$y+\tfrac14 x=C$

[soft zealot]

Remlis

[tranquil pine]

$-2+\tfrac14(3)=C$

[soft zealot]

Remlis

[tranquil pine]

$C=-\tfrac54$

[soft zealot]

Remlis

[tranquil pine]

in cartesian form it would be

$y=-\tfrac14 x-\tfrac54$

[soft zealot]

Remlis

[tranquil pine]

multiply by 4 to remove fracs

$4y=-x-5$

[soft zealot]

Remlis

[tranquil pine]

$x+4y+5=0$

[soft zealot]

Remlis

[tranquil pine]

this is wrong though

[obtuse flame]

is the correct answer y = 4x - 14

you seem to be finding the line perpendicular to the given line

[tranquil pine]

yeah

[tranquil pine]

oh

so there was no need to find the normal to the slope

$y=4x+C$

[soft zealot]

Remlis

[tranquil pine]

$-2=4(3)+C$

[soft zealot]

Remlis

[tranquil pine]

$C=-14$

[soft zealot]

Remlis

[tranquil pine]

$y=4x-14$

[soft zealot]

Remlis

[tranquil pine]

$4x-y-14=0$

[soft zealot]

Remlis

[tranquil pine]

ah okay that's it

thanks

.close

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[lost drift]

Can someone help m e on this please? how is q in meters, if w=1s and t = s, then the quanitites dont match

[tranquil pine]

$4(x+4)=17

[lost drift]

w=1/s

[solid cobalt]

You're right, I'm afraid.

If w=1/s and t=s, then e^iwt is not dimensional so unit is 1

Then ay+q has to be m/s so a=1/s and q=m/s

That's my proposed solution

[light wren]

I agree

[lost drift]

thanks guys,

.close

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[full halo]

how

the ()^6 part how is that it

[scarlet sequoia]

e^x - 1 is equivalent to x

[full halo]

isnt it 0

1-1

just like he got the 2

1+1

[scarlet sequoia]

?

you can't be equivalent to 0 unless you're equal to 0 in a neighborhood of the limit point

[full halo]

how he got that 2 then

[scarlet sequoia]

because 2 is a non zero constant

the definition of equivalence :

$f\sim_a g \Longleftrightarrow \frac{f(x)}{g(x)}\xrightarrow_{x\to a} 1$

[soft zealot]

rafilou2003
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[full halo]

thx !close

!close

.close

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[final saddle]

Please don't occupy multiple help channels.

[clear moon]

the sin(theta) term is missing a square root in the denominator btw

i'm pretty sure you can't uniquely find theta from this

unless you know values of a and b

[sonic bloom]

oh yeah

mb

[clear moon]

otherwise the best you could really do is like arccos(a/sqrt(a^2+b^2))

[sonic bloom]

sin(theta)=b/sqrt(a^2+b^2)

[fathom walrus]

or arctan(b/a)

[sonic bloom]

yes but b/a need to be in ]-pi/2 ; pi/2[

[fathom walrus]

.

[clear moon]

this problem only makes sense if they give you a,b

[sonic bloom]

we don't know values of a and b

it's in general

[clear moon]

otherwise you just write theta in terms of arccos or arcsin or arctan

[sonic bloom]

I gotta do a disjunction of case

example

if $b > 0 : x \in \displaystyle\bigcup_{k \in \mathbb{Z}}[\arctan(\dfrac{b}{a})-\dfrac{\pi}{2}+2k\pi ; \arctan(\dfrac{b}{a}) +\dfrac{\pi}{2}+2k\pi]$

[soft zealot]

Akti

[rocky tusk]

this is just the definition of sin and cos lol

let a^2+b^2=c^2

you can find theta with arctan

like stephen said

c=sqrt(a^2+b^2)

[sonic bloom]

you have the system sin(t)=.../... and cos(t)=.../...
So sin/cos=b/a (those squares get simplified)

[rocky tusk]

mhm

so theta=arctan(b/a)

[sonic bloom]

no

u can only do sin(x)/cos(x)=tan(x) if x in ]-pi/2;pi/2[

[clear moon]

pretty sure you can always do this

tanx = sinx/cosx is an equality regardless of x

unless cosx is undefined

[rocky tusk]

well no yea

arctan is just restricted to -pi/2 to pi/2

[rocky tusk]

still holds it’s just that tanx isn’t continuous

at pi/2

and -pi/2

and just odd multiples of pi/2 of course

[sonic bloom]

so I gotta do a disjunction of cases if cos(theta)=0 ie ... ±pi/2+2kpi... and the rest of x values ie x=arctan(b/a) ?

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[hushed apex]

How do I solve this problem?

[hushed apex]

Nevermind, got it

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[restive badger]

It says Factor the polynomial by factoring out the greatesy common factor, x+2. how do I factor a polynomial?

[restive badger]

im a bit lost when it comes to factoring

[dense crystal]

try letting y = x +2 so you can visualize how the factoring is done

,,x^2y-4y=0

[soft zealot]

Renz

[dense crystal]

y is common so factor that out

[restive badger]

wait

[dense crystal]

,,y(x^2-4)=0

[restive badger]

where did you get that equation from?

[soft zealot]

Renz

[dense crystal]

and y is x+2 so you just put that back in

[restive badger]

im still confused.

wait so to make it easier for myself I just imagine this into the equation.

no its still not making sense to me

[solar glade]

Maybe start with something easier

Do you know how to factor ax-bx?

[restive badger]

(a)(x)-(b)(x)

I did not take intermediate algebra, I jumped into a college algebra class after a 5 year hiatus.

[solar glade]

You factor as x(a-b)

[solar glade]

If you expand this you will get the initial

x * a - x * b = ax-bx

you remember how to expand right?

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@restive badger Has your question been resolved?

[restive badger]

barely

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[drowsy epoch]

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[tranquil pine]

curious as to why it's positive 3

and not negative

[gritty solar]

what is positive 3

[tranquil pine]

dot product between (x,y)1 and (x,y)2

[gritty solar]

it is negative though

[tranquil pine]

yeah

but why is it

[gritty solar]

oh the cosine

[tranquil pine]

yea

[gritty solar]

yeah should be negative

[tranquil pine]

but it isn't apparently

same thing here

[gritty solar]

lol

are these vectors

or are they direction cosines

[gritty solar]

/ratios

[gritty solar]

yeah okay acute angle

[tranquil pine]

seems wherever it's negative they flip it positive

[gritty solar]

then it's fine

and they are lines not vectors

[tranquil pine]

but wherever it's positive they keep it positive

[tranquil pine]

but isn't that relevant

for the theta value

[gritty solar]

if you draw two intersecting lines you will find one acute and one obtuse angle

you're only being asked the acute angle

[tranquil pine]

acute means positive?

[gritty solar]

the cosine is positive yes

[sharp goblet]

<90

[sharp goblet]

Yeah correct

[tranquil pine]

oh. i see

alr thx

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[quiet ivy]

Basically I have to find the volume bounded by the function Z, bounded by y=5 and they also bounded by the first octant

Can anyone verify my double integral?

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[empty wharf]

i think the ans key is wrong, i got 5 as the answer

but i didnt get any sleep last night so i might have done it wrong

[robust mulch]

show your work please

[empty wharf]

im on my pc, but ill try to send it one moment

[pseudo knot]

that's correct

[empty wharf]

5 is the answer?

[pseudo knot]

yeah

that integral is 1/2 so 10I is 5

[empty wharf]

yes thats what i got, the answer key says 0

[hybrid heath]

yeah it should be 5

I'm assumimg answer key did pi instead of pi/2 by mistake, if I had to guess

[empty wharf]

hm yeah makes sense, thanks for the help

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[proper yoke]

"Calculate the flow integral ... where S is the mantle surface to the paraboloid and N is the inwardly directed (with positive z-component) unit normal. Is it possible to close the closed surface and use Gauss's theorem as in the previous problem?"

[proper yoke]

i have calculated ∇ * F = 0 and the top and bottom disks are z = 24 and z = 12. i have calculated 1/(x^2+3y^2) [-3y, x, 1] [0, 0, -1] = -1/(x^2+3y^2)

and we have x = rcos(theta) and y = rsin(theta). and from x^2+3y^2 = 12 and x^2/12 + y^2/4 = 1 we get the permetarization (2sqrt(12r)) and we have the bounds 2pi to 0 and 1 to 0

so what exactly am i supposed to integrate?

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@proper yoke Has your question been resolved?

[proper yoke]

<@&286206848099549185>

[final saddle]

@proper yoke Has your question been resolved?

[final saddle]

@proper yoke Has your question been resolved?

[drowsy epoch]

Why did you calculate ∇ * F = 0?

[proper yoke]

that what is became -6xy/(x^2+3y^2) + 6xy/(x^2+3y^2) + 0 = 0

F(x,y,z) = (-3y/(x^2+3y^2), (x/(x^2+3y^2), 1/(x^2+3y^2)

[drowsy epoch]

hmm

[feral ginkgo]

no you can't just 'close' the surface, cuz then it's a different surface. And since it's not closed you can't use Gauss Theorem. The way to evaluate the integral is by parametrisation:
$\int \int_{S} \textbf{F} \cdot \hat{\textbf{N}} \ dS = \int \int \textbf{F}(\textbf{x}(r, \theta) \cdot \left( \frac{\partial \textbf{x}}{\partial r} \cross \frac{\partial \textbf{x}}{\partial \theta} \right) \ dr \ d\theta$

[soft zealot]

lgkoo

[drowsy epoch]

yea

[feral ginkgo]

I don't think that answers the question why you calculated $\nabla \cross \textbf{F}$

[soft zealot]

lgkoo

[proper yoke]

Yes okay

Well I am not sure why

[feral ginkgo]

also, note that the top and bottom of S is an ellipse, not a circle, so the parametrisation for (x,y) is not rcos(theta) and rsin(theta)

[proper yoke]

Hmm okay

I got help a few days ago that’s why I used rcos and rsin

How about this

[feral ginkgo]

the method looks right, you just need to fix the parametrisation

[proper yoke]

Oh really what wrong

[feral ginkgo]

as I said, rcos(theta) and rsin(theta) is the parametrisation of a circle, not an ellipse. The diagram should also be ellipses

[proper yoke]

Oh okay

[drowsy epoch]

well it just changes by some factor 3

still looks like a mess to solve

fuck

,,\vec{s}(r,\theta) = \begin{pmatrix} \sqrt{3}r\cos\theta \ r\sin\theta \ 3r^2\cos^2\theta + 3r^2\sin^2\theta \end{pmatrix} = \begin{pmatrix} \sqrt{3}r\cos\theta \ r\sin\theta \ 3r^2 \end{pmatrix}

sqert(3)

[feral ginkgo]

yh

lol before I was gonna say it

[drowsy epoch]

I feel so stupid

😭

[feral ginkgo]

lemme make you feel worse; you also forgot the 3 in front of r^2 sin(theta)

[drowsy epoch]

yes

[soft zealot]

𝔸dωn𝓲²s

[proper yoke]

Ohh

[drowsy epoch]

Do we need to normalize the the normal vector?

[feral ginkgo]

no

[drowsy epoch]

is there a reason

[feral ginkgo]

yes, lemme see if I can find, cuz I've shown that proof to my friend before our amv test

[drowsy epoch]

so the norm cancels out

[feral ginkgo]

yeah

[drowsy epoch]

nice thanks buddy

[feral ginkgo]

np

[proper yoke]

What is an amv test?

[feral ginkgo]

amv stands for analysis in many variables in my uni

which is just second year calculus

[proper yoke]

Ohh that does not sound fun

[feral ginkgo]

no it's a lot of fun

way better than theoreotical modules like

probability and algebra

those are so pure

[proper yoke]

Hmm okay

Well thank you for the help!

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[empty wharf]

!status

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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.
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7. None of the above

[empty wharf]

I got -10 and -1 as the answer, so number of solutions is 2

I need someone to confirm that -10 and -1 are the answers please

[sharp goblet]

,w sqrt(log_10(- x)) = log_10(sqrt(x ^ 2))

@empty wharf Correct

[soft zealot]

Results provided by WolframAlpha

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

Requested by itzmecairo

[empty wharf]

well apparently -1 isnt valid according my textbook

[tulip coyote]

Well, it should work (unless they're doing something weird with their notation), after all sqrt{x^2} is just |x|, so the right hand side becomes log10(1) = 0, as does the right side become sqrt{log10(1)} = sqrt{0} = 0

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[molten roost]

im not sure how to do this

[tulip coyote]

How are points of inflection defined?

[molten roost]

second derivative = 0

[tulip coyote]

Alright, and can you find the second derivative of g here?

[molten roost]

no

oh right

[molten roost]

no

[tulip coyote]

Do you know what the first derivative of this g is, at least?

[molten roost]

well it shows an integral sign so after you take the integral you have to take the derivative again?

[tulip coyote]

Well, I mean you could, but you should ideally remember the FTC

[molten roost]

wahts that

[tulip coyote]

That if you have the function $F$ defined by $F(x) = \int_a^x f(t) \dd t$, that $F'(x) = f(x)$

[soft zealot]

@tulip coyote

[molten roost]

well wouldnt that be the case if thre was a d/dx next to the integral

[tulip coyote]

How do you mean, as in that $\dv{x} \int_a^x f(t) \dd t = f(x)$?

[soft zealot]

@tulip coyote

[molten roost]

okay im getting confused

[tulip coyote]

Alright, as a tl;dr of the FTC, it's saying that if you're differentiating something like your original g in the question, you just take the function you're integrating, replace the t's with x's, and then take that as the derivative of the "integral function"

[molten roost]

i see

kinda atleast

[tulip coyote]

Basically but remove the integral sign too, you just have $g'(x) = x^2 - 5x - 14$

[soft zealot]

@tulip coyote

[molten roost]

right

then do it again

i assume

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[light violet]

how do i do this

[tulip coyote]

So for the lower tail, you found X <= 5 as the critical region, right? And the probability of being in that region as 0.0232578...

[light violet]

yes

im pretty sure

[tulip coyote]

What about for the upper tail? Did you find anything for that?

[light violet]

uhhh