help-17

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[vast shale]

why is the answer d and not second picture ?

[vast shale]

(sum of geometric)

[vale snow]

Do you know the formula for geometric series?

[karmic mica]

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

[vast shale]

yes it's 1-q^(number of terms)/1-q

[karmic mica]

r = 1/e

[vast shale]

it's that ?

[karmic mica]

i suppose you set r = e

[vast shale]

r=e^(-1)

[karmic mica]

then you made a mistake somewhere

d is correct

[vast shale]

how did you go from line 1 to line 2

[karmic mica]

1-(1/e) = (e-1)/e

use this

[vast shale]

can you write it in latex or something I can't understand i'm trying to write it on paint but

the lines

[karmic mica]

uhh

maybe someone else can help with latex

[vast shale]

okay

[vocal sleet]

@vast shale Has your question been resolved?

[vocal sleet]

@vast shale Has your question been resolved?

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[vast shale]

it's not but I'll say yes

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[lofty hornet]

ok im bad at set theory

[lofty hornet]

can i get some guidance as to how to go about such a proof

all i know about proper subsets

is that

A is a subset of B

there is at least on element in B not in A

idk how to prove a subset or set tbh

[empty frigate]

well A is a subset of B if every element of A is also an element of B

so
"let x be an element of A
...
therefore x is an element of B"

once you know it's a subset, to show that it's a proper subset, ...the obvious way would be to give an element of B that's not an element of A

[lofty hornet]

this is the part i do not know how to do

i dont know how to i can represent the tuples with arbitrary values

just say that both the values are even, therefore its in B??

[empty frigate]

well you could just make some variables for each value

`let $(s,t) \in S \times T$' or whatever

[twin meteor]

bee

[empty frigate]

and then, since s is 0 mod 4, it's even, therefore it's in E

and t is 2 mod 4 so it's even so it's in E

therefore (s,t) \in E \times E

[lofty hornet]

man im rly bad at the write up part of discrete maths

or maths in general

im just not very good ahaha

ok

so

i understand representing the tuple as some arbitrary (s, t) for S x T

but if i break down the values in each half of the tuple

and link it to the values in the tuple it is being compared with (evens, evens)

then it means that the tuple exists in the other set

what do i say for the part with being a proper subset

(s != t), but e can equal e?

[empty frigate]

well find an element of E \times E that isn't an element of S \times T

[lofty hornet]

therefore element of A not in B

therefore proper subset

ill try a write up

see how it goes i guess

@empty frigate how does this look as a rough proof?

[empty frigate]

...i don't understand any of this

what does $(s, t) \subseteq (e_1, e_2)$ mean?

[twin meteor]

bee

[lofty hornet]

sorry

uhh

im rly tired

that should be

(s, t) subset of E

or actually

s, t in E

s and t are both individually in E

so (s, t) is in E x E

[empty frigate]

that sounds a lot more reasonable

...so then the other thing i don't get is what you did to conclude that it's a proper set

why not just come up with an actual element of E \times E that isn't an element of S \times T

[lofty hornet]

ok so i dont know what notation

[lofty hornet]

is that rigorous?

is it not better to show every element not shared

[empty frigate]

...what do you mean by "show every element not shared"

[lofty hornet]

like

for example

2,2

is in ExE

but cannot be in SxT

bcoz 2 is only in T not S

or does it literally not matter

and its fine to just show a single element

[empty frigate]

why would you show multiple elements

if B contains even one element that A doesn't, then they're not the same set

${1, 2} \subset {1, 2, 3}$ for example

[twin meteor]

bee

[lofty hornet]

ok fair enough

can i say

supposed n = 2, 2 in E, and ExE = (2,2)

2 not in S, but 2 in T, therefore (2,2) not in SxT

therefore SxT proper subset of ExE

[empty frigate]

notation objection: (2,2) is in ExE, it's not equal to ExE

[lofty hornet]

sorry my bad

[empty frigate]

also mentioning that 2 is in T is a bit redundant, the fact that 2 isn't in S is sufficient to conclude that (2,2) isn't in SxT

also what's n for here

[lofty hornet]

they used n

[empty frigate]

but yeah the basic idea is good
(2,2) is in ExE, but it's not in SxT, therefore SxT is a proper subset of ExE

[lofty hornet]

so i thought i should

ok sick

ill write it up nice tmw morn

i cant keep my left eye open rn, too tired, but ty for help

appreciate alot

have a good night/day!

.close

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[vital oracle]

simple but i dont understand

[opal raft]

have you tried to visualize it?

[vital oracle]

my brain doesnt work like that 😭

[lime gorge]

Ok, then draw it

Have u tried doing that

[vital oracle]

no i haven't i will right now tho

[opal raft]

i tried on graph paper, you can see the squares there

[vital oracle]

could u guys use a random number as an example and try to explain to me how to convert?

[lime gorge]

Let’s just do it using the drawing

Have u drawn it?

[vital oracle]

yep

[lime gorge]

Alr

So what do we know about the sides of a square

[vital oracle]

ummm there are 4 sides??

[lime gorge]

Yes, and how about a property about their length?

Relative to each other

[vital oracle]

they are equal

[lime gorge]

Correct

So let’s call one side “s”

Give me an expression that represents the perimeter of a square with a side length of “s”

Remember that the perimeter is the sum of the lengths of the sides

[vital oracle]

i dont follow, does expression mean equation??

p=4 x s idk how to write it

[lime gorge]

Yes, p = 4s. For future reference, an expression is without an equals sign.

[vital oracle]

ohh okay

[lime gorge]

So we now have the value of the perimeter and that the p=4s

Solve for s

[vital oracle]

i forgot in algebra u dont need a multiplication symbol

[lime gorge]

Lmk what u get

[vital oracle]

p/4=s ?? idrk

[lime gorge]

Ok

p = 4s

What’s the perimeter given in the problem

[vital oracle]

4s

oh

56

[lime gorge]

Yes

Now solve for the length of one side

[vital oracle]

is it 56 divided by 4?

[lime gorge]

56 = 4s, so yes, divide both sides by 4 to solve for s

[vital oracle]

56dividedby4 = 14

[lime gorge]

Ok

So s=14

Now we have the length of one side of the square

And we want the area

What’s the formula for area of a square?

[vital oracle]

side x side?

[lime gorge]

Yes

[twin meteor]

Stephen

[vital oracle]

i dont understand

[lime gorge]

What don’t u understand?

[vital oracle]

the dot in the middle

[lime gorge]

Just means multiplication

x

[vital oracle]

ohh

wait i stll dont get it

is A for area

[lime gorge]

Yes

[vital oracle]

and does s=14 rn

[lime gorge]

U alrdy got it here

[vital oracle]

sooo 14 squared?

[lime gorge]

Ye

[vital oracle]

is this the same thing?

orr

[opal raft]

this is for the previous one

if it's correct...

another way to solve it

[vital oracle]

this method is confusing to me

[opal raft]

you can check for simple versions of the same problem, and predict what will happen with more complex versions

4 is the four sides, s is the length, so 1 for the first square, it has an A area of 1

[opal raft]

have you tried to visualize it? 🙂

[opal raft]

if you simplify the problem you can often understand how it works

and study how for example sides and areas and perimeters connect

[vocal sleet]

@vital oracle Has your question been resolved?

[vocal sleet]

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[vital oracle]

.reopen

[vocal sleet]

✅

[vital oracle]

still confused on this

[soft walrus]

call $b=9.8$ we know $A=bh$ and that $P=2b+2h$ so is there someway you can solve for $h$, then by extension get the perimeter just using the formula?

[twin meteor]

XxMrFancyu2xX

[vocal sleet]

@vital oracle Has your question been resolved?

[zinc patio]

what the f

[languid berry]

?

[trail mesa]

trolling

his only messages are

[vocal sleet]

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[vocal sleet]

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[vast shale]

What is the slope of (6, 90) and (11, 165)

[vocal sleet]

Please don't occupy multiple help channels.

[vast shale]

i dont know how to find the slope

[flat whale]

you have a channel already

.close

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[solemn spear]

how should I integrate this?

these terms look difficult

will I be using something from here?

[eager walrus]

juse use power rule

@solemn spear

[eager walrus]

those too

😊

[solemn spear]

so ill have t^4 / 4

what about the 4/5?

[flat whale]

.

what do you think c equals in your first integral

[solemn spear]

I'm using power rule and whatever that rule at the same time??

[eager walrus]

yes

you can pull out constants

$\int 5x dx = 5 \int x dx$

[twin meteor]

sills

[languid berry]

i think maybe you just have a flawed understanding of how fractions work

i remember you from earlier trying to teach you how to subtract a fraction from a number

[solemn spear]

no

not here

dont bring that up

so i'll have 4/5 * x

[flat whale]

you can combine these two into a single step yes. or leave them as separate

[solemn spear]

will the x, or in this case t, be with the 4 in the fraction or right next to the / sign?

[flat whale]

what is this supposed to equal in your problem

[solemn spear]

the integral of 4/5

[flat whale]

where are you being asked to integrate just 4/5?

[solemn spear]

here

[flat whale]

i only see a 5

not a 4/5

[solemn spear]

that was just an example to show me the x is there when we find the antiderivative of a "normal" number

or am I missing something?

[flat whale]

Still don't know where the 4/5 is

[solemn spear]

idk what you want me to do. theres a t^3 attached to the 4/5. I thought that person earlier was trying to say I can find the antiderivatives individually

[eager walrus]

you can

@solemn spear you can either seperate everything and pull the constants out or you can just starting doing power rule on each term

[languid berry]

you can just start doing power rule

what's the power rule on t^3

[solemn spear]

what do you recommend?

t^4 / 4

[eager walrus]

start by doing power rule

[languid berry]

multiply that by 4/5

[solemn spear]

oh god

[languid berry]

as i said

[eager walrus]

just stay organized

[languid berry]

i believe you have a lack of understanding of how fractions work

[eager walrus]

make sure your work is easy to understand

[solemn spear]

fractions cross multiply correct?

[eager walrus]

yes

[solemn spear]

ok so that gets me 5t^4 / 16

[eager walrus]

you didn't simplify correct

ly

we'll do the first term together

[languid berry]

wait

i suggest reviewing how fractions add, subtract, divide, and multiply

[solemn spear]

ok t^3 -> t^4 / 4

[languid berry]

you had issues with subtracting fractions earlier and now multiplying

[eager walrus]

es

yes

[languid berry]

https://www.youtube.com/watch?v=rl7e0djo9Go&ab_channel=MathwithMr.J

[eager walrus]

$\int \frac{4}{5} t^3 dt = \frac{4}{5} \cdot \frac{t^4}{4}$

[twin meteor]

sills

[languid berry]

from earlier

[solemn spear]

ope

wipe those logs

[eager walrus]

@solemn spear does this make sense

[solemn spear]

yes

[languid berry]

watch this video you're gonna need to do a lot of fraction operations going into further math

[eager walrus]

do you know how to simplify from that

[solemn spear]

well I can either make it decimal, which is bad. or a secret 2nd thing

[eager walrus]

what is the secret 2nd thing

[flat whale]

Don't use decimals. Just watch the video and practice fractions

[solemn spear]

idk why I thought we multiplied like this

[flat whale]

That's why this video is important

[solemn spear]

I vaguely remember something happening that way with other problems tho

other problems as in years ago

[eager walrus]

i agree with the other ppl

u dont have ur algebra skills down

[solemn spear]

ok I get it now 4t^4 / 20

t^4 / 5

[languid berry]

you gotta learn to crawl before you can walk

you're trying to do calculus with a flawed understanding of fraction operations

[solemn spear]

i dont have time to crawl

school is almost over

[eager walrus]

ok well you are crawling rn because you are stuggling to do pre-pre-calculus tasks

[languid berry]

pre-pre-algebra

[eager walrus]

true

tbf its not easy though. we were struggling to do lcd in calc 2 the other day

[solemn spear]

ok so my next terms antiderivative is t^3 / 4

[languid berry]

[languid berry]

wait

lcd?

[eager walrus]

least common denom,inator

[solemn spear]

not me

[languid berry]

isnt lcd always 1

[eager walrus]

i apoligize

i meant gcd

[languid berry]

ah

[eager walrus]

wait

it was just adding fractions

ok time to stop griefing this channel;

[languid berry]

do u mean

lc

lcm

[solemn spear]

ok so my last term is t^2 / 5

I have 3 terms, 1 with a 4 in denom and 2 with a 5 in denom

in order to add or subtract them i need common denom

20

but I can't just put a 20 in denominator, I need to also ...

something something the numerator

[languid berry]

^

[zenith belfry]

guys imma start being political 😡

[languid berry]

huh

[solemn spear]

get outta here jimmy

[zenith belfry]

nah bruh stop screenshotting until its good

[karmic imp]

Stop interrupting someone's channel

[zenith belfry]

$0^0$ is undefined

[twin meteor]

Jimmy Neutron

[zenith belfry]

nah fr mb

[solemn spear]

this is the 3rd time in 24 hours i've been intruded on

[zenith belfry]

3rd times the charm am i right

alr sorry bye

[solemn spear]

ok so I now have 4t^4 - 5t^3 + 4t^2

all over 20

do I now start putting in my 4 and 0 to do f(4) - f(0) ?

[solemn spear]

am I at this point?

[eager walrus]

ye

[solemn spear]

since the f(0) will pretty much "void" that part of the process, I really only need f(4) and thats the answer?

it all just multiplies with 0/20

very cool

[vocal sleet]

@solemn spear Has your question been resolved?

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[peak herald]

unsure about 5 and 9

[mild herald]

for 5 it should be root 3

[peak herald]

oh how?

[mild herald]

do you know how the ratios work in similar triangles?

[peak herald]

yes

[mild herald]

for volume, area and side length

okay uh

ill draw something one sec

imagine these lengths are l and z

[peak herald]

ok

[mild herald]

wait sorry i meant to cover the whole base

[languid berry]

dont give answers give methods

[mild herald]

like this

[mild herald]

i am explaining how i reached to that answer

[peak herald]

ok

[mild herald]

now what would be the area for triangle dog

in terms of x and l

1/2 x base x height

right?

[peak herald]

yea

[mild herald]

so that would be

1/2 xl yes or no?

[peak herald]

yes

[mild herald]

okay similarly can you do this for cat

tell me whats the area

[peak herald]

1/2 xz

yz

mb

[mild herald]

close

it should be yz

yep

correct

okay so what is the ratio of the areas

in terms of x,y,l and z

just divide the two areas

what do you get

[peak herald]

uh 1/2 xl divided by 1/2 yz

[mild herald]

simplify it

we can rid of the 1/2

[peak herald]

ok

[mild herald]

since its there in both the numerator and denominator

okay is everything clear so far

theyve mentioned in the question that the ratio of areas is = 3

[peak herald]

yea

[mild herald]

so that implies that

so we know that xl/yz = 3

[peak herald]

yea

[mild herald]

now keep that aside

in similar triangles

my side lengths are doing to be proportional

so what that means is

if i find the ratio between l and z

that should be the same as the ratio between x and y

think of triangle dog as a sort of larger version of triangle cat

[peak herald]

ok

[mild herald]

everything clear?

[peak herald]

yes

[mild herald]

so i can write l/z =x/y

yeah?

[peak herald]

yeah

[mild herald]

know let me assign some random variable to l/z

let me call this rratio as

uh

P

now go back to our first equation

the ratio of areas

this is what we had right

[peak herald]

yes

[mild herald]

i can see that ive got both l/z and x/y

so substitute the ratios for P

does this make sense?

[peak herald]

uh let me see

oh I see now

P is our answer

[mild herald]

sure let me know if you have any doubts

[mild herald]

and weve just now found that

P^2= 3

[peak herald]

yea

[mild herald]

so that would mean that P = root 3

and we had to find P which was our ratio of sides right

[peak herald]

yes

[mild herald]

so our final answer is root 3

and not 3

this is the reasoning behind why our answer is root 3

if you want to skip this

you could just remember the result

[peak herald]

alright

[mild herald]

there are a couple of youtube videos

that just give the results

so maybe watch one of those and remember the result

[peak herald]

alright

but im totally confused about 9

[mild herald]

let me check

[peak herald]

here

[mild herald]

oh

okay so

basically

do you know what a maximum and minimum is

[peak herald]

yes

[mild herald]

this right

[peak herald]

yes

[mild herald]

so whats the maximum value

[peak herald]

1.75

[mild herald]

and minimum?

[peak herald]

1.5

[mild herald]

yep

now the questionn has mentioned that

to find the amplitude

we have to find half of the difference between these values

[peak herald]

yes

[mild herald]

this line

[peak herald]

yea

[mild herald]

so whats the difference between the max and min values

[peak herald]

.25

[mild herald]

now whats half of that

[peak herald]

.125

[mild herald]

so what does this value give us

[peak herald]

so the amplitude is basically 1.625?

[mild herald]

where did 1.5 come from

[peak herald]

the minimum

[mild herald]

why did you add it

the question states that to find the amplitude

all you have to do

is find the difference

and divide it

theres no need to add the minimum value

[peak herald]

oh

[mild herald]

make sure you read the question carefully

[peak herald]

guess I confused it with smth else

so basically .125

[mild herald]

could be

[mild herald]

yep

[peak herald]

alright ty

[mild herald]

np

[peak herald]

have a good day

[mild herald]

you too

[peak herald]

.close

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[thin vale]

<@&268886789983436800> "I'm a ten yr old" ....

[vocal sleet]

Channel closed

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[lost yarrow]

dealt with

[vocal sleet]

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[blazing lintel]

Hello, I'm trying to find a vector B such that the distance from vector A to vector B is equal to some distance.

A is given, and the distance is given. How would I solve this?

[paper depot]

http://xyproblem.info/ ?

[blazing lintel]

I don't know how to set it up, would i do $d = \sqrt{(x_{a} - x_{b})^{2} + (y_{a} - y_{b})^{2}}$ and solve for $x_{b}$ and $y_{b}$?

[twin meteor]

beanbeanjuice

[blazing lintel]

doing it that way gives 2 unknowns though

[paper depot]

as is, the possible solutions to your problem form a whole circle of points.

so i would like you to show the original problem exactly as it was given to you or came up in your project or whatever.

[blazing lintel]

I don't really have an original problem... I'm working in Unity using C# and have vectors that I'm trying to do.

I guess if I were to make my OWN question about it is,

I want the vector B to be at the position where vector A is facing with the same distance

Distance(A to B) = d such that B is facing the same direction as A

[paper depot]

oh so you want B to be in the same direction as A as well

and you want |A-B| = d

[blazing lintel]

Yes

[blazing lintel]

I'm thinking I use this equation and somehow solve for $x_{b}, y_{b}$ and multiply by the unit vector of position A

[twin meteor]

beanbeanjuice

[blazing lintel]

but I am probably wrong

[paper depot]

you're overcomplicating things

since B is in the same direction as A you know that B = kA for some scalar k

for some positive scalar k

and |B - A| = |kA - A| = |k-1| |A| = d

so |k-1| = d/|A| and now you have k = 1 ± d/|A| which may give you one or two possible values depending on if d is shorter or longer than |A|

[blazing lintel]

I see an |A| is the magnitude of A right?

okay let me try this..

[paper depot]

yes

[blazing lintel]

I got it, thank you!

.close

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[tiny obsidian]

Hello, Kepe. What have you tried?

[river minnow]

What about connecting two big circles' centers with the small circle's center?

[merry python]

It is

Use trig

[tiny obsidian]

You're headed in the right direction. Try drawing lines from the vertices to the center of the smaller circle and see what observations you can make from that.

[merry python]

Join centre of small circle and large circle

Then use cos30

And hypotenus is R+r

While base is R

[lapis marten]

If you connect the center of the small circle to the meeting point of 2 bigger circles you get a 30 60 90 triangle

[river minnow]

Either use law of cosines or draw a line from the small circle's center perpendicular to the triangles' sides

[vocal sleet]

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[vast shale]

,rotate

[vast shale]

<@&286206848099549185>

[errant briar]

Denominator is basically the same thing

[vast shale]

Yeah

it is

but what after that

let me show my work

i can't think of anything

[errant briar]

What is the answer?

[vocal sleet]

@vast shale Has your question been resolved?

[vast shale]

idk who cares

i didn't check yet

[vocal sleet]

@vast shale Has your question been resolved?

[late sky]

What do they mean by AP?

[strange crater]

arithmetic progression maybe?

[vast shale]

Advanced Placement

[late sky]

Audemars Piguet buss down

[vocal sleet]

@vast shale Has your question been resolved?

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[crimson sedge]

im a bit confused

[crimson sedge]

if i say

$f(x) = e^x$

[twin meteor]

yomiko

[crimson sedge]

$f(x) = e^x-4$ \ $f(x) = lnx$ \ so, if i do $ln(x) -e^x+4$ it would = 0 right?

[tiny obsidian]

not quite

[spiral radish]

nop

it would be +4

[crimson sedge]

yh +4 i mean

[twin meteor]

yomiko

[tiny obsidian]

yes

[spiral radish]

considering youre referring these equations as 2 different equations

[desert hornet]

that doesn't make any sense

[short nymph]

yo can someone help me

[desert hornet]

!help

[vocal sleet]

Please read #❓how-to-get-help

[short nymph]

i have a thread no one is replying

[desert hornet]

then wait

[soft walrus]

$\ln(x)$ and $e^{x}-4$ are two different function's, why would subtrating them yield 4?

[twin meteor]

XxMrFancyu2xX

[spiral radish]

i mean, the sign of -4 should be +4 instead

[soft walrus]

have pinged helpers after 15 mins?

[soft walrus]

ohhh ok

[tiny obsidian]

i guess it could for some x

[crimson sedge]

its f(x)??

[soft walrus]

what is your question, are you finding the zeroes of the function?

[tiny obsidian]

what exactly is your question?

[crimson sedge]

im trying to prove that lnx = e^x - 4 have a root between 1.4 and 1.5

but i need to make it into f(x) =

[spiral radish]

that doesnt make sense

it shouldnt have a root

[tiny obsidian]

it does

[spiral radish]

Oh my bad, i mixed up y and x

[crimson sedge]

,w plot y = lnx and y = e^x -4

[twin meteor]

Results provided by WolframAlpha

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

Requested by yomiko#1028

[karmic mica]

nah

[tiny obsidian]

what methods are you instructed/ permitted to use?

[karmic mica]

let g(x) = lnx - e^x + 4

now put 1.4 and 1.5 and check if they are positive or negative

[crimson sedge]

to sub in x = 1.4 and x = 1.5.... if there is a change of sign it would mean that there is at least 1 root between the points

[tiny obsidian]

indeed

[crimson sedge]

i get that part

its just the part to make the two function into one

[tiny obsidian]

you're not making them one. u're equating them to obtain a new function.

[crimson sedge]

right

i think i got it

[spiral radish]

this is big brain, i didnt think of such before, nice one

[crimson sedge]

f(x) = lnx = e^x -4
f(x) = e^x-4-ln(x)

its the easier part that I struggled on lol

[tiny obsidian]

don't call e^x - 4 - ln(x) f(x), call it something else. otherwise you're abusing notation and it risks being deemed erroneous.

[crimson sedge]

its reversed

[tiny obsidian]

it's the same

[crimson sedge]

alr

[crimson sedge]

my answer i got negative for 1.4 and positive for 1.5

but i guess it doesn't matter

[tiny obsidian]

indeed, it does not. wht matters is that the signs flip.

[crimson sedge]

because i just want to see that there is a change in sign

[crimson sedge]

yh

Thanks

[tiny obsidian]

Please remember to close the channel with .close if your question has been resolved.

[crimson sedge]

.close

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[azure star]

membership notation

why sometimes we write x "\in A" on the left hand
side of the set building notation, sometimes we
write x "\in A" in the right hand side of the set building notation? Thank you.

[vocal sleet]

@azure star Has your question been resolved?

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[peak shard]

how do I show that these are equal?
$\ \sum_{k=1}^\infty (-1)^k \cdot \binom{-\frac{1}{2}}{k} \cdot x^{2k} = \sum_{k=1}^\infty x^{2k} \cdot \frac{(2k+1)!}{2^{2k}(k!)^2}$

[twin meteor]

tobi.

[vocal sleet]

@peak shard Has your question been resolved?

[vocal sleet]

@peak shard Has your question been resolved?

[vocal sleet]

@peak shard Has your question been resolved?

[azure star]

❌

[vocal sleet]

@peak shard Has your question been resolved?

[placid lake]

u still neeed help?

[peak shard]

actually I solved this

[placid lake]

grand

[peak shard]

just now

[placid lake]

i jus saw it now lol sorry

[peak shard]

!close

[placid lake]

.close

[strange crater]

.close

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[proper karma]

Why isn’t the answer (E)? The correct answer is C (3) but I don’t get it

Like isnt the limit DNE

[peak matrix]

no

there is x^2

and x^2 will be always positive. So for both -0.000...1 and 0.00...1 it will be f(1-small positive number)

[proper karma]

Yeah i follow

Ohhh

So its always gonna be

F from the left

[peak matrix]

yep, it's same as limit of f(x) as x approaches 1 from left

which is just 3

[proper karma]

Wow that’s hard

Thanks for the help

This calc bc exam is gonna whoop my ass

[placid lake]

bc?

[proper karma]

Yeah

[placid lake]

whats that

[proper karma]

It’s a math course

Basicallt calc II

[placid lake]

inrterseting ok

[proper karma]

Thanks t’all

.close

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[north venture]

I actually don’t know what I messed up on here

[dapper meteor]

2nd line

2nd integral

is from -3 to 2

not -5

So you end up with $\int_{-5}^5f(x)dx$

[twin meteor]

ByShaDowZ

[north venture]

Weird

[dapper meteor]

you don't see it ?

[north venture]

No…

[dapper meteor]

ok

when you go from line 1 to 2, you are dividing the left integral into 2 parts

[north venture]

Okay.. so my confusion lies on the number on the bottom.. and why it’s incorrect

[dapper meteor]

$\int_{-5}^2f(x)dx = \int_{-5}^{-3}f(x)dx + \int_{-3}^2f(x)dx$

[twin meteor]

ByShaDowZ

[dapper meteor]

That's what you should have written

do you understand that ?

[north venture]

Actually no..

[dapper meteor]

that's this rule : $\int_{a}^bf(x)dx = \int_{a}^cf(x)dx + \int_c^bf(x)dx$

[twin meteor]

ByShaDowZ

[north venture]

So from that second line I should combine like terms?

Oh alright.. nvm.. thank you

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[visual frost]

hello

[visual frost]

can anyone help me with this integral?

integral of x^4*e^(x)^3 dx

I have thought ab substituting

but so far,to no avail

😦

[midnight rapids]

Maybe try by parts?

[visual frost]

hmm

the only derivative that i can use is x^5/5

and then

what would e^x^3 be>

lemme try

no

still doesn't work

[flat whale]

Did you type this wrong

[visual frost]

no

[flat whale]

$e^{x^3}$ or $(e^x)^3$

[twin meteor]

riemann

[visual frost]

the first

[flat whale]

,w int x^4e^(x^3)

[twin meteor]

Results provided by WolframAlpha

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

Requested by riemann#2529

[flat whale]

Welp, good luck with that

[visual frost]

well

[flat whale]

Series could make sense

[visual frost]

let me backtrack

so

the integral I am looking to calculate is

definite integral from 0 to 2 of x^4*e^(x)^3+2 dx

[flat whale]

Show the original question. Screenshot or picture

[vocal sleet]

@visual frost Has your question been resolved?

[vocal sleet]

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[wary mantle]

Regarding quadrics, if you were given something like x^2 + y^2 + z^2 = 1, how would you sketch it quickly?

[wary mantle]

Or is that not usual and you would let a graphing calculator do that?

[pallid zenith]

there are a few things you can draw quickly

like spheres, ellipsoids

some conics

in each case youd just highlight the important bits

and be interpretive with the rest

[wary mantle]

Okay

.close

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[pallid zenith]

that was easy

[wary mantle]

I didn't have a concrete question, I will learn about quadrics in two weeks or so in a student seminar, I will learn ahead about them

[pallid zenith]

it can help to think of slices

like, imagine the xy-plane, the yz-plane, etc

and imagine the 'slice' of the surface or whatever it is that passes through that plane

[wary mantle]

Alright

[pallid zenith]

this is really helpful with conics

lets you draw like, a wireframe

[vocal sleet]

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[vast shale]

Given N = 8^8 + 9^9 + 10^10

[vast shale]

Find remainder

when divided by 18

No clue where to start

No knowledge of modulus

[midnight rapids]

Have you learnt binomial expansion?

[vast shale]

yes

[midnight rapids]

Okay you could try using that

For 9^9, the remainder will be 9

[dark kiln]

how what

[midnight rapids]

Cause all even multiples of 9 are divisible by 18

So odd multiples of 9 have to leave a remainder of 9

[dark kiln]

how to expand 9^9

[midnight rapids]

You don’t have to

Its gonna be an odd number and its a multiple of 9

So itll leave a reminder of 9 when divided by 18

Now 8^8 = (18-10)^8

Do the binomial expansion

All terms are divisible by 18 except for -10^8

Now (-10)^8= 100^4 = (90+10)^4

Binomial expansion, everything gets divided except for 10^4

=100^2

=(90+10)^2

Binomial expansion leaves you with 10^2 =100

Which leaves a remainder of 10

[dark kiln]

extremely fun, thank you

[midnight rapids]

Lol

Try the same thing with 10^10

So for 8^8 the remainder is 10

For 9^9 its 9

For 10^10 (try it out) its 10

So total remainder is their sum = 29

Now 29 divided by 18 leaves a remainder 11

that is the answer

[dark kiln]

(18−8)^10, leaves −8^10

then idk

is that a bad first step?

[midnight rapids]

You don’t have to split it tbh if ure comfortable dealing with 10

10^10 = 100^5 = (90+10)^5

Binomial expansion and divide, we have 10^5

[dark kiln]

right 10^5

[midnight rapids]

= 10*(100)^2

=10*(90+10)^2

We’re left with 10*10^2

[dark kiln]

i see

[midnight rapids]

10(90+10)

900 gets divided

We’re left with 100

And tada, 10 is the remainder

Btw the reason I took 90 and 10 is because 90 is divisible by 18 and the same thing gets repeated for 10

Usually the problems are easier than this

[wraith venture]

It's -64 * 10 = -640 = -100 = 8 mod 18 ?

[dark kiln]

basically 10^ anything is 90000 +90000 +9000 ... + 10
and 8^8 is (−10)^8

the 9 thing was really cool though

"it's odd and multiple of 9 must be 9"

cash money

[wraith venture]

No it's + 8^10

[dark kiln]

yeah

+8 you;re right

[wraith venture]

Then you get 10

[vocal sleet]

@vast shale Has your question been resolved?

[dark kiln]

no it's −8 what

it's the same thing as +8 but you do get (−8)^10

[wraith venture]

Which is positive

• 8^10 < 0 and is 8 mod 18

[dark kiln]

yes

i meant (−8)^10, me bad

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[jagged flax]

What would the exponential function be

[pallid zenith]

they tell you in 4?

[vocal sleet]

@jagged flax Has your question been resolved?

[jagged flax]

Its telling me to explain what the variables represent

I dont know what it would be though

[pallid zenith]

have you tried desmos?

[vocal sleet]

@jagged flax Has your question been resolved?

[jagged flax]

Not sure how it’d help

[pallid zenith]

youd understand what they control

[jagged flax]

?

[vocal sleet]

@jagged flax Has your question been resolved?

[jagged flax]

Ohhh, I know what they control but im asking what would this specific problem be because idk

[pallid zenith]

write out the number of people in each step that have the bread

[vocal sleet]

@jagged flax Has your question been resolved?

[jagged flax]

I know but what would the exponential function be

[pallid zenith]

well at x=1 what should the value be

how about at x=2

[jagged flax]

Since this starts from 1 then goes to 4 instead of 1 to 3

[pallid zenith]

i think they want it to go like

3 9 27 etc

[jagged flax]

Yea but the 3 part is where im confused

Because wouldnt the original bread owner also be included

So 4?

[pallid zenith]

well the problem isnt clear

so why not make your life easy

[jagged flax]

True

Thx

.close

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[pallid zenith]

[vocal sleet]

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[vast sparrow]

hey I'm just a bit confused

[vast sparrow]

like ?.. so the answer is two which made sense before, I split up the functions, gettings 0/0 and 2/0 you use L'Hopitals rule to get -1, but then like I realized the LIMIT as you approach 0 like for the cos functino doesn't exist, so does the X go away somehow?..

[formal rock]

Hmm

I suggest you rearrange the function into

[mellow dock]

Can somebody please help me how to solve this
"Determine the area of the triangle with vertices ( _ , 5, 2) ; (1, _ , 6); and (2, 1, _) by evaluating the magnitude of a vector product."

[formal rock]

$f(x) = \begin{cases} \frac{\sin(x)}{x} + \frac{\sin(x) \cos(x) }{x}, ,,,,,x \neq 0 \ \
p, ,,,,,, x = 0 \end{cases}$

[formal rock]

#❓how-to-get-help

[twin meteor]

VulcanOne

[formal rock]

And when you take the limit of the first portion, you can use the special limit

limit as x tends to 0 of sin(x)/x = 1

[vast sparrow]

the special limit?..

[formal rock]

Yeah

[vast sparrow]

is that like a special exception

and then the second part is sinxcosx?..

like I know sinx + cosx = 1 but it's 1 + cos x

[formal rock]

$\lim_{x \to 0} \frac{\sin(x)}{x} \to 1$

[twin meteor]

VulcanOne

[heavy plover]

you can split the (sinxcosx)/x one into (sinx/x)cosx and take the limit of them separately

giving you 1*1

[formal rock]

That's not true. sin^2 x + cos^2 x = 1 is valid tho

[vast sparrow]

yeah that's what I did but P ends up being two which is killing my brain

wait so are there any other random limit rules I should be aware of?.. Is that special limit thing because of L'Hopitals or?

oh and the REASON it's sin(x)cos(x) is cause you're multiplying?.. but where does the + 1 go?

[formal rock]

No it's because of a geometric proof that proves the derivative of sin(x) and enables you to use l'hopital in the first place

[heavy plover]

sinx/x can be proved hundreds of ways its a very well known limit you dont even need lhopitals you can prove it geometrically

[vast sparrow]

but I thought like the SINE Function AT 0 is 0?.. hold up lemme type it into desmos

[heavy plover]

you arent evaluating it at 0 tho

ur taking a limit

sin(0)/0 is undefined but the limit isnt

[vast sparrow]

ohh shit

[formal rock]

The limit approaches 1

From the left and right

[vast sparrow]

wait so if you graph sin(x)/x it looks like at zero the Y value is 1?.. this is wack

[heavy plover]

yes

at then at 0 there is a hole in the graph

[vast sparrow]

damn I see sorry my brain is imploding

and the reason it's undefined is cause math just breaks as per usual

[formal rock]

Yeah

0/0

[heavy plover]

its like doing sin(0.000000001)/0.000000001

[formal rock]

But limit of this is 1

[vast sparrow]

cause like ON THE GRAPH it's definitely approaching 1 bruh, but evaluating sin(0)/0 is obviously 0/0, so that got disproved by a dude and THEN HE CAME UP WIHT L'Hopitals?!?

what a g

[formal rock]

Well

[heavy plover]

l'hopital didnt prove sinx/x i dont think

[formal rock]

L'Hopital came from derivatives

[vast sparrow]

;(

[formal rock]

And derivatives were proven by limits

[vast sparrow]

yeah yeah

[formal rock]

So yeah

So you can't properly use l'hopital without limits

[vast sparrow]

okay yeah so that first one be making sense, what about the second part?.. Im just having a hard time udnerstanding how it goes from (1+Cosx)/(x) to (sinxcosx)/(x)

[formal rock]

We distributed the sin x to (1+cos(x))

[vast sparrow]

oh shit

yeah okay, then like what does that equal

[formal rock]

sin(x) + sin(x) cos(x)

All divided by x

[vast sparrow]

OH wow we split it up in the sense that we distributed the sin(x) from the first part INTo the second one MAKING IT AN ADDITION PROBLEM now

so that explains the 2 but if you evaluate that is it not just like 2/0 since x is 0

then like cosx/x is just undefined? so it ends up being 1 * 1 bceause of the sins equaling 1

or 1 + 1

[formal rock]

Wait

Using the limit we had above

The sin(x)/x likit

Limit

We know it is 1 right?

[vast sparrow]

yeah ?..

[formal rock]

Okay and now we have cos(x) multiplied by it

And we also know that cos(0) = 1 right?

[vast sparrow]

yeah?..

[formal rock]

So 1*1

[vast sparrow]

yeah

[formal rock]

For the sin(x)cos(x)/x part

[vast sparrow]

yeah definitely

[formal rock]

And for the sin(x)/x part, we already did that

So in the end

The limit is 1+1 = 2

[vast sparrow]

yeah, it's just 2. Honestly I appreciate it a ton, it makes lots of sense now, but I do have one last thing, why is cosx/x just equal to 1?.. WHERE does the x in the denom. go

cause obviously cos(0) = 1, but the denom has an x, and like the limit approaching 0 there is a vertical aymptote

[formal rock]

Actually this is not the case

cos(x)/x as x tends to 0 is undefined

From the left, cos(x) = 1, but x = -0.000... so it becomes -inf, and same reasoning from the right

Right not equal left

So you get a non-existing limit

[vast sparrow]

wiat so what does that mean

[formal rock]

sin(x)/x has a limit

sin(x)*cos(x)/x also has a limit

But cos(x)/x doesn't have a limit

As x tends to 0

[vast sparrow]

ahh it makes sense, so it just kinda goes away?

[formal rock]

Which?

[vast sparrow]

the cosx/x part

of the sinxcosx/x

[formal rock]

No the cos(x) is multiplied by the sin(x)/x in the limit

So it gives a net 1

[vast sparrow]

I see so those are multipled over the x, so the x goes with the sin?..

hmm yeah, I really appreciate it. I hate to bother you more but can you help me with this super easy word problem?.. I honestly think I may just be too tired at this point. "At time t = 0, an object is thrown upwards from the top of a building that is
100 meters above the ground. The object is thrown with a velocity of 12 meters/second.
Find its height above the ground after 4 seconds. Use that the acceleration due to gravity
is 10 meters/second2 downwards"

[formal rock]

Okay so you know v_0 = 12

[vast sparrow]

yeah definitely

[formal rock]

And you know the acceleration = 10

And you know that acceleration is the derivative of velocity

[vast sparrow]

for sure

[formal rock]

Okay so

a(t) = g

= 10

Then undoing the derivative, we get

v(t) = 10t + constant

[vast sparrow]

damn 10x + 12?

qhh

[formal rock]

Plug in v(0) = 12 and find the constant

You get v(t) = 10t + 12

:)

[vast sparrow]

ahh I see wow, so then you do the same for displacement function

[formal rock]

Yep

You integrate again

[vast sparrow]

like so it'd be like d(t) = 5t^2 +12t + Constant, and solving for at time = 0 it'd be C = 100