help-10

[heavy bobcat]

<@&286206848099549185>

Im so stuck, i just need someones help

[obtuse pebble]

@heavy bobcat Has your question been resolved?

[heavy bobcat]

<@&286206848099549185>

ive been waiting for like 45 mins

lol

[amber dirge]

Let me take a look, just got to 11.8 so I can explain that one, method will be identical for the other problem

Also, the bases are not the same

If they were the same it would be statically indeterminate

Not that it affects our calculations here, but you shouldn't be having the same base when drawing out the problem, there's a pinned support and a roller support, the roller one has no horizontal reaction

I use method of joints to solve these, are you familiar or have you only done method of sections?

[heavy bobcat]

no but thank you so much taybee

[heavy bobcat]

your the first person to help

I would preferably like to see what method you used and how you did it.

[amber dirge]

Okie dokie, I'll talk through the one that comes out as 11.8kN and then you can try to apply it to the problem you pasted (the answer is about 7.03kN so you know when you get it right)

Step 1 is to calculate your support reactions

[heavy bobcat]

perfect thanks bro!

lets do this

actually taybee do you do calls?

[amber dirge]

Sometimes but not at the moment as on a call with someone else lmao, bear with me I'll type and paste some screenshots as I go

[heavy bobcat]

coolio

[amber dirge]

Support reactions are done by looking at forces and moments, we know the sum of forces vertically and horizontallyhave to be 0 as it's in equilibrium, then similarly the sum of moments (clockwise and CCW) have to be zero

Gives reactions at A and D of 40/3 and 20/3 respectively

[heavy bobcat]

this is my question?

i thought we were doing the 11.8 one?

[amber dirge]

Oops yeah opened the wrong page and used the wrong numbers lol

One sec

Reaction forces remain the same though as the distances are proportional

[heavy bobcat]

nah ur a legend bro honestly the first person to help me

[amber dirge]

Fixed, same reactions as mentioned

Onto the next bit, take a joint, easiest to start at the point of the support, so we'll look at A

Now you equate vertical and horizontal forces:

[heavy bobcat]

how did u do that step?

[amber dirge]

Like so

The only vertical force we have in the joints is the vertical component of Fe, so we know that that has to be 40/3 (as in the triangle on the right)

So you can use trig to work out the hypotenuse and adjacent (i.e. the magnitude of Fe and the horizontal component of Fe)

And using the horizontal component of Fe you know that Fb has to be the same in the opposite direction to maintain equilibrium

[heavy bobcat]

did u get my answer of 7.03 kN for the other question?

[vague knot]

gang

[heavy bobcat]

to 3 significant figures

[vague knot]

wsg

[amber dirge]

Do the same at joints E and B to get 11.8

[amber dirge]

Yup

[amber dirge]

Also that calculation at the bottom is equating vertically not horizontally btw wrote it next to the wrong label lol

[heavy bobcat]

all gd, also taybee i wanted to ask another question if thats alright

[amber dirge]

Fire away, I'll see if I can take a look before I vanish lol

We might all be, who knows

[heavy bobcat]

so this was the same as the other but the values for other question Q is 5 kN and P is 12kN with a height of 4m they got -13.42 as theyre final answer

also this question is really doing my head in but im just baffled at what the hell to do

i selected to member AK is a zero member but wanted to double check

[amber dirge]

Okay I definitely won't have time to complete those before I disappear sorry XD But for both those structures, work out your reactions and then method of joints if required as before - just make sure you pick a joint that doesn't have too many unknowns, generally you can only pick a joint that has two unknowns coming out of it

Lemme look at the MCQ one quickly tho

That might be fast just need to read it lol

[heavy bobcat]

bro i desperatly need you assistance

please could you have a go at the top question for me im lit begging you

if you could get a final answer for it and ill attempt it and try get the same

[amber dirge]

I came out with B in the MCQ

Haven't checked if I've done something stupid tho so do check but I checked moments to work out reaction forces and came out with that

[heavy bobcat]

perfect

bro

im begging you to try the other question

[amber dirge]

Also, in case your engineering (assuming you're studying that and not just randomly doing statics lol) coursemates haven't sent this your way, https://skyciv.com/free-truss-calculator/

[amber dirge]

There's a node limit to that calculator but unless they've reduced it since I was using it a couple years ago, this problem you sent should be solvable using it

The nodes are the joints just input them like coordinates

Then connect them through members

Pick your supports appropriately

Then input your point loads

And it should just solve them

If you pay then it even gives step by step explanations lol

[heavy bobcat]

could u try and get a final answer for it and i could double check bro

please

[amber dirge]

Family is calling me but I can only hope and assume there are more engineers than just me on this server XD

[amber dirge]

I'll do the analysis on skyciv but I dont have time to work it out properly

One sec

[heavy bobcat]

thank you!

Legend honestly

[amber dirge]

They've reduced the member limit since I used it, you might be able to get past it with a free account, or its literally nothing like 10/month to subscribe and it'll step you through and remove the node limit

I tried to approximate it in such a way that it was stable and statically determinate which was tricky and made it quite simplified

But it came out with this

Don't have the brain capacity to think about how true to your diagram that will be, but some of it should be correct like the reactions at the supports and some members

But honestly, if you anticipate doing a lot of these or it will feature on your coursework, subscribe to skyciv, it's a lifesaver

[heavy bobcat]

its asking for CH so 8.944

[amber dirge]

Maybe

Hopefully lmao

[heavy bobcat]

give it a go lol

[amber dirge]

But yeah I was a mechanical/medical engineer not a civil engineer and I still got saved by skyciv

[amber dirge]

Nah I can't sorry, gotta go

[heavy bobcat]

all gd bro

[amber dirge]

Get skyciv lmao

Or there should be another engineer buried among the helpers on here XD

[heavy bobcat]

ive sent u a fr so if ur on later could u help me out bro?

[obtuse pebble]

@heavy bobcat Has your question been resolved?

[obtuse pebble]

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[ripe hazel]

ik theres a +2 and a +3 but how do i incorporate that into the given function

[pastel wren]

(x+2)^2-2(x+2)+3

[ripe hazel]

well isnt a reflection in the y axis supposed to be (-x) and not -(x)?

[pastel wren]

yeah but you asked only about +2 and +3

[pastel wren]

put -x instead of x here

[ripe hazel]

ok

like so? (x+2)^2+2(-x+2)+3

[pastel wren]

why isnt there -x at the start?

[ripe hazel]

wait nvm

ur right

it should be (-x+2)^2+2(x+2)+3

[pastel wren]

(-x+2)^2+2(-x+2)+3

[ripe hazel]

hmm ok

[pastel wren]

its f(-x)

so you put -x instead of ALL x's you see

[ripe hazel]

yea it applies to both

i simplified it to x^2 -6x+11 and it was wrong

wouldnt this just be like f([-x+2]^2 + 3) = x^2-2x?

[obtuse pebble]

@ripe hazel Has your question been resolved?

[obtuse pebble]

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[red pendant]

Hi. I'm not sure how to do this. I've already attempted it 10 times. I'd take pic of my work but it's in onenote (I'm on laptop typing this but my notes are on my ipad), but I'd be happy to upload it here if you'd like. I'm just not really sure how to do any of it and my professor isn't the greatest so I honestly have no clue what's going on (But I really want to understand).

[strong pike]

do you understand how the comparison test works

[red pendant]

Kind of. I known there's bn > an > 0 where bn is like the big one and an is the small one. And then if one of them converges or diverges so does the other

But I don't really know how to use it here. And it's really throwing me off when they have multiple n's in the numerator and denominator

This is like my work thus far

[strong pike]

Kinda. If you have a series that you know diverges, and can find another series that's larger than it. That series diverges. If you know a series converges, and can find a smaller series, it converges.

[red pendant]

ohh

imma write that down lol

[strong pike]

These all look like the limit comparison test, which you use instead of the standard comparison test if you can't prove it with the former.

Looking at the first one, for example.

Factor out an n from both sides, you end up with 1/(n+1+n^-1)

Intuitively, this series looks like the 1/n series, given that n is the dominant (greatest at large n values) term in the denominator. We know the 1/n series diverges.

But we need to show that it diverges.

First step (in an exam situation) is to try to apply the direct comparison test

Is the series 1/n smaller than the series 1/(n+1+n^-1)?

Unfortunately not, the extra terms (despite not being dominant), makes 1/(n+1+n^-1) smaller than the 1/n series

Instead, we use the limit comparison test, where we get the limit of the quotient of the two series (1/n and 1/(n+1+n^-1)) as n goes to infinity.

If L is a nonzero number (and it will be), then both series diverge or converge. In this case, both diverge since 1/n does.

Apply the same process for the other questions, with an added caveat that if L = 0 and your known limit converges, so does the other.

[red pendant]

How do you know which one to choose to use?

[strong pike]

first you try direct comparison, if it fails you try limit comparison

You should "know" the answer (diverges or converges) before attempting to solve though

[red pendant]

okay so like looking at 2 for example

Would it be reasonable to choose 1/n^3

because they both have n^3

[strong pike]

Yes

Exactky

n^3 is the dominant term of the denominator

so you "know" the series will act like 1/n^3

which you know converges, so you try to prove it

[red pendant]

Okay cool

I'm going to try 3 & 4 real quick

okay wait so

because they start at n=2

Do I need to reindex to start at n=1

.close

[obtuse pebble]

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[timid silo]

Would setting |x-5| < 6 also be valid to making |x-5||x+2| < delta.(6)

sorry some parts of missing

[minor bone]

fix delta in terms of epsilon

[timid silo]

yeah I can do that by setting delta = epsilon / 8 but would delta = epsilon / 6 also be valid?

[minor bone]

uh I haven't really worked it out in my mind lol

[timid silo]

Im mainly asking about the blue writing, will |x-5| < 6 also hold

instead of |x-5| < 8

[minor bone]

oh uhh

it doesn't really pass the vibe check but I'm also rusty on this

[timid silo]

real

[minor bone]

you just need a delta such that like I think (-2-delta)^2 -3(-2-delta) - 10 = 4 + 4delta + delta^2 + 6 + 3delta - 10 = 10 + delta^2 + 7delta, so delta^2 + 7delta just needs to be less than epsilon

[timid silo]

where on earth are you getting those values

I found the answer to my question from someone else but I respect your help @minor bone

.close

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[feral acorn]

CAn i have help with this q?

[cyan kite]

which one

[feral acorn]

i have done this so far

not sure if it is right

.close

[obtuse pebble]

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[grizzled stone]

,rotate

[grizzled stone]

how do I answer this algebraically

[timid silo]

manipulate the equation to solve for them

[grizzled stone]

what

[strong aspen]

go to desmos, that way u can visualize the line

[timid silo]

thats cheating

first find the expression for x in terms of y

[strong aspen]

that's just for reference imo

but if it's cheating, don't do it

[grizzled stone]

wdym

[strong aspen]

put x as the subject

x = something

[grizzled stone]

whats something

[strong aspen]

from y = -2x+5

find a way

to make x alone

in the left hand side

[strong aspen]

while u do that you will find that "something"

I'm going to sleep 👋

[frank monolith]

take some values of x of your choice, and find the corresponding values of y. form a table using them and plot those values on graph to get the graph of the line.

[grizzled stone]

brah

can someone go over this step by step and make it easy

please?

[vernal pollen]

My English ain't that good, but what you're trying to do is isolate x to find a value for y?

[frank monolith]

read the 'Plotting straight line graphs' section in https://thirdspacelearning.com/gcse-maths/algebra/straight-line-graphs/

[timid silo]

dyssrupt

[grizzled stone]

yea

no no

we already have y

wait nevermind

[vernal pollen]

I recommend what Dysrupt has linked above, what may also help is that in order to isolate x you need to do the opposite, so + becomes a - on both side, the * becomes a / at each side

you have the equation in the form:
y = ax+b
and you want to isolate x so, +b becomes -b on both sides, and same for the multiplication

[grizzled stone]

I just need to find y and x intercept algebraically using y=-2x+5

[vernal pollen]

Read out the equation fully, the last thing you read is the one you start with

That's where you start removing the +5 on the right side and putting it on the left side, because plus becomes minus

[grizzled stone]

what after

[vernal pollen]

I'll help you with the first step
y = -2x+5
y-b = -2x
cause you did -b at both sides, removing the +b at the right side and getting -b at the left side

Then you come with 2x, which is a multiplication and in order to get that to the left side you divide it

[grizzled stone]

ok

[vernal pollen]

All good?

[grizzled stone]

ok but then

I get -2.5y = x

[vernal pollen]

uh

no no

you have b-y = -2x

How do you get rid of the -2*x?

[grizzled stone]

divide 2 each side

[vernal pollen]

Yes

[grizzled stone]

I have this

-5y = 2x

then I divide both by 2

[vernal pollen]

You have 5-y = 2x

When you divide each side you take the minus with the 2 so you divide each side by -2

But you don’t divide the 5

Cause that would mess it up

It would make

(5-y)/-2 = x

Do you see how I got to this answer?

[grizzled stone]

5-y?

[vernal pollen]

You started with this

from y = -2x+5

The plus becomes a minus

[grizzled stone]

yes

[vernal pollen]

Hence 5-y

[grizzled stone]

shouldn't it be -5y though??

[vernal pollen]

If you do -5y that would mean it is -5*y

[grizzled stone]

ok

[vernal pollen]

And that won’t work cause it’s not a divider at the right side

Cause dividing becomes multiplication

And plus becomes minus

[small lance]

isnt it suppose to be -5+y

5-y is just (+5)-y

[grizzled stone]

I'm so cooked bro

[vernal pollen]

No cause you do minus 5 at the left side

[small lance]

then it becomes y=5

[vernal pollen]

And not plus five at the left side

[small lance]

y-5*

not 5-y

[vernal pollen]

It can be both to be honest

[small lance]

no

5-y is

(+5) + (-y)

[grizzled stone]

ok so y=2x+5

[small lance]

where did the y get its negative

[vernal pollen]

If whatever you fill in for y is in brackets it can be both

[grizzled stone]

the +5 becomes negative making it y=2x-5

[small lance]

lemme make this clear

whats the original formula?

[grizzled stone]

please

[vernal pollen]

Math solver gives me this

[grizzled stone]

I need to get x and y INTERCEPT ALGEBRAICALLY

[small lance]

thats what i mean

you got y already?

[vernal pollen]

He has the formula

y = -2x+5

[small lance]

honestly i only know math in dutch

lemme translate the question

[vernal pollen]

yea me too

Hij moet x losmaken van de formule

so the only logical answer I can get = x = (5-y)/2

[small lance]

snijpunt met het

[vernal pollen]

Cause + 5 becomes - 5 both sides, the -2x you divide

[grizzled stone]

yes its 5

[small lance]

snijpunt met het as?

@vernal pollen

[vernal pollen]

?

M.b.v. de formule kan hij de snijpunten berekenen maar dan moet hij eerst de formule om kunnen zetten van y = naar x =

I don't do this often anymore, cause computers lmao but I put it in math solver from microsoft and it's giving the same answer as I got which is x = (5-y)/2

[small lance]

isnt it just 0=2x+5

[grizzled stone]

bro

[vernal pollen]

Not in this case since it's with a graph and if you have 0 you can't do much with the graph

unless you mean y =

[small lance]

then its y-5 = 2x

[grizzled stone]

I need someone, to help me understand this step by step to get the x and y intercept ALGEBRAICALLY

[small lance]

not 5-y= 2x

[vernal pollen]

microsoft gives me 5-y

And I got 5-y aswell

[small lance]

in this formula the y is negative not the 5

[vernal pollen]

the thing that you fill in for y should be put in with brackets

In that case the y isn't negative

[small lance]

@grizzled stone

ill do it

basically what you need to do is

x = 0
y = 0

[vernal pollen]

That's a different method

[small lance]

y = 2*0 + 5

y = 5

[grizzled stone]

.close

[obtuse pebble]

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[obtuse pebble]

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[verbal totem]

f(x)=(x^3-2x^2)^2 how can we solve it without l'hopital

[verbal totem]

I tried to make it in derivative form but i couldn't we need something like (f(2+h)-f(2-3h))/h right?

anybody have idea?

[obtuse pebble]

@verbal totem Has your question been resolved?

[obtuse pebble]

@verbal totem Has your question been resolved?

[verbal totem]

knk çözümünü biliyon mu

eyv knk sana da

[obtuse pebble]

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[fervent wind]

What does \epsilon and \delta represent for a boundary?

[obtuse pebble]

@fervent wind Has your question been resolved?

[lime wolf]

do you mean concerning limits?

[fervent wind]

yes

[obtuse pebble]

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[fervent wind]

.reopen

[obtuse pebble]

✅

[lime wolf]

they're used to represent the ideas of being near a limit or a point

if you have a limit

[fervent wind]

what there exact def ?

[lime wolf]

$\lim {x\rightarrow x{0}}\left( f\left( x\right) \right) =f\left( x_{0}\right)

why won't this bot work urgh

[fervent wind]

$\lim {x\rightarrow x{0}}\left( f\left( x\right) \right) =f\left( x_{0}\right)$

[warm shale]

Voltaire

[lime wolf]

yes so the statement says that for any positive number epsilon

it can bea ny size

there will be another positive number (delta)

such that if the distacne between x and x0 is less than delta

then the distance between f(x) and f(x0) is less than epsilon

This will be true if the limit exists / is true

take this limit, if you reducing the value of delta to keep getting closer to the limit horizontaly

you'll notice the values of epsilon can't continue getting smaller

in this visual example the epislons would be "blocked" by the points

[obtuse pebble]

@fervent wind Has your question been resolved?

[fervent wind]

i don't understand

[obtuse pebble]

@fervent wind Has your question been resolved?

[obtuse pebble]

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[languid harness]

no way

i just diod that question

[bright pagoda]

wtf

[languid harness]

deadass

[bright pagoda]

no way

[languid harness]

oimfg

what the actual

[bright pagoda]

HUH?!?!

[languid harness]

YEAR 11 METHODS???

[bright pagoda]

yep

[languid harness]

HAHA WTF

[untold marten]

dam that hits hard

[languid harness]

THERE IUS NO WAY

OK ILL EXPLAIN

[bright pagoda]

thats insane

[untold marten]

lol i did that last week

[languid harness]

i cant believe it

ANOTHER METHODS

WTH HAPPENING

[untold marten]

you prolly dont believe me

[bright pagoda]

wtf

[languid harness]

jesus christ

god put us here

i got stuck on c since i re arranged wrong lmao

ok so

firstly

find the x intercepts

tell me what you get

[bright pagoda]

anyway i got (x:-3<x<2)U(-4>x) (note all > or < signs are meant to have the equal thing under them i just dont know the symbol)

i think that answer only works if it = 0 tho right?

[languid harness]

yes

you should put it in interval notation

[bright pagoda]

yeah thats my issue

idrk how

[languid harness]

ok do yuou know how to graph it

[untold marten]

sorry to interrupt lads do any of you's do specialist

[languid harness]

no lol

[untold marten]

• S shape

[bright pagoda]

yes

[languid harness]

ok so ill teach u

[bright pagoda]

i might next year

[languid harness]

let me draw one sec

[untold marten]

good luck all i can say

[bright pagoda]

real

it cant be that much harder than methods

[languid harness]

so you have this

[bright pagoda]

yep

[languid harness]

you kinow how the graph goes onto infinity

[bright pagoda]

yep

[languid harness]

on the left side u have the negative quadrants

so to the left is negative infinity

on the positive right side it goes to positive infinity

so

when writing interval notation, you go from left to right

and you always use a normal ( bracket when including infinity

so firsty from the far left you have $-\infty$

[warm shale]

Lebob

[languid harness]

do you follow so far

[bright pagoda]

yeah

[languid harness]

tell me

what are all the points on the graph where it is greater than or equal to 0

[bright pagoda]

-3<x<2

-4>x

[languid harness]

the red right

yeah

[bright pagoda]

yeah

[languid harness]

so

the red on the right can be written as

$(-\infty,-4]$

[warm shale]

Lebob

[languid harness]

so from the negative infinity

you move along the graph until you reach the final point where it is greater than or equal to zero which is negative 4

then, this square bracket is put in when you have a greater than or equal to sign, it basically means that you are including negative 4, because you want where it is also equal to zero

[bright pagoda]

yeah [ inclusive
( non-inclusive

[languid harness]

exactly

so if it was just >0

[bright pagoda]

it would be (

[languid harness]

then it would be a ) instead of ]

[bright pagoda]

yep

[languid harness]

yep

ok cool

then

we put a

$\union$

[warm shale]

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

[languid harness]

nvm i suck with latex

$\bigcup$

[warm shale]

Lebob

[languid harness]

this is the union sign

and it basically means "and"

[bright pagoda]

👏 got it eventually lol

so the other one is like

[languid harness]

so far

$(-\infty,-4]\bigcup$

[warm shale]

Lebob

[languid harness]

we have this

[bright pagoda]

[-3, 2]

[languid harness]

well yes exactly

u got it

[bright pagoda]

cheers for the help

[languid harness]

$(-\infty,-4]\bigcup [-3,2]$

[warm shale]

Lebob

[bright pagoda]

same question is crazy 💀

[languid harness]

that union is huge

[bright pagoda]

haha

[languid harness]

but yeah u just go from left to right

and remember infinity always has a normal bracket

no matter what

[languid harness]

ikr

before u go

ill show u one other thing thats also important

say you had this, and its asking the interval that is greater than or equal to zero

greater

sorry

you have this whole zone but also that singular point

$[-2.-1]\bigcup{2}$

[warm shale]

Lebob

[languid harness]

man

i cant get curly brackets

anyway

with that singular 2

[bright pagoda]

so 2 is an intercept in this senario

[languid harness]

you need to put it in {}

[languid harness]

yes

but if it asks greater than or equal to zero

2 still counts

so you include it in curly brackets

that is only if a singular point satisfies the interval

[bright pagoda]

so thats what the curly ones are for

to show its a loner

[languid harness]

yeah

basically

[bright pagoda]

no friends

L

[languid harness]

because 2 is at the point y=0

so it still counts in the interval

just use curly brackets

and also

only use the union if u have two points otherwise it the interval will just sit on its own

something like this

[bright pagoda]

yeah

[languid harness]

u should be good now

[bright pagoda]

cheers

[languid harness]

do u have a test soon

[bright pagoda]

i have a sn tmr

[languid harness]

mine is wednesday

sn?

[bright pagoda]

2nd sn task

[languid harness]

oh

s/n

[bright pagoda]

yeah

[languid harness]

good luck with that

have a good one

[bright pagoda]

goodluck to you

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[obtuse depot]

"Determine the number of ordered pairs of integers (m, n) that satisfy m^n = 7^108"
I know that you have to assume 7 as positive and 7 as negative

108 = 2^2 * 3^3

[obtuse depot]

I think i'm supposed to use m^n = 7^kn

but i'm stuck there

[strong atlas]

Yes

[obtuse depot]

yeah m = 7^k

[strong atlas]

Yea, now k can be any factor of 108

So we need to find the number of factors of 108

[obtuse depot]

is there a formula for that?

[strong atlas]

Yes

Look it up

[obtuse depot]

For a number N, whose prime factorization is Xa × Yb, we get the total number of factors by adding 1 to each exponent and then multiplying these together. This expresses the number of factors formula as, (a + 1) × (b + 1), where a, and b are the exponents obtained after the prime factorization of the given number.

3 * 4 = 12

is that right?

[strong atlas]

Yes

Also look up the derivation of this for better clarity about this formula

[strong atlas]

So the final answer will be 12

Do you understand why the final answer is 12?

[obtuse depot]

because the total factors of 108 is 12, and ...

yeah no idea

oh 7 is prime so there's basically no replacement for it as the base

[strong atlas]

No

Basically nk = 108

[obtuse depot]

btw shouldnt we also assumes m = (-7)^k so we will just find the factors of 108 that is odd, it will create a new pair

[strong atlas]

Oh yes shit,mb💀

[obtuse depot]

please continue

._.

[strong atlas]

So let's break the question into 2 parts, when the base is 7 and when it is -7

In this first part

nk = 108

For every value of k(which is a factor of 108), the value of n gets fixed right

So (7^k)^n gets fixed as we know both n and k

So m gets fixed

So our part 1 gets converted into : the number of ordered pairs is equal to the number of values of k

Which is the number of factors of 108

Now as for -108

We need to find number of even values of k

[obtuse depot]

Wtf.. my keyboard suddenly broke

On my phone now

[obtuse depot]

So we have to find every factor of 108 first?

@strong atlas

<@&286206848099549185>

[obtuse pebble]

@obtuse depot Has your question been resolved?

[obtuse depot]

.close

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[strange pagoda]

why is this incorrect?

[obtuse pebble]

Please don't occupy multiple help channels.

[haughty hamlet]

Someome help with this

[strange pagoda]

.closd

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[plain nebula]

why do they sub in 2?

[obtuse pebble]

@plain nebula Has your question been resolved?

[obtuse pebble]

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[rustic sentinel]

what is nominal, ordinal, interval and ratio?

[plain nebula]

ordinal has order while nominal has not

idk about interval and ratio

[rustic sentinel]

can you give an example?

[plain nebula]

2, 4, 6, 8, 10 is ordinal

2, 4, 12, 123, 6, 2

is nominal

[rustic sentinel]

so nominal doesn't have any orders

while ordinal has orders

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[amber oak]

Hey, could anyone explain how they went from -2 * 2(1 - lambda) to just -4?

[median whale]

factoring

by (1-λ)

[amber oak]

Would you maybe know of a video that explains how to do this? When i search for vectoring i get broad explanations that just use x and y and such

or site

or wtv

[median whale]

I'll show you wait a sec

[obtuse depot]

(1-Λ)(1-Λ)^2 -4(1-Λ)
= (1-Λ)[(1-Λ)^2-4]

wait imagine
(1-Λ) = x

x(x^2) - 4x = x((x^2 )- 4)

is this what you mean?

[amber oak]

It is yeah, but i still dont fully get how 4(1-λ) isnt 4 - 4λ

[obtuse depot]

ohh you don't have to multiply it, just leave it seperated

-4(1-Λ) = -4 + 4Λ

[obtuse depot]

you forgot the minus

[amber oak]

oh yeah true

[obtuse depot]

it is the same

[amber oak]

but then where does the (1 - λ) go?

[obtuse depot]

it's going outside of the brackets

(1-Λ) [ (1-Λ) ^2-4 ]

if we multiply it back inside

it's going to be the same

(1-Λ) [ (1-Λ) ^2-4 ] = (1-Λ)(1-Λ)^2 - 4(1-Λ)

it is the same thing

uhh @median whale help me out here

._.

[amber oak]

aaaaaahhhh nonono i get it

HAHAHAH i feel so dumb every time i ask smt

but you throw it into brackets with the (1-Λ)^2

and then it's the same thing

right?

[obtuse depot]

yes

[median whale]

with what, sorry I didn't follow I thought you took care of it

[obtuse depot]

nvm nvm

[timid silo]

Could anyone help?

[obtuse depot]

sry for the ping

[amber oak]

he did! In that case I understand, ty for the help man

[obtuse depot]

okok

[amber oak]

AYO IM IN HERE bro

love u

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[timid silo]

ah, sorry for ping and interference, didn't check

[obtuse pebble]

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[timid silo]

I need to make y the subject

[wise lava]

!status

[obtuse pebble]

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

[timid silo]

2

Well its like half way through a question

, I just need to turn it into y= f(x)

[wise lava]

What is ln1?

[timid silo]

0

When you e^ all,
is it e^ asd + e^ 0 + e^ asd

x+1 +1 + 1-y^2

but its not sqrt (x+3) = y

this guy

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[zenith orbit]

given function f : X → Y.
it's said f(x) is a function from X on Y, if for every y ∈ Y, ∃x ∈ X, making y=f(x).

an example of this case for R → R is with the function y=x³, and an example of this case not happening is y=x². can someone explain why it doesn't happen in y=x²?

[zenith raft]

is there anything that maps to -1?

[zenith orbit]

no

[zenith raft]

is that a satisfactory explanation

[zenith orbit]

I get that X → Y for y=x² is R → [0, ∞), but

[zenith orbit]

one second I need to make my statement clearer

[zenith raft]

sure thing

[zenith orbit]

ty, I got it now

.close

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[leaden socket]

Hi, it is possible for double integral be lower than 0 ?

[viral blade]

??

Yes, a double integral can evaluate to a negative number

[leaden socket]

even if we calculate the integral of the area lying in the first quarter?

[viral blade]

Area shouldn't be negative

[leaden socket]

okay, i get it, thanks

[sleek escarp]

integrals are for signed area

[leaden socket]

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[mint drift]

Hi, currently working on a high school lab report and need to derive an equation. I am having trouble simplifying an equation which involves nested frctions (I think?) and am getting conflicting answers on the web and using calculators. Here is the equation that needs to be simplified :

[teal turret]

ok, id say, apply the square

then go from there

[mint drift]

Yeah I did, I am just struggling with the 2g in the bottom part of the fraction. I dont understand if the 2g goes to the numerator or the denominator if that makes sense?

[teal turret]

ah ok

so look at the length of the fraction lines

whichever fraction line is bigger, thats the main fraction

so if u have

$\frac {\frac ab}c = \frac ab \div \frac c1 = \frac ab \cdot \frac 1c = \frac a{bc}$

[warm shale]

Stephen

[mint drift]

ah ok this makes alot more sense

thanks for the clarification

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[timid silo]

whats wrong with that

its just like how it goes

4/10 is 0.4 the same thing applies here

it is.

oh so you mean repeating

do you know like infinite series by any chance

because you can showcase this from the opposite direction

[worn yoke]

not an inherent property of 9 per se but a consequence of base 10 number system (9 = 10 - 1)

[twin sandal]

1/99 = 0.01 (repeating), so multiply both sides by 62 and you get 62/99 = 0.62 (repeating)

similarly 1/999 = 0.001 (repeating), so multiply both sides by 104, etc

[obtuse pebble]

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[junior vine]

Hey, so Im taking a Linear Algebra Course as a uni sophmore and Id appreciate if someone could help clarify this confusion for me about basis and dimensions.

So basically, I know that to find the dimension of a vector space I have to find its basis and the number of vectors in the basis is the vector space’s dimnesion.
But for example in this example,
‘Find a basis for the given subspace of V and deduce its dimension. V= R^3
a- plane 3x-2y + 5z = 0’

So i get i move one of them to the otherside, factorize, giving me 2 vectors, but isnt the dimension given 3? What does V = R^3 supposed to mean here?

Ill send a pic too

[remote skiff]

You are given a subspace of R^3, you are told its a plane, and asked to find its dimension

[junior vine]

ahh

okay thats make sense thank you

i thought that plane is for V itself which was mainly whats confising me

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[fleet stag]

i dont understand how it = to -1 for part ii

[obtuse pebble]

@fleet stag Has your question been resolved?

[timid silo]

@fleet stag hi

have u got answers to a and b

basically

it’s 11 - the RcosAlpha stuff

the max height is will be when the RcosAlpha stuff is negative with its greatest magnitude

as - - becomes a positive

[fleet stag]

hi soz for the late reply

i stiwll dont understand 😅

[fleet stag]

ohh on the graph it show the wave above the graph

but that dosent matter

as it about magnitude

[timid silo]

ok

[fleet stag]

so we can use under the graphg when its at its peak

[timid silo]

let’s ignore the graph

say you had the equation

A = 1 - cosX

[fleet stag]

yh

[timid silo]

what are the values that cosX can take

between 1 and -1

right

[fleet stag]

yh

[timid silo]

so A is at a maximum when?

[fleet stag]

1 and -1

[timid silo]

no

if cosX = 1

[fleet stag]

ohh

[timid silo]

then A = 0

but

[fleet stag]

so its -1

[timid silo]

if cosX = -1

what does A =

[fleet stag]

2

[timid silo]

rirght

but itll never be greater than that

say i had B = 1 + cosx

B is at a maximum when

[fleet stag]

cosx=1

[timid silo]

yes!

its the same principle

[fleet stag]

its easy to tell here but from the equation

how can u tell in this 1

ohhh

casue -1 time - wil become positive

[timid silo]

yeah

[fleet stag]

and add that to 11 u get a bigger number

[timid silo]

yep

[fleet stag]

ok thank you i understand

ty

[timid silo]

its just one of those things

once you see it a few times it just becomes common sense

no prob

[fleet stag]

yh i need to do more practise

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[tacit sable]

good day all

[tacit sable]

can someone help pls

[obtuse pebble]

@tacit sable Has your question been resolved?

[strong pike]

!status

[obtuse pebble]

What step are you on?
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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

[strong pike]

Set up your system of equations, plug in arbitrary variables (a, b, c) for vector X

[obtuse pebble]

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[latent walrus]

youll need to find the equation of the plane containing A B and C first, 3 points is enough to do so

finding two vectors in the plane is the first move

[obtuse pebble]

@timid silo Has your question been resolved?

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@timid silo Has your question been resolved?

[obtuse pebble]

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[wanton radish]

Sam needs to reach the supply center from his campsite. The supply center is 300 miles away from the campsite with a bearing of N35E. However, Sam must first drive 180 miles S20E to reach a well for water. Then, he drives towards the supply center. Find out the bearing and distance Sam needs to drive from the well to the supply center.

Im just confused with the whole question tbh

[gilded needle]

hmm, have you tried drawing a picture?

[wanton radish]

yeah

[gilded needle]

what does the notation N35E mean, is that 35 degrees clockwise from north?

[wanton radish]

ye i think so

first quadrant

[gilded needle]

yea

and then S20E i guess is 20 degrees clockwise from south?

can you show your picture?

[wanton radish]

no it would be 20 degrees counter clockwise from south

[wanton radish]

my parents took my phone sorry

[gilded needle]

harsh

[wanton radish]

yup

[gilded needle]

lemme sketch something, one moment

[wanton radish]

ok thanks

[gilded needle]

ok

not drawn to scale and kinda sloppy but hey

i think your goal is to find x and e

x is the length of the long side

and e is the angle between the long side and the horizontal dotted line

i labeled some angles that might be useful

the known ones (20 and 35)

and some unknown ones (a,b,c,d)

[wanton radish]

ok

[gilded needle]

is this consistent with how you pictured it?

[wanton radish]

yeah i think so

[gilded needle]

hmm, so the fastest way to find x is probably the law of cosines, do you know it?

[wanton radish]

yeah

[gilded needle]

for that, you need the angle a

but you can easily get that since you know that 35 + 20 + a = 180

(see the image if that's not obvious)

[wanton radish]

yea a is 135

[gilded needle]

125

[wanton radish]

oh shi mb

[gilded needle]

ok so calculate x

i think you have everything you need for law of cosines

[wanton radish]

wait what

so a is 125 right

dont i also need b

[gilded needle]

yea you'll want b

in order to get e

you can get b using the law of cosines also

(after you find x)

[wanton radish]

so which law of cosine i use?

cuz i know like 3 lol

[gilded needle]

this one here

[wanton radish]

oh

uh so

what would i put for b lol

[gilded needle]

well find x first

oh, do you mean the b in that screenshot or in my original image

[wanton radish]

the b in the formula

[gilded needle]

ah

well you want the unknown side x which corresponds to c in that formula

and your angle a corresponds to their gamma

and so their a and b are the sides on either side of the angle
those would be your 300 and 180

so you would do:
$$x = \sqrt{300^2 + 180^2 - 2(300)(180)\cos(125)$$

[warm shale]

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

[gilded needle]

don't forget to use degree mode for cosine haha

[wanton radish]

lol alr

is it 429.355

or something close to that

[gilded needle]

that's what i got

so now you need e

which you can get if you find b

and d

d is easy since it's part of a right triangle

for b you can use the law of cosines again

[wanton radish]

uhhh

i think i found the equationm

and the other one

n = 2cos((2π/12)(t-6)) + 12

and n = 13.73

[obtuse pebble]

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[timid silo]

hey can someone help me to understand why this expression works for finding r in the larger calculation? like wouldn't subtracting (x^2+2) result in the added area to the right of the red line? Is that fine?

[timid silo]

(its also not that I'm confused about the entirety of finding the expression, just the fact that it doesnt seem to be bounded by y=6...)

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

<@&286206848099549185> just calc, asking to pinpoint my flaw in reasoning

jesus christ

.close

[obtuse pebble]

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[topaz sail]

#help10

[obtuse pebble]

Please don't occupy multiple help channels.

[topaz sail]

I need someone to look over my homework

And tell me if I have anything wrong

[silk geode]

Hmm, i dont we do that here

[topaz sail]

Sorry

Oh

I mean this is important

Takes less than 3 mins

Just tell me if I have anything wrong and I’ll fix it

Hello,

Ah

.close

[obtuse pebble]

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[timid silo]

I need help with Mathematica

[timid silo]

Is there any way I can create/define a new domain? Mathematica has Z, R, and Z/nZ, but I'd like to be able to use the Solve function under the ring of polynomials with integer coefficients

This question might be more fitting for math stackexchange but whatever

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

<@&286206848099549185>

[obtuse pebble]

@timid silo Has your question been resolved?

[latent walrus]

i believe that looks something like Z[x]

[timid silo]

Yea

[latent walrus]

im not overly familiar with mathematica though

[timid silo]

Yea

But Mathematica doesn't have Z[x] as a domain

So I was wondering if anyone knew how to implement it or smth

[latent walrus]

https://reference.wolfram.com/language/ref/FromCoefficientRules.html would this be of any use

its the closest thing my search has given me

though not a set

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

.close

[obtuse pebble]

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[obtuse pebble]

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[plain nebula]

ok im kinda braindead rn

[plain nebula]

but what can 500x500x500x500x500 be simplified to

is it not 500^5

[spare gale]

then?

[last pilot]

yes

[astral silo]

yes it can

[last pilot]

it can be

[spare gale]

maybe try writing it in terms of primes?

[last pilot]

$500 \cdot 500 \cdot 500 \cdot 500 \cdot 500 = 500^5$

[warm shale]

ren

[plain nebula]

why tf is my calculator giving something else

[last pilot]

not necessary?

what is it giving?

probably the evaluation of 500^5

[spare gale]

then?

[astral silo]

the calculator doesnt simplify i think

[plain nebula]

3.125*10^13

wait im a fkn idiot

[astral silo]

yea its scientific

[plain nebula]

does that equal to 16k

[last pilot]

no?

[astral silo]

it evaluated

[plain nebula]

ok holdup

i got a question

[spare gale]

no?

[last pilot]

this is correct

[whole dock]

You can simplify it as $5^5 \times 100^5$

[warm shale]

ColdTe²

[astral silo]

same thing

ah

[last pilot]

this is easy

[astral silo]

bacteria questions

[last pilot]

P_0 = population at time 0

[spare gale]

then write 5^5 × 10^10

[astral silo]

8 minutes?!

[last pilot]

k = growth rate per hour

[plain nebula]

this is the working out for it but i just did 500*2^5

am i gonna get marked down or some shit

[last pilot]

?

[astral silo]

did u have calculators?

wait thats pretty simple tho

did u not have time?

[spare gale]

This is correct I suppose

But why not write it as 16000

[plain nebula]

nahbthis is tech free

[astral silo]

ik i realised it was easy enough for it to be tech free

[spare gale]

fr

500×2×2⁴ = 1000 × 16

[astral silo]

yea

32*500 is simple enough too

[plain nebula]

whats e*ln

[spare gale]

ln is the log base e

[plain nebula]

ok ty

.close

[obtuse pebble]

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[obtuse pebble]

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[stone radish]

,rotate

[oblique zodiac]

No 13.. pls

[stone radish]

!status

[obtuse pebble]

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

[oblique zodiac]

7

Or 1

[stone radish]

wdym

oh

[oblique zodiac]

Like idk the formulas and stuff

[stone radish]

so there are n+1 terms

[oblique zodiac]

!rotare