help-28

[glossy valve]

Simon

[weary vector]

Can someone help me idk hownto do it

[onyx glen]

ok wait

is the sqrt(9/2) meant to be in the denominator?

you should use \frac{}{} to make actual fractions

$\frac{\sqrt{0.5}}{3\sqrt{\frac{9}{2}}} \cdot \left( -\sqrt{\frac{81}{4}}\right)$

[glossy valve]

ann.in.a.teacup

[onyx glen]

is the original problem like this?

@weary vector

[weary vector]

Yes

it is

Actuallt the last one is root of 81/4

the 4 is not in the root

[onyx glen]

sqrt(81)/4 then?

"root of 81/4" written like this is ambiguous

[weary vector]

Yea as long as 4 is not root 4

Idk how to do itt

[onyx glen]

ok let's try this

do you know that sqrt(ab) = sqrt(a) * sqrt(b)

[weary vector]

Yuo

Yup

[onyx glen]

do you also know what sqrt(81) is?

[weary vector]

9

[onyx glen]

this will be easier to show and write on paper

do you understand what i did here?

[weary vector]

Why 1/2

Oh nvm i got it

A/B = a * 1/b

I get it what u did

But what now

[onyx glen]

math is case sensitive

[onyx glen]

well there are two things you can do here

one is that sqrt(9) is also a value you can work out

the other is that the sqrt(1/2) cancel out

[weary vector]

Oh i see them

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[torn plinth]

how do we solve -sqrt(x) <= sqrt(x)

[torn plinth]

i tried squaring both sides, but i get x <= x

[onyx glen]

you can't square both sides if you don't know that both sides have the same sign

(and here they don't)

[torn plinth]

what do i do then? is this unsolvable

[hollow wharf]

What operations can you do on inequalities that don’t have strings attached?

[torn plinth]

add sqrt(x) to both sides?

[hollow wharf]

Try it

[torn plinth]

0 <= sqrt(x) + sqrt(x), 0 <= 2sqrt(x), 0 <= sqrt(x), 0 <= x

[torn plinth]

why is that

[hollow wharf]

-4 < 3 but 16 > 9

[hollow wharf]

yay!

[torn plinth]

lol

[torn plinth]

i see

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[final wing]

Hey can someone look at this proof for me?

[honest hill]

what is a?

[final wing]

oh i never put that an is a sequence ;-;

[honest hill]

a_n is just a sequence

that is convergent?

[final wing]

well i must proof that if the sequence is monotonic decreasing and is bounded that it must be convergent

alright gtg for now so ill just open a new one later today

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[tired blade]

do I find the equation of the tangent of y, then set it to 4x-y+11 = 0

[sharp flame]

yeah

a parametric tangent

[tired blade]

ok so to get the equation

I need a

point

any point will work

?

on y

so 1 maybe

[sharp flame]

Assume a general point (t, 16/t^2 - 1)

And find the tangent at that point

[tired blade]

ohh okay

yeah i misunderstood

ok thanks

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[drifting walrus]

-1 microcoulomb * (20-10)
gives a value of
-1.0x10^-7 or -1*10^-5?

[brittle reef]

$-10^{-6}\cdot10=-10^{-5}$

[glossy valve]

GoldBarley

[drifting walrus]

when im plugging my answer in the calculator im getting 10 raised to the power of 7

[drifting walrus]

10 ^-6 x 10 ^ 1
-6 + 1 = 5
i understand this

[brittle reef]

That's only if you divide by 10. Are you sure you're doing the operations correctly?

[drifting walrus]

yes
-1.0µC * (20-10)

[brittle reef]

Well, check again.
It should be -5 not -7.

[drifting walrus]

OHHHHHH

tysm

my bad

[brittle reef]

😅

[drifting walrus]

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[wanton talon]

hi guys need some help w my hw

[surreal oracle]

when completing the sqaure is the number outside always going to be negative or can it be positive

eg like in the picture there is a -6squared is that always a to the minus

[wanton talon]

need some help

[onyx glen]

!occupied

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Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[onyx glen]

(the other person came like a second earlier...)

[surreal oracle]

its alr i got an answer

[wanton talon]

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[dapper bobcat]

1. Suppose that the amount of product used in one day, represented by $Y$, has a gamma distribution with $\alpha=1.5$ and $\beta=3.0$.

a. Find the probability distribution for $Y$.

b. What is the mean of the distribution?

c. What is the variance of the distribution?

d. What is the probability that an earthquake striking the regions will be a minor earthquake (note: the earthquake will fall between 2.0 and 3.0 on the Richter scale)?

e. What is the probability that an earthquake striking the regions will be a strong earthquake (note: minimum value is a 6.0 on the Richter scale) or worse?

[glossy valve]

Mads

[dapper bobcat]

I only need help with d. I am allowd to use R-studio and I would love to be able to use that to help with this problem

This is what I also got for the other values.

[leaden ermine]

you can rewrite it

P(2 < Y < 3) = P(Y < 3) - P(Y < 2) = (1-P(Y > 3)) - (1-P(Y > 2))

[dapper bobcat]

AHHH! I think this is what we were trying to get to last night but I did not understnad it.

But with a fresh set of eyes, that makes more sense. I will try to use R-studio to help me out and then come back to check.
Thank you!

Do you think it matters that it is not >=?

[leaden ermine]

here it's not a discrete distribution but continuous so it doesnt make a difference

[dapper bobcat]

Thank you! This is what I found and I feel confident in this answer

I appreciate the help! I am going to work on a couple others and come back if I get stuck.

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[fallen shell]

how do i do this

[dry arch]

you can use log rules

2logx+logy=5
logy-logx=2

now you can solve for logx and logy

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[edgy epoch]

Hello! I am someone who is really into mathematics, and wants to self-study mathematics all over again to build a solid foundation, the reason I want to self study math isn't really for the academic purpose, although it'd probably help, but I just love mathematics purely. I have self-studied math in the past using tools such as youtube and khan academy, but I've felt like Khan Academy was bland and boring over the years, and both youtube and khan academy made me feel holed up in said subjects, so I'd really appreciate it if someone shone light on the path I am supposed to take, like with a roadmap, or methods on how to self study :D any questions would be fine with me if you want to know how far I got, and thanks for reading!

[hallow vault]

there is a roadmap on math stack exchange

but still you have to modify it accoring to your level

[edgy epoch]

Ah, alright I see, you got any idea about other methods I can follow?

[hallow vault]

https://math.stackexchange.com/questions/4994330/roadmap-for-self-learning-pure-mathematics-from-high-school-level-onward

for books go to #book-recommendations

for high school math refer khan academy

[edgy epoch]

thank you for the help and yer time :D

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[torn jolt]

.

does anyone remember

[torn jolt]

the thing for the sin cos tan angles

where for example sin theta = 36

we find the reference angle

and find the other angle by adding 180 or something

and if 36 was negative we do all the rules other than the reference and the sin

can someone tell me the rules

[limpid junco]

Im guessing you are asking for this :
$$\sin \theta = y \implies \theta = \sin^{-1} (y) + 2k\pi $$
$$or$$
$$\theta = \pi - sin^{-1} (y) + 2k\pi$$
$$\cos \theta = y \implies \theta = \pm \cos^{-1} (y) + 2k\pi$$
$$\tan \theta = y \implies \theta = \tan^{-1} (y) + k\pi$$

[glossy valve]

casework

[torn jolt]

wasnt sin +180

tan -180

and cos +360

[limpid junco]

I dont quite get what you mean. But what i sent should be correct if that is what you wanted.

[torn jolt]

alr thanks

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[cold trail]

what is the answer to this question

[dry arch]

you can use subtraction

subtract the area of the small rectangle from the area of the large rectangle

[cold trail]

hi

sorry I am back

so I need to do 2-1

right?

are you afk axe

[gritty rose]

what are you calculating

[cold trail]

this

[gritty rose]

did you find these areas

2-1 doesn't equal anything in his hint

[cold trail]

would I need to do 4x3 for area

[median yoke]

You'll have to calculate both areas, and then substract the white area from the gray area.

[cold trail]

ohh so 2x1=1 then 4x3=12

[median yoke]

yes!

but 2x1 = 2

https://cdn.discordapp.com/attachments/1251788052590886964/1332344319981523006/SPOILER_icon_5.gif

[cold trail]

that's what I meant

[median yoke]

Do you understand why we do 12 - 2 after?

[cold trail]

yes

I do

[median yoke]

Okay perfect

So what's the answer?

[cold trail]

so the answer would be 10

right?

[median yoke]

10 what?

apples?

[cold trail]

cm squared

[median yoke]

check it again

[cold trail]

cm cubed

[median yoke]

meters squared

cubed would be volume

[cold trail]

cm 2

[median yoke]

no

[cold trail]

m squared

[cold trail]

10 meteres squared

[median yoke]

yes!

yes

Understood everything we did?

[cold trail]

yes

[median yoke]

okay, then close both channels now

🐺

[median yoke]

ah another one

[cold trail]

for this one would I need to do 12x2=24 then would I neeed to do 7x6 divided by 2

[median yoke]

Almost!

[cold trail]

wait

I would need to do 7 minus 2

[median yoke]

You have t he formula for a triangle right, but we won't do 7-2.

How will we calculate the lenth of the 'invisible' side?

[cold trail]

1/2x (a+b

[median yoke]

no no no just the length of the circled part

we need that length to fill in our formula for the area of the triangle, right?

but we don't have it yet

[cold trail]

so you would need to

was I kinda right at the start

[median yoke]

yeah, but we have to substract different lengths

which ones do you think we'll use?

[cold trail]

7 minus 6

[median yoke]

Not really...

[cold trail]

ohh wait

[median yoke]

We have to use the 12

[cold trail]

12-6

[median yoke]

Close!

I think you meant '7' right

12-7

[cold trail]

yes

which would be 5

[median yoke]

Yes!

Now we'll use this to fill in b+l / 2

[cold trail]

was I right at the start

with the base

[median yoke]

FIll in the fornula

[cold trail]

12+2

I thoguht it is 1/2x (a+b)

[median yoke]

1/2 (a*b)

we multiply

just like with a rectangle

[cold trail]

ohh yea

[median yoke]

https://tenor.com/view/the-rock-nod-nodding-agreeing-agree-gif-15564795990526873010

[cold trail]

1/2 (12x2)

[median yoke]

So, what do we do to get our area?

Huh

Why 12x2?

We just calculated that the length is 5, right?

[cold trail]

so I need to do 1/2 (5x6)

[brittle zenith]

yes

[cold trail]

can you help me with more please

[median yoke]

I think it's best that you understand what we just did, and then use those techniques for yourself

Did you undderstand what we just do?

[cold trail]

would the answer be 15

[median yoke]

Yes!!

And then PLUS the other area, which is 24

12 x 2

24 + 15 = 39

[cold trail]

I thought thw answer is 15

when did the 24+15 come in

[median yoke]

we need all area

[cold trail]

ohh so I need to do 24+15 then that would be the answer

or not?

[median yoke]

yeah

[cold trail]

so the answer is not 15

so the answer would be 39 cm squared

??

apparently the answer is 45 cm squared

is that right

Are you afk

[hot herald]

who's saying the answer should be 45

[cold trail]

is the answer 45 cm squared

co piolet

[median yoke]

39 cm^2, we add both areas

[hot herald]

don't use those ai for math help

[cold trail]

is it wrong

[brittle zenith]

just saying don't trust ai with math its not very reliable

[sacred rivet]

!nogpt

[full forum]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[brittle zenith]

yeah the 45 cm^2 is wrong

[cold trail]

could you help me with some other questions please

[brittle zenith]

i can help in a bit sure (not sure how long)

[cold trail]

how comes

[brittle zenith]

but btw try to understand how we got to the answers on the other questions

[median yoke]

calculate all white areas

then substract

from gray area

[cold trail]

so 5x4

wait I mean 10x12

or not

[hot herald]

where are those coming from?

[cold trail]

wait

I need to times the white

4x2=8

so all of the middles ones would be 8 then I add them up

or not

[hot herald]

that would be part of the stuff nreded

[full forum]

@cold trail Has your question been resolved?

[brittle zenith]

do you remember what you did previously with shive

[median yoke]

@cold trail

1. Calculate all white areas individually
2. Calculate the area of the whole rectangle, gray and white
3. Substract the total white areas from the total

[cold trail]

sorry I am back

what is going on

[brittle zenith]

alright let's try to solve that question together

[cold trail]

ok

sounds good

[brittle zenith]

alright so remember the first question you did with shive, you calculated the area of the grey part minus the area of the white part right?

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[cold trail]

yea

[brittle zenith]

btw type . reopen

so you can still use this channel

[cold trail]

reopen

[brittle zenith]

don't forget the dot ".reopen"

[cold trail]

''.reopen''

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[cold trail]

so I need to do 10x8

[brittle zenith]

yes

for the grey area

what would that equal to?

[cold trail]

so that would be 24

then I do 80-24

[brittle zenith]

yes

and that's gonna be?

[cold trail]

Give me a sec

[brittle zenith]

alright take your time

[cold trail]

56 cm squared

[brittle zenith]

correct

good

you understand it

[cold trail]

could you help me with some others

please n

[brittle zenith]

sure

[median yoke]

franky

[cold trail]

yes

[median yoke]

think for urself a bit now

after all we did just now

[cold trail]

The thing is

[median yoke]

this is the exact same exercise but different

[cold trail]

I did think all of it by myself

Didn't I Hintrill

[brittle zenith]

yeah the previouse question you answered correctly by calculating yourself

but we did say the steps to get to the answer

[cold trail]

I kinda did know what to do

[brittle zenith]

ah alright so what do you think we gotta do to know the area of this one?

[cold trail]

for this one I need to

7x5=35

then I have to do 9x12=108

then I need to do 108-35

[brittle zenith]

not exactly

[cold trail]

noooo

mann

[brittle zenith]

look i think your trying to the same strategie like you do to calculate the grey area

but for this one you do something els

[cold trail]

yes I am

[violet lion]

so ez

[cold trail]

mate for you it is

so be quite

not everyone is good at maths

[violet lion]

what grade r u in

[cold trail]

yea 11

[median yoke]

joe show some respect bruh

[brittle zenith]

^

[violet lion]

which country has that lvl of math in 11th grade

[violet lion]

no im curious bruv

[cold trail]

United kingdom we do GCSE

[violet lion]

ahhh gcse

ok

[brittle zenith]

anyways let's continue with the question

[cold trail]

for this I need to minus

one of the skies

sides

[brittle zenith]

no no

[cold trail]

ok

[brittle zenith]

your making to difficult for yourself

[violet lion]

yes you do

[violet lion]

hes right

in order to find the base of the triangle subtract 7 from 12

[brittle zenith]

first split up the figure with what kind you know like rectangles traingles and etc

[cold trail]

would I do 12-7

which is 5

[violet lion]

thats the base of the triangle

good

area of right angle triangle is 1/2 * base * height

height of triangle is 9-5 thats is 4

[cold trail]

so I would do 1/2x (5x4)

[violet lion]

so area of triangle = 1/2 * 5 * 4

[violet lion]

good

that would give you 10

10 cm^2

[cold trail]

so the answer is 10

[violet lion]

nope

now you got to add the area of rectangle

to the area of triangle

area of rectangle = length * breadth

thats gives you 5 * 12

which is 60

there fore area of figure = 60+10

[cold trail]

then I do 60-`0

[violet lion]

= 70 cm^2

[cold trail]

10

[violet lion]

60+10

add the areas

of triangle and rectangle

[cold trail]

so the answer is 70

[violet lion]

right

correct

[cold trail]

could you help me with some more

please

[violet lion]

like what?

[cold trail]

give me a sec

here

so in the middle I do 5x3

[violet lion]

yes

[cold trail]

which would be 20

[violet lion]

15*

[cold trail]

my bad

that's what I meant

[violet lion]

its fine

happens

[cold trail]

then I do 10x12

[violet lion]

now subtract 15 from the area of triangle

[cold trail]

then would I do 10x12=120 then I do 120-15

[violet lion]

the information i think is inadequate

regarding the triangle

[cold trail]

so I am wrong

[violet lion]

is it a right angled triangle?

yes it is

[violet lion]

you do 1/2 * 10 * 12

[cold trail]

ohh yea

[violet lion]

so that is 60

60-15 = 45

there you go

[cold trail]

can I show you some more questions

wait

[violet lion]

i gotta go now

[cold trail]

ok

[brittle zenith]

if you want we can go through them

[cold trail]

ok

so the answer for this is 45cm squared

[brittle zenith]

yes

got another question?

[cold trail]

wait

sorry my bad did not mean to sound rude

[brittle zenith]

nah dw

[brittle zenith]

oke for this one i always try to look for figures that i'm familiar with

[cold trail]

ok

[brittle zenith]

what you can see is a rectangle and a triangle right?

[cold trail]

wait

can I show you another question quickly

previous one

[brittle zenith]

sure

[brittle zenith]

this one?

[cold trail]

is this right

[brittle zenith]

yes

[cold trail]

how would one of it be 4

that's what I am trying to work out

[brittle zenith]

ah i will draw in on pain rq

[cold trail]

ok

thx

ohh so I did 9-5

to work that out

[brittle zenith]

so to calculate a traingle you need the base and the hight so the blue part and the red part

so yk the total length is 9 right from top to bottom but we want the blue part only

we know that 5 + blue = 9 right?

so that means blue should be 4

because 5 + 4 = 9

[cold trail]

Sorry I am back

[brittle zenith]

welcome back

btw if i'm explaining too difficult you say so i will try to explain it in a different way

[cold trail]

so to get 4 I did 9-5=4

[brittle zenith]

yes

[cold trail]

that makes sense now

[brittle zenith]

and for the red part you did

7 + red = 12
so
red = 12- 7

red = 5

[cold trail]

Which one are we doing

[brittle zenith]

i was explaining this one further but now for the red part

[cold trail]

for quesiton 7

[brittle zenith]

anyways we can now go back to this question if you want

[cold trail]

ok

so I would do

100x75

[brittle zenith]

yes

and that would be?

[cold trail]

7500

[brittle zenith]

alright and now try to calculate the triangle

here you go to visualise it better

[cold trail]

to celculate a triangle I would have to do 1/2 (something times something)

calculater

wait I would do 100-30

[brittle zenith]

yes

so your height would be?

[cold trail]

70

[brittle zenith]

good now try the blue part

what would the base be?

[cold trail]

I would need to do 160 divided by 2

I mean 70 divided by 2

[brittle zenith]

no no

we do 160 - 75

do you know why

or should i explain it

[cold trail]

yes please

[brittle zenith]

oke we know that the total length is 160 yeah?

[cold trail]

yea

[brittle zenith]

and we also know that the green part is 75 yes?

so that means 75 + blue = 160

so to only get the blue part it would be

blue = 160 - 75

[cold trail]

85

[brittle zenith]

nice

good

[cold trail]

blue= 160-75=85

[brittle zenith]

now we know the base and the height

so we put it in the formula (h * b)/2

[cold trail]

wait

is the formula 1/2 (85x70)

wait that is way to big

[brittle zenith]

yes

[cold trail]

so that is right

[brittle zenith]

yeah

base is 85 and the height is 70

so its correct

and what would the answere be?

[cold trail]

2975

[brittle zenith]

nice

now we have area of traingle and area of the rectangle

but we want to know the full area of this figure

so what do you think we do?

[cold trail]

so I have completed that question or not 4

not

[brittle zenith]

not yet

[cold trail]

ok

[brittle zenith]

gotta do the final step and that's adding them up

7500 + 2975

[cold trail]

ok

[brittle zenith]

and that would be? (don't forget the unit)

[cold trail]

10475m squared

[brittle zenith]

nice good job

that's the answer

so is everything clear? or do you still have questions about this exercise

[cold trail]

I am on forming and Solving equations

seems quite hard it is a grade 4

whereas the other one was a grade 3

[brittle zenith]

look you got this and the more and more you exercise you will become better at it

just gotta keep on trying harder and harder questions

and eventually everything will snap and become easier

[cold trail]

let me send you the question

[brittle zenith]

sure

[cold trail]

how long can you stay and help me

[brittle zenith]

idk i will see and let you know when i gotta go

[cold trail]

here

is it ok if I go for a min I will be back soon

[brittle zenith]

sure go ahead

[median yoke]

look franky, you should try some exercises yourself, we cant do them all for u

[cold trail]

she isn't

I have got quite a lot of them right

[median yoke]

he

im his friend

he ditched me and i shelping here

[brittle zenith]

bro i wanted to help him and its fun helping ngl

anyways let me know when your back

[cold trail]

I am back

[brittle zenith]

alright lets try it

yk how to calculate a perimeter right?

[cold trail]

adding

[brittle zenith]

yes

[cold trail]

the outside of the shape

[brittle zenith]

alright and we know that the total perimeter is 31 cm

so that means if we add all sides up it should equal to 31 cm

[cold trail]

when it says 3x-5

we do the opposite

3x5

which would be 15

[brittle zenith]

uh wait are you seeing the x as multiplying?

[cold trail]

wait

can I show you to see if this is right so I did 15x2=30 then I did 30-1=29 after that I did x+1 which is 30 because I did 29+1=30

[brittle zenith]

i'm a little confused on what you did ngl

how did you get 15 * 2?

[cold trail]

never mind am I meant to divide not times

[brittle zenith]

wel anyways how i would do it is as followed

we know that if we add up all sides it would be 31 cm

first side = 3x-5
second side = 2x-1
third side = x + 1

just saying x is not multiplying its your unkown value

so we gotta figure out the unkown value

[cold trail]

this is quite hard

[brittle zenith]

just think unkown values are like what we did previously like 5 + blue = 9

but instead of me writing blue its just x

soo:
5+x=9

x=9-5

x=4

we solved an unkown value

you understand this right?

[cold trail]

not really

is this algerbra

[brittle zenith]

yes

have you seen algebra before or is this new to you?

like solving for x and stuff like functions?

[cold trail]

I have seen it

[brittle zenith]

ah alright want me to refesh for you with some simple ones before we do this one so you can see how solving for x works?

[cold trail]

ok

[brittle zenith]

x + 20 = 30 let's try to solve this one

what we always want to do is to get the unkown on one side of the equal sign and the numbers on the other side

[cold trail]

ohh

[brittle zenith]

we bring the + 20 to the other side

so that would make it - 20

you can remember like this if we bring something to the other side of the equal sign it will become the opposite like

• becomes -

• becomes +

• becomes /
  / becomes *
  ^n becomes n-th sqrt()

[cold trail]

ohh mann

I know

I have been doing this before

[brittle zenith]

alright so you remember now right

[cold trail]

yes

[brittle zenith]

do you think we can solve the exercise or want me to say something more about them

[cold trail]

we could try

[brittle zenith]

alright

we know that if we add all sides it should be equal to 31 cm right?

[cold trail]

yea

[brittle zenith]

so that's legit what we gonna do we will write the sides and add them up and equal the to 31 like so

3x-5 + x+1 + 2x-1 = 31

[cold trail]

sorry what am I meant to write on my paper

[brittle zenith]

so for question a) you will just write down what i just wrote

for question b) we will try to work out and find what x should equal to

do you think you could try to solve b) for me i will tell you if your doing good or not

[cold trail]

ok

are you around tomorrow

I am getting tired

[brittle zenith]

idk i will see i might be busy tomorrow tho

[cold trail]

that's fine

wait could you try and help me

and complete the sheet

I am afk quickly

back

[brittle zenith]

well i can't just solve it for you and say "here is your answer"

[cold trail]

lol

[brittle zenith]

alright could you try to solve b)?

[cold trail]

ok

my mind has gone blank

[brittle zenith]

like type or write the steps and i will tell you if you did it correct or not

[brittle zenith]

you got this

[cold trail]

do I need to minus the 31

[brittle zenith]

just do this.

[cold trail]

so I need to divide it

[brittle zenith]

yes you will need to divide at some point

here is a random example that shows how to solve for x:

2x + 3x + 2 = 17
2x + 3x = 17 - 2
5x = 15
x = 15/5
x = 3

[cold trail]

my mind is still blank

[brittle zenith]

uhh is there maybe something you don't understand about it? or are stuck at?

[cold trail]

could you give me the answer then I can work my way around on how you go tthe answer

[brittle zenith]

sorry but i can't just tell you the answer

yk let's try to solve it together

[cold trail]

ok

[brittle zenith]

3x-5 + x+1 + 2x-1 = 31

alright this is what we got first we gotta bring the numbers to oneside and the unknown to the other side
let's bring the numbers to the right and the unkown will stay at the left

[cold trail]

ok

[brittle zenith]

3x**-5** + x**+1** + 2x**-1** = 31

the numbers that i have put in bold gotta go to the other side, could you try do it?

remember this

[cold trail]

3x+5

x-1

[brittle zenith]

no no

[cold trail]

2x+1

[brittle zenith]

remember this example

[cold trail]

10

[brittle zenith]

oke yeah the answer to that is 10 but i want you to try that method to the this

3x-5 + x+1 + 2x-1 = 31

[cold trail]

coudl you write me the step to the answe

could

[brittle zenith]

oke i will bring one number to the other side so you could see it

[cold trail]

so for the bottom one I should write out

[brittle zenith]

yeah and now you do the same for -5 and +1

what would that be then?

[cold trail]

so for b the answer is the bottom

[brittle zenith]

no

i wrote one step

[cold trail]

so should I write it out on my paper

[brittle zenith]

wel i myself would not write that step but its your paper you can do what you want

[cold trail]

what would you write out

[brittle zenith]

i would write the step where i brought all the numbers to the right side and all the unkowns to the left side

[cold trail]

could you write it down for me pleas e

[brittle zenith]

no try to understand this concept of how solving for x works

if i just write the step you would probably not know how you got that answer its always best if you find it yourself with a little bit of help yk

[cold trail]

yea

sorry I am really stuck

[brittle zenith]

alright let's try it another way

instead of bring the numbers to one side and the unkowns to the other side

let's just simplify the left side

3x-5 + x+1 + 2x-1 = 31

just add up the numbers without the x to them like -1 up
and add up the unkowns like 3x

[cold trail]

should I write the that down

3x so on

[brittle zenith]

well sure but first show me what your going to write

so i can see if its correct or not

[cold trail]

I am going to write this 3x-5 + x+1 + 2x-1 = 31

[brittle zenith]

oke sure write that first

[cold trail]

so I would have to do 31+1

[brittle zenith]

well yeah but wait how would you like to solve this first by simplifying and then solve for x or solve instantly for x?

[cold trail]

Sorry I have to go I am extremely tired

it is so hard I need to pass my GCSE

[brittle zenith]

ngl idk what a GCSE is

never had it

[full forum]

@cold trail Has your question been resolved?

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[shrewd hamlet]

real quick, for the third question, i wanna find the map. do i need to express the basis vectors of B in terms of B' or the basis vectors of B' in terms of B?

[shrewd hamlet]

'in terms of' = a linear combination of

[nimble crane]

should be the former

I think

[fast peak]

imo its easiest to think about what should happen to the vector (1,0,0)

[shrewd hamlet]

ah i c, lol cuz i asked chatgpt and it kept alternating its answer

[shrewd hamlet]

hmm

[nimble crane]

LLMs can be a little inconsistent

[shrewd hamlet]

why that vector specifically?

[fast peak]

cause its simple

the standard unit vectors in general

also remember that the columns of a matrix are the result of multiplying that matrix by the standard unit vectors

[shrewd hamlet]

ok i see

so to find the coordinates of (1,0,0) in the basis B', we want to multiply it with the map phi

but how does that help me get phi

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[storm hinge]

honestly nothing i don't really get this section

[leaden ermine]

any thoughts what integration technique you can use here?

[narrow path]

if they got rid of the 3x i wouldve been very scared for you

[honest hill]

u-sub

[storm hinge]

do i need to make it look different first? like put the e^x in the denominator?

[leaden ermine]

no

[honest hill]

.

[storm hinge]

okay then i have u=x^2?

[honest hill]

yes

continue

[storm hinge]

du=2x dx so dx=du/2

[honest hill]

dx=du/2?

[storm hinge]

du/2x sorry

[honest hill]

yes

now fill it in

(remember the bounds)

[storm hinge]

i never remember how to change the bounds can you plz remind me

[honest hill]

you wrote u in terms of x

u(x)

and our lower bound is 0, so our new lower bound will be u(0), which in this case is also still 0

similarly for the upper bound inf

[storm hinge]

okay sick

[honest hill]

in this case, the bounds dont change

i was kinda trickign you

but good to remember for other integrals

anyway, lets continue

what does our integral become after the u-sub?

[storm hinge]

this is what i have so far

[honest hill]

no.

just keep it as e^-u

[storm hinge]

aw damn okay

[honest hill]

the derivative of ln(e^u) is not 1/e^u

d/du ln(e^u)=1

[storm hinge]

okay so then do i need to do a power rule?

[honest hill]

you havent had the antiderivative of e^x?

[storm hinge]

thats just e^x

[honest hill]

yup

now what about e^-x?

[storm hinge]

-e^x?

[honest hill]

almost

-e^-x

so then what is $\frac32\int_0^\infty e^{-u}\dd u$?

[glossy valve]

Bonk

[honest hill]

why is it a 4 now?

[storm hinge]

hahah because i cant copy numbers down or read them without switching them up

okay i changed it back to 2

[honest hill]

okay haha

anyway

continue

youre almost there

[storm hinge]

now i plug in 0 and inf for u

[honest hill]

you should be able to do this without problems

from here

[storm hinge]

ya i got 1.5

[honest hill]

keep it as a fraction

its cleaner that way

,w int_0^inf 3xe^(-x^2)dx

[glossy valve]

Results provided by WolframAlpha

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Requested by .bonk__

[honest hill]

perfect

[storm hinge]

thanks

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[storm hinge]

do i need to do partial fractions for this one?

[honest hill]

no

u-sub

[storm hinge]

damn

[native stratus]

Also this thing is odd, so you don’t even need to integrate

[storm hinge]

how can you tell?

[native stratus]

Plug in -x

[leaden ermine]

i am not sure if this is applicable since the bounds are infinity

unless you justify it with the cauchy principal value

[storm hinge]

so you get (-3x)/(1-x)

[honest hill]

its especially applicable if the bounds are infinity, no?

or is that for even functions?

[honest hill]

no

[native stratus]

The integral is defined as the limit from n to infinity of the integral from bounds -n to n

[storm hinge]

well it says if its divergent the type D as the answer

(-3x)/(1+x)?

[honest hill]

still no

where did x^2 go?

[storm hinge]

(-x)^2 is just x

[honest hill]

no.

[storm hinge]

so it would be +x

[native stratus]

??

[honest hill]

no.

$(-x)^2$

[storm hinge]

uggg😩

[glossy valve]

Bonk

[honest hill]

$((-1)(x))^2$

[glossy valve]

Bonk

[honest hill]

$(-1)^2(x)^2$

[glossy valve]

Bonk

[honest hill]

need i continue?

[storm hinge]

no ur right

[honest hill]

go tit now?

*got it

[storm hinge]

is this right (-3x)/(1+x^2)

or am i just completly fucked

[leaden ermine]

I don't think that works unles the integrals from 0 to inf and -inf to 0 exist

else the integral of x from -inf to inf would be 0 too

[native stratus]

that's not true

[leaden ermine]

you are assuming inf-inf = 0

[native stratus]

,, \int^\infty_{-\infty} f(x)dx := \lim_{n\to\infty}\int^n_{-n}f(x)dx

[glossy valve]

Bair

[native stratus]

by definition

[leaden ermine]

source?

[storm hinge]

okay im going to my profs office hours bye

.close

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[native stratus]

alright, maybe you are right that it can be defined differently, but in this case i think we can infer that this is the correct interpretation

[slender onyx]

thats one of two main defs

[leaden ermine]

i was wondering cause i read this

so you were prob referring to the cauchy principal value

[slender onyx]

my complex analysis brain prefers principal value

but ive seen some calc 2 problems refer to the double limit def

prob best to clarify with prof

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[craggy socket]

not sure what this is called, just want to ask if it's inclusive or exclusive

[craggy socket]

like would it multiply with four and zero included

[robust slate]

Yeah, you would include the terms corresponding to n=0 and n=4

So $\prod^{4}_{n=0} a_n=a_0 a_1 a_2 a_3 a_4$

[glossy valve]

Civil Service Pigeon

[robust slate]

Btw this is called the product symbol

It’s the capital version of pi in the Greek alphabet

Alright imma dip now 👋

[craggy socket]

thanks!

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[proven wolf]

I'm having problems trying to demostrate that f is integrable in [a,b] if and only if exist a sucession of partitions of [a,b] Pn such that $$\lim_{n\to\infty}[U(f,Pn)-L(f,Pn)]=0$$

[glossy valve]

Gui

[proven wolf]

my first attempt was using that f is interable in [a,b] if for each $\epsilon>0$ then $U(f,P)-L(f,P)<\epsilon$

[glossy valve]

Gui

[proven wolf]

then since U>=L implies that U-L>0

[fathom saddle]

What's your definition of "integrable"? This ends up being book dependent

[proven wolf]

they gave me two diferent

the one that I put there

and

f is integrable in [a,b] if and only if $$\sup(L(f,P))=\inf(U(f,P))$$

[glossy valve]

Gui

[proven wolf]

tbf, now I think the definition that will help me the most is this one, since it already have an epsilon and I can use it to define the convergion of the sucesion I need to create

what I' currently trying if make a regular partition and make the sucession about the norm, but I'm not sure how to defying it, or even if my idea is correct

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[whole merlin]

I can't figure out the order of operations to solve this and I am stuck

[tidal rain]

,rccw

[whole merlin]

that stuff in parentheses is (t-3)

[whole lily]

Divide by pi/6 then add 3 to all sides

[whole merlin]

can you show me how that would look?

i am absolutely mentally fried rn

[tidal rain]

u can split it into two equations

0<=(t-3) and (t-3)<=2pi

smth like this

[whole merlin]

and then multiply by 6?

this is for plane trig i am trying to get the beginning and ending points for a sin wave

[tidal rain]

make t the subject

[whole merlin]

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[whole merlin]

thank you

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[devout lion]

Could anyone help me with this? x is in the interval [0, pi/2]

[gritty rose]

Taylor's remainder theorem

[devout lion]

thank you

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[restive violet]

could someone please help me with this problem

im not really sure where to start

wait hold on

ok im stuck nvm

i plugged in g(x) into x of the f(x) function

[gritty rose]

f(g(x)) = x implies g is the inverse function of f

[restive violet]

oh

ok then give me a sec

thanks for helping

thanks i got it now

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[thorny shuttle]

how do i eval this if ti has to be cont and equal to 3 doesnt look like 1^inf

[thorny shuttle]

not sure if taking logarithm helped me either

[leaden ermine]

you can force it into 1^inf

[thorny shuttle]

how?

[leaden ermine]

well you need something in the form of (1+x)

split the fraction

[thorny shuttle]

i did

that c term stays tho

[leaden ermine]

well choose it wisely

[tall lily]

if its 1^inf you can change have f^g as e^[g*(f-1)]

[full forum]

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[unique valve]

could someone help me or check my answer if i did it right

[unique valve]

heres the problem

deaths of men

summary data of living of men

The average weight of living men ranges between 83.9 kg to 89.1 kg.
The average age varies between 50.42 years to 55.41 years.

[unique valve]

here is my answer

[full forum]

@unique valve Has your question been resolved?

[unique valve]

<@&286206848099549185>

[high fjord]

tough questions these days

[next crypt]

You only provided data for 2 things, you are to compare for all, for each

1. Weight Comparison
2. Age Comparison
3. Socioeconomic Class (SEC) Comparison
4. Eye Color Comparison
5. Key Observations and Insights

For the living and the dead. 5 is, articulating your understanding of the information gained

[unique valve]

how do i compare it

[next crypt]

Find the mean, median, mode, of living normal distribution against the deceased

[unique valve]

ok wait ill get them quickly

@next crypt did i do it right?

[full forum]

@unique valve Has your question been resolved?

[unique valve]

<@&286206848099549185>

[soft dew]

Looks right based on just looking at it…

[next crypt]

You need to compare for both, alive, and deceased. Then draw your conclusions, such as people of with X are more likely to die at Y age etc etc

@unique valve

Also, you may not find mean, median, mode for each. For some, you need to use, median, and for some, mean, and so forth for mode. Example, in your case, SEC is qualitative, so you will use mode, because you find to find some correlation, middle and lower are both at 40% so you pick the upper bound, say middle, as the mode, andd phrase it as middle and under are more likely to die at X etc. You can also use "incomplete results argument" Or that the data is skewed, or you can normalize to draw better conclusion. I don't know how advance the expectations are, but I'd recommend you do all this to ensure you get a maximum grade, as either it counts and you get a better score, or it doesn't, and it wouldn't matter. But doing it will ensure both outcomes are in your favour.

[unique valve]

idk how to start i got the mean median and mode

how do i draw my conclusion

[cunning vessel]

all you can really do, and is the best, is compare among groups

with this form of data

[unique valve]

can u teach me @cunning vessel

[cunning vessel]

I guess you know how to compare number

[unique valve]

i do

[cunning vessel]

or you can find relation

of two factor

I can't tell you much beside statistics as the qestion asks about your analyzation skills rather than math

[unique valve]

how do i find the relation of 2 factor

i just wanna get this over dri

ive been stuck for the past 3 days on this problem

@cunning vessel please help me to get over with this problem

[cunning vessel]

for example, see if higher weights leads to earlier death by looking at district data

[unique valve]

can u give me an example

like answer number 1 for me

and i can follow up

i can compare but idk how to start

its very difficult for me

[cunning vessel]

I can't really help because I dont understand "Isolate two table of data"