help-10

[crystal stream]

like Dirichlet

[spiral knot]

ah ok

wait so this would be allowed

a function that is x for all x>=8 but 1/(8-x) for all x<8 would be defined for all real numbers but you cant take the limit at 8-.

[crystal stream]

why doesnt it exist

I mean

yeah

why?

@spiral knot

[novel knoll]

Try and calculate the limit and you get the answer

But that isn’t a counterexample since its not increasing

[crystal stream]

Did you find a point where it does not increase?

[novel knoll]

,calc 1/(8-7.9)

[warm shale]

Result:

10

[novel knoll]

f(7.9)>f(8) for example

So not increasing

[crystal stream]

yup

ugh..

nice

Do you have an Idea where can I try to start proving it?

I tried assuming towards a contradiction

then I have 2 options either inf or -inf

[spiral knot]

When is the only time a limit doesn't exist?

[novel knoll]

f is increasing

[spiral knot]

it can't be at inf because f is increasing, and you can go higher than inf

I'll let you figure out why -inf doesn't work either

if you need my help ping me

[crystal stream]

so if I assume

that it is not exists

it does not

so I can get the contradiction by the fact that from 7 to 9, the function has to be increasing

there exists x_1 and x_2 between 7 and 8 lets say

maybe x_1 is smaller than 8 and bigger that 8-delta

and that x_2 is bigger than 8

and then see that the value of f(x) when x -- > 8-

is inf

no?

thats hard...

[novel knoll]

What is definition of having limit infinity?

[crystal stream]

for every k > 0

[novel knoll]

(No real reason to use contradiction at all though)

[crystal stream]

really?

[crystal stream]

there exists an M>0, which for ever x>M f(x) > k

[novel knoll]

Ah

For all N>0 exists a delta>0 such that f(x)>N whenever 0<8-x<delta

if f(x)>N for all N when x is close enough to 8 how can f be increasing?

[obtuse pebble]

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[earnest gull]

Hey so basically I’m not even sure what concept this question is asking me to apply? Any idea how to approach this or what I can read up on in relation to this? This is for a undergrad probability theory class

[obtuse pebble]

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[shut marlin]

Every 5 minutes a train travels along a straight, isolated track between stations A and B in a constant speed and direction.
(1) Show that for any points between stations A and B, a train must pass through each of the points exactly once every 5 minutes.
(2) For the same set of points in (1), show that how many trains must pass through each of the points every minute.
(3) For the same set of points in (1), show that how many trains must pass through each of the points every 11 minutes.

[obtuse pebble]

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@shut marlin Has your question been resolved?

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@shut marlin Has your question been resolved?

[forest sinew]

first ones not so bad right?

what class is this for

or what subject

i think that "constant speed and direction" are your heavy lifters for these problems, as its a strong assumption

[obtuse pebble]

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[calm comet]

Let ABCD a square and X a circumference which go throug A and B but C and D are outside X. Suppose there is a tangent at X with tangent point T and DT=2AB, any hint to find de radius of X in terms of AB?

[obtuse pebble]

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[calm comet]

<@&286206848099549185>

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[strong delta]

APB is isosceles right?

AP is r and the height must be perpendicular

[obtuse pebble]

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[timid silo]

How do you simpifly a radical example square root of 162

[autumn remnant]

from what i know, you simplify it by making the numbers smaller
like 9 to 3 and likewise

you turn the square root of 162 to a smaller number

[timid silo]

But when I try to simplify square root 50 the answer was 5 square root of 2

[autumn remnant]

can i recommend a video? it might help explain it in detail

[mild ocean]

sqrt(ab) = sqrt(a)*sqrt(b)

pull out numbers that you know are perfect squares

[timid silo]

What’s sqrt

[timid silo]

Yea

[autumn remnant]

square root

[mild ocean]

for example, sqrt(50) = sqrt(25)*sqrt(2) = 5sqrt(2)

sqrt(x) = square root of x

[timid silo]

ok but sqrt(162) why is it 9 sqrt(2)/2

[mild ocean]

it’s not

it’s 9sqrt(2)

9sqrt(2) = sqrt(81)*sqrt(2) = sqrt(162)

[timid silo]

I put 9 sqrt(2) and got it wrong

[mild ocean]

they’re wrong then

[timid silo]

How

[mild ocean]

,w 9sqrt(2)

[warm shale]

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[mild ocean]

,w sqrt(162)

[warm shale]

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[mild ocean]

same thing

,w (9sqrt(2))/2

[warm shale]

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[mild ocean]

different

[timid silo]

I must of got x wrong then

,w (9sqrt(2))/2

,w sqrt(12)

[warm shale]

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Requested by KXG#1763

[obtuse pebble]

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[lament vapor]

does anyone know how to explain negative sqaure roots?

like the purpose of i

i think its mean -1 from online resrearch

[mild ocean]

i = sqrt(-1)

[lament vapor]

but when we calculate would it be -4 +6 not -6??

im confused on that part

since negative times negative is possitve when compeltin g the answer -4 + 6/1 is correct or incorrect?

[winged coral]

no, its imaginary. so you can't factor out a -i or anything.

[lament vapor]

oh so its just there and we just follow the +/- beside it

[winged coral]

You would leave you solution in the form -4 -3i, -4+3i

[lament vapor]

oh okay that makes sense

so after -4+- 6i/2, then it becomes -4 + 6i/2 and -4 -6i/2, which then becomes what is shown in the image above?

so -4+6/2 and -4-6/2?

[spiral maple]

yeah you dont just randomly drop the i

[lament vapor]

oh alright thankyou for explaining 🙂

wait so in the final answer why do you drop it?

[spiral maple]

you dont

[lament vapor]

or does it stay??

[spiral maple]

the roots are complex

[lament vapor]

-4+6/2 and -4-6/2 so this is incorrect if we used i?

[spiral maple]

$\frac{-4\pm 6i}{2}=-2\pm 3i$

[warm shale]

Mosh

[spiral maple]

so your roots are -2**+3i and -2-**3i

[lament vapor]

ohh okay that makes sense, but since we know i is just an imaginary one the final answer can we write it as 1 or not?

for -2 + 3i as an example

[spiral maple]

no

[winged coral]

No i is a special number

[spiral maple]

-2+3i is just -2+3i

the far right stuff on your SS is complete nonsense

[lament vapor]

oh alright thanks

[obtuse pebble]

@lament vapor Has your question been resolved?

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[valid kestrel]

I believe it’s incorrect because of step 3

If the triangles are congruent by SSS then why are two angles congruent because of CPCTC?

Doesn’t make sense to me

The options for this one are:

The proof is (correct, incorrect) and step number blank is the first unjustified step due to a (wrong reason, missing prior step)

<@&286206848099549185> It’s been 15 minutes, I want to see if I’m correct on this

[obtuse pebble]

@valid kestrel Has your question been resolved?

[valid kestrel]

Nvm it’s not step 3

[obtuse pebble]

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[dim trench]

Is this legal ?

[mild ocean]

probably not

[junior inlet]

where did you put the minus sign in e^{-x}?

outside the integral?

[mild ocean]

actually definitely not

yeah what’s with that?

[dim trench]

Yes

[junior inlet]

you can

t do that

(pressed enter instead of apostrophe)

[mild ocean]

that makes no sense

$e^{-x} \neq -e^x$

oh gosh

[junior inlet]

we have $e^{-x} = \frac{1}{e^x}$

[warm shale]

Kanga Gang Adorped Adavocowana

[dim trench]

I mean this

[warm shale]

quantum

[junior inlet]

$e^{-x} = \frac{1}{e^x}$

[warm shale]

Kanga Gang Adorped Adavocowana

[dim trench]

I probably misunderstood your question

[dim trench]

Yeah, I know that, but I wasn't touching it

[junior inlet]

then what's your substitution?

[dim trench]

$$u = - x$$
$$du = - dx$$

[warm shale]

Andrey_K

[junior inlet]

OH

i thought it was u=x

my bad

it's probably correct

[dim trench]

KK, thanks!

.close

[obtuse pebble]

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[clear girder]

so you want show that 7^2n - 4^n is divisible by 5 for all n ?

n elements of natrual numbers?

the principal of induction is clear right?

the induction step?

well wait i will sending something

sorry another question: you mean 7^(2n) or (7^2)*n?

ok

yes i am, but i am stuck too

XD

<@&286206848099549185>

well you can do the following for the first $4^{n+1} = 4^{n-1} \cdot 16 = 4^{n-1} \cdot (4+4+8) = 4^n+4^n+8 \cdot 4^{n-1}$

[warm shale]

Armin

[clear girder]

this could be a start

now you have $4^n$ you only still need $7^{2\cdot n}$

[warm shale]

Armin

[obtuse pebble]

@verbal comet Has your question been resolved?

[obtuse pebble]

@verbal comet Has your question been resolved?

[upbeat island]

hello if i understand you just need to show that if 5 | 7^(2n) - 4^n then we have 5 | 7^(2n + 2) - 4^(n+1)?

@verbal comet

5 | n means n is divisible by 5

read "five divides n"

well we have our inductive hypothesis, 5 | 7^(2n) - 4^n.
so we just need to simplify 7^(2n + 2) - 4^(n+1)

that's the plan?

[obtuse pebble]

@verbal comet Has your question been resolved?

[upbeat island]

k go ahead and simplify then

[upbeat island]

hey i gotta go but a hint would be that 5 | m + n if and only if 5 | m and 5 | n
use that after you reduce 7^(2n + 2) - 4^(n+1) and that's pretty much it

[obtuse pebble]

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[obtuse pebble]

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[tough stream]

Hey, how would one go about figuring the difference between one number and another's nearest quarter?

[fierce siren]

do you mean, like, rounding a number to a quarter and finding the difference between it with another number?

[tough stream]

No, like, if y = 100, i want to know the difference between x and 0, 25, 50, 75 or 100 which ever difference is smallest

Ive tried the absolute value of (y / 4) - (x mod 4)

I think you could use a triangle wave 🤔

[timber quest]

hi

can someone help me plsss💀

[tough stream]

Hey

[timber quest]

i wasn’t here for one lesson

and i wanna cry

if possible that is

[fierce siren]

it's 7i times 7i

i = sqrt of -1

it isn't a real number on it's own, because it has no real value

but when you square it, or multiply it by itself, you get -1 by definition

[timber quest]

wym no real value

[fierce siren]

there isn't a number in the whole universe that when squared gives us -1

[timber quest]

ohhh that’s what it means

[fierce siren]

we make up for that by writing the sqrt of -1 as i

you know that classic expression, e^pi * i = -1?

i in that expression is the square root of -1, and it's one of the few equations with i that has a real value

[timber quest]

i don’t think so

i know that i=-1

[fierce siren]

nah, i is the square root of -1. like $\sqrt{-1}$

[warm shale]

alias

[fierce siren]

i^2 = -1

[timber quest]

ohhh so that’s why i was confused

thanks

anyways u can keep going sorry

[fierce siren]

alright no problem lol. if you need help you can ask me

[timber quest]

só that’s the only one

7i(7i)

can i get help with 3 more pls

sorry

[fierce siren]

yeah, you can perform algebra on it normally to get $49 \times i^2$

[warm shale]

alias

[fierce siren]

and that would give you -49

[fierce siren]

of course, just make your own question channel

[timber quest]

how do i do that

[fierce siren]

go to one of these channels, and type your question there

[timber quest]

oh i thought this was one of those

mb

[obtuse pebble]

@tough stream Has your question been resolved?

[obtuse pebble]

@tough stream Has your question been resolved?

[tired shell]

@tough stream minimum(x (mod y/4), y/4-x (mod y/4) )

although this only works if 0<x<y

if x is outside those bounds it's just -x if x<0 and y-x if x>y

[tough stream]

Thanks, ill give it a go 🙂

[obtuse pebble]

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[timid silo]

How can I quickly find the equation for a sequence like this 21, 32, 45, 60, 77

[timid silo]

Where 21 is figure 1

I noticed that they added by 11, 13, 15, 17

So 21+(11*n)+2(n-1)

that's wrong tho

21+11(n-1)+2(n-1)

[mild ocean]

still wrong

[timid silo]

21+13n-12

13n+9

[spiral maple]

2nd differences are constant, so it's a quadratic rule.

so you can just get a 3x3 system for the coefficients

[timid silo]

I am not sure I follow, I don't know what these terms mean

Didn't take math in English

[spiral maple]

$f(n)=an^2+bn+c$ since f(n) is quadratic

[warm shale]

Mosh

[spiral maple]

then use the fact $f(1)=21, f(2)=32, f(3)=45$ to get a system of equations for a b and c

[warm shale]

Mosh

[spiral maple]

or note the terms are all 4 less than a perfect square

much easier

[timid silo]

So it should be (n+4)^2 -4

[spiral maple]

yep

[obtuse pebble]

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[urban summit]

for this im wondering if the equation x+2=(480/x)-8 would work

i might have made a mistake somwhere but im not sure

<@&286206848099549185>

.close

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[quaint glen]

Ratio test?

yw

[obtuse pebble]

@sudden condor Has your question been resolved?

[spiral maple]

If you're done with the channel... dont keep it open.

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[silent robin]

anyone?

[obtuse pebble]

@silent robin Has your question been resolved?

[obtuse pebble]

@silent robin Has your question been resolved?

[radiant gale]

<@&286206848099549185>

[obtuse pebble]

@silent robin Has your question been resolved?

[glass granite]

@silent robin I can try to help if you want

[radiant gale]

were the same person, so yea we still dont have a soln

can u?

@glass granite

[glass granite]

Alright

Let me try

[radiant gale]

thanku tyt

[glass granite]

Im not a helper or anything, just want to help people lol

This is geometry, correct?

[radiant gale]

ye

similar triangles

[glass granite]

Alright

I’ll solve first then tell you how to do it

Hmm

This is a difficult question

[limpid roost]

Let's set a = RC = BQ.\
We know triangle QCR is similar\
to triangle BCA, so we also know\\
$\frac{BQ}{QC}=\frac{AR}{RC}$\\
We can then substitute with $a$, $x$, and $y$.\\
$\frac{a}{y-a}=\frac{x-a}{a}$\\
Cross multiply.\\
$a^2=(y-a)(x-a)$\
$a^2 = xy - (x + y)a + a^2$\
$(x + y)a = xy$\
$a = \frac{xy}{x + y}$\\
And then you can just use that to
find what you need.

[glass granite]

Oh wow

[warm shale]

@limpid roost

[limpid roost]

Slightly better spaced.

[glass granite]

Nitrogen you are a genius

[limpid roost]

Not really. I almost gave up without trying. :P

[glass granite]

Lol

[radiant gale]

wait so then

for say, whats AR in terms of x and y then?

[glass granite]

TuckeR is this honors geometry or something?

[radiant gale]

just hs stuff

[glass granite]

This is hs stuff lol

[limpid roost]

AR = x - a.

= x - (that thing for a).

And then maybe simplify.

[radiant gale]

god thank u for the help

appreciate it man

[silent robin]

thanks for the help!

.close

[obtuse pebble]

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[magic cairn]

How can I prove that 2cotxcsc2x = 1+cot^2(x)

[limpid roost]

Try using a double angle identity.

[magic cairn]

sin2x?

[limpid roost]

Mhm.

[magic cairn]

left side i got to 2cosx

[limpid roost]

For the entire left side?

[magic cairn]

yea

[limpid roost]

$2\cot(x)\csc(2x)\= 2\cdot\frac{\cos(x)}{\sin(x)}\cdot\frac{1}{\sin(2x)}$

[warm shale]

@limpid roost

[limpid roost]

Did you get this far?

[magic cairn]

yes

thenthe right fractioin becomes 1/2sinxcosx

[limpid roost]

$=2\cdot\frac{\cos(x)}{\sin(x)}\cdot\frac{1}{2\sin(x)\cos(x)}$

[warm shale]

@limpid roost

[limpid roost]

Mhm.

And then you simplify it.

[magic cairn]

maybe i simplified it wrong idk lol

[limpid roost]

The 2 cos(x) cancels out.

[magic cairn]

o i didnt do this

[limpid roost]

$=\frac{1}{\sin^2(x)}$

[warm shale]

@limpid roost

[magic cairn]

it works ahhh thxxxxx

@limpid roost thanks :DDDDDDDDDD

have a good day

[limpid roost]

You too. :>

[magic cairn]

😄

.close

[obtuse pebble]

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[crystal stream]

hi

[crystal stream]

I am having some trouble trying to understand which number is the biggerst number that divide by 8 or 6 or 7 ..

lets say starting numbers. [10000, 70000]

how can I know which one is the biggest one that divide by something

and the smallest one

[obtuse pebble]

@crystal stream Has your question been resolved?

[crystal stream]

<@&286206848099549185>

[limpid roost]

Is [10000, 70000] a range?

[crystal stream]

yeah

[limpid roost]

Do you know what floor division is?

[crystal stream]

umm

not really

[limpid roost]

You could do 70000 // 8 (or whatever factor) to get the largest number divisible by 8 in that range.

[crystal stream]

I dont have a calculator

[limpid roost]

Floor division is basically where you get the whole number. For example, 5/4 = 1.25. If you only take the whole number (remove the decimal places completely), you would get 1.

[crystal stream]

I must do it by my self

[limpid roost]

Oh. D:

Long division, maybe?

I don't know if there's an easier way or not.

I would just do 70000 divided by 8 (or another factor) and stop when I get to the decimal part.

[crystal stream]

lik

like

long divisiopn

divisio

[limpid roost]

Mhm.

[crystal stream]

and stop when getting to the .?

[limpid roost]

Mhm.

[crystal stream]

my teacher said something

[limpid roost]

So if you're doing this, you can just stop once you get to 11666.

[crystal stream]

about looking at the digits number

and see if it divides by the number

or something like that

idk

didnt understand that

[limpid roost]

And then multiply 11666 by 6.

[limpid roost]

Ohh, that's for some specific numbers, I think.

Well, for 7, you kind of already know 70000 is divisible.

7 * 10000.

For 8, it was if the last three digits are divisible.

For 2, it was if the last digit was even.

For 3, it's if the digits add up to a number divisible by 3.

Do you mean things like this?

[crystal stream]

yeah

exactly

like

how would I know that 69992 is the biggest number that divides by 8

[limpid roost]

In that range? It looks like your range is inclusive of 70000, which is divisible by 8, so would 70000 be the greatest number?

[limpid roost]

But if you were looking at this, 992 is divisible by 8.

[crystal stream]

damn

it is...

I missed it

how would you know that?

[limpid roost]

That's because 8 * 125 = 1000, and 992 is only 8 away from 1000, so 8 * 124 = 992.

[limpid roost]

The 70000?

[crystal stream]

yeah

[limpid roost]

If it's inclusive or if it's divisible by 8?

[crystal stream]

divisable

divisible

[limpid roost]

The last three digits are 000. 0 is divisible (I think) by 8 because there's no remainder.

[upbeat island]

https://en.wikipedia.org/wiki/Divisibility_rule

[obtuse pebble]

@crystal stream Has your question been resolved?

[obtuse pebble]

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[royal locust]

So I have been struggling with choosing the correct bounds/equations for finding the area in between two polar graphs.

[royal locust]

I have this graph, where the red graph is $$r(\theta) = 2+2cos(\theta)$$ and the blue graph is $$r(\theta) = 6cos(\theta)$$ and I am trying to find the area of their shared insides.

[warm shale]

SilverTongue

[royal locust]

I understand that you need to set them equal to eachother and solve

and when you do this you get pi/3 and 5pi/3 as their intersection

and I know to draw the radial line from 0 to pi/3

(I'm not using 5pi/3, as I'm just doubling the area above the axis)

So essentially my question is

which of these is correct

and why

[royal glacier]

This one

[royal locust]

Ok but why?

[wary vigil]

wouldn't that count the triangle twice?

[royal locust]

Where tf is there a triangle???

[royal locust]

Wait what

[plain owl]

No it wouldn't.

[royal locust]

I swear I have never seen a triangle used with these before

[wary vigil]

nvm

i didn't visualize the integration right

these are the two integrations

if i'm not mistaken

i thought it was different

[royal locust]

Ok yeah

Ok wait so

I’ve always had trouble visualizing

Like I know how to visualize it when I am doing a rectangular area

Like with rectangular coordinates I mean

Cause it’s just to the x or y axis

But with polar

Is it basically just like

Radial like to radial line?

[wary vigil]

yeah

a line from the origin

[royal locust]

Ok yeah

[wary vigil]

instead of lines to the axes

it

s

lines to the origin

but i was doing it wrong

[royal locust]

Ok

[wary vigil]

thats why i said the triangle

[royal locust]

Aight

Ok so like if

Using the same graph

I’ll resend

If I wanted the area

Inside blue

Outside red

It would be

0 to pi/3

Of the blue squared?

[wary vigil]

if you want the inside of the blue but not inside the red

you'd calculate the inside of the blue total

and subtract what you just found

[royal locust]

Why subtract what I found not the red?

[wary vigil]

because this bit

is part red part blue

you could also start at the point where the two functions cross

and only subtract the red bit

if thats what you asked

[royal locust]

here give me a second

It’s taking a while to send sorry lol

Both of these would be correct?

[obtuse pebble]

@royal locust Has your question been resolved?

[royal locust]

@wary vigil sorry for the ping but does this work?

[wary vigil]

i don't think the boundaries are correct

pi/2 in the blue one is y = 2

no sorry wait

that's the red one

one sec

you're completely right

i'm dumb

this is the blue one from 0 to pi/2

so you're right completely @royal locust

[royal locust]

Ok thank you for the help

You really helped me understand that it's the origin/pole that the area comes from

not like

in relation to an axis

.close

[obtuse pebble]

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[sour fox]

How would one prove that f(x) = 10x is surjective if f: A --> B? or is it even surjective?

I know that it is surjective for R but since we have different sets, i dont know what to do.

[timid silo]

the way this function is defined leaves a bit to be desired

[obtuse pebble]

@sour fox Has your question been resolved?

[obtuse pebble]

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[little ruin]

I am having an issue understanding what these problems want me to do exactly. Up until this point I have not encounter a question like this in my previous work, and when I try to reference the textbook I see no similar problems to try and reverse engineer.

[little ruin]

What do A and B mean in this context?

[stable dawn]

Well there is a rule about logarithms saying you can bring the power of the argument to the front

[upbeat island]

A and B are just numbers

[stable dawn]

B is not though

because there's no right hand side of the expression

[upbeat island]

oh sure, i guess an expression

[stable dawn]

so B is just x

[upbeat island]

(x is a number to me 😉 )

but yes not digits per se

[stable dawn]

sigma male attitude 😄

[little ruin]

So if x is B, is A the power of 8?

[stable dawn]

reminds me of my teacher who announced he won't write the arrows upon the vectors because it takes too long

[stable dawn]

in the first case yes

[warm shale]

citrusmunch

an example $\log_3 \sqrt{x} = \log_3 x^{1 \over 2} = \frac{1}{2} \log_3 x$

[upbeat island]

same tho

[brave bramble]

Get a table of log properties in front of you. Here's one I got from Google

Note rule 3 is the rule you'll use for this q

[little ruin]

So for the second problem, would it be ^2√x as A, and then x again as B?

If i've understood

[upbeat island]

not quite

[upbeat island]

i wasn't going to be pedantic here, but it seems it matters now
A = 8 here, no power involved anymore

your second problem is very similar to my example

the first step is just interpreting radicals as exponents
the second step there is the new log stuff you're learning

[little ruin]

I understand what to do now! Thanks for the help

.close

[obtuse pebble]

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[sharp orbit]

Why is it a -2 and not a 2?

[plain owl]

We have (4-x)^{-1/2}. Sign of x inside the bracket is negative

[sharp orbit]

AHHHHHHHHhhhh

OMG

thanks

how do i open this channel again?

.close

[obtuse pebble]

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[plain owl]

You mean this specific channel? Open??

[sharp orbit]

I think close did the trick?

[obtuse pebble]

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[timid silo]

Would appreciate help with a problem I had on an exam which I couldn't solve: A car drives with v=25 m/s, suddenly, at 58m in the distance there's an obstacle, the car then decelerates with a=-8 m/s². Calculate the final velocity with which the car will hit said obstacle. Take a reaction time of 1s into account.

[forest sinew]

at what distance does the car start braking?

[timid silo]

57m due to the 1s reaction time

[forest sinew]

i dont think thats correct

it doesnt take the driver 1 meter to start braking

it takes them 1 second

[timid silo]

oh wait yeah

right my bads

bad

[forest sinew]

youre chill dude

[timid silo]

welp then idk

just ignore the reaction time thingy, that's not as important

[forest sinew]

it is sorta important

well its an easier sub problem

im running 5 m/s at a wall

if i have a 2 second reaction time

and i suddenly wish to stop

it will take me $\frac{2 \cancel{sec}}{1} \cdot \frac{5 m}{\cancel{sec}} = 10 m$ to start slowing down

[warm shale]

jan Niku

[forest sinew]

in your problem, were plodding along at 25 meters per second

so if something takes me a second to do, how far have you traveled in that second?

[timid silo]

25 if I believe

25m

[forest sinew]

yea

so how far away is the object once we begin to react

takes us 25m to realize whats going on

it was 58 meters away

but now its..

[timid silo]

58-25

[forest sinew]

33

[timid silo]

yeah

[forest sinew]

33 meters away, we begin to decelerate

from 25 m/s to some final speed

at -8 m/s^2

what equations do you have to use?

i mean what sort of equations is it assumed you are given or know

[timid silo]

oh wait I got it

[forest sinew]

[timid silo]

got t, therefore I just use $v=a*t+v_{0}$

[warm shale]

Lambdalo

[timid silo]

which is $(-8)m/s^2*1.89...s+25m/s$

[warm shale]

Lambdalo

[timid silo]

that correct?

[forest sinew]

yea

i think you could go with just one formula

well im not sure if its correct

tbh

but it seems right to me

[timid silo]

man dang, I was really hoping my prof made a mistake

[forest sinew]

?

why

[timid silo]

because that would mean that I'd be excused for not getting the result

darn, welp thanks for the help though

[forest sinew]

good luck

maybe i should double check

[timid silo]

nah you're good

just checked

[forest sinew]

okay

[timid silo]

es correcto, mas

[forest sinew]

[timid silo]

mal*

welp you live and you learn, next time!

[forest sinew]

😄

[timid silo]

reminds me of a poem by piet hein:

The road to wisdom? Well, it's plain
And simple to express:
Err
and err
and err again,
but less
and less
and less.

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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@willow trellis Has your question been resolved?

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[lofty quiver]

Hello I need help to calculate the limit at 0 of this function please

[kindred elk]

pls help

nvm

[obtuse pebble]

@lofty quiver Has your question been resolved?

[obtuse pebble]

@lofty quiver Has your question been resolved?

[obtuse pebble]

@lofty quiver Has your question been resolved?

[obtuse pebble]

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[dusky shore]

Given the complex function f(z) =z|z|, how would i prove its not analytic anywhere. In a previous part of the problem, i was able to prove its differentiable at z=0, so not entirely sure how to go about this

[brave bramble]

Analytic = Complex differentiable

So you just need to check the CR equations

Don't check only at z = 0

[obtuse pebble]

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[dull terrace]

heh?

[forest sinew]

did you write out some terms

whatd you get for the first 5 terms

[dull terrace]

0,8,0,-8 then it repeats like that to infinity

im confused on what its asking for?

[forest sinew]

sound like something that has a limit?

[dull terrace]

yh but what do write?

[forest sinew]

depends on the class

id write something like

[dull terrace]

its a series and sequences lesson

[forest sinew]

a_n = {..,0,8,0,-8,...}

periodic, limit dne

[dull terrace]

ok ill try that thanks

[obtuse pebble]

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[gilded viper]

hi can someone help me solve this?

[high lily]

what have you tried?

[gilded viper]

I don’t know i’m rlly confused

i decided to choose left side though

[high lily]

consider using conjugates and/or pythagorean trig identies

[gilded viper]

yea i haven’t learned conjugates yet

so i guess its pythagorean trig identiies

[hot mist]

it's not hard

assume that they are equal

[gilded viper]

ok i just started

[hot mist]

then use a/b = c

=> a = bc

since we have (a-b)(a+b) = a^2 - b^2

[high lily]

best not to mix sides

[hot mist]

and sin^2(a) + cos^2(a) =1

[high lily]

teachers don't like that

[hot mist]

it's using algebra

[high lily]

they insist that you use stuff like lhs rhs proofs

[hot mist]

oh

[gilded viper]

yes

[hot mist]

but my way is simpler(maybe)

can u use assumptions?

[high lily]

why would you need assumptions

[hot mist]

it's easier

[gilded viper]

don’t know what that is so no

[high lily]

there's a fine line you could walk that your teacher may or may not approve of

[hot mist]

oh

my teachers only want me to prove it in any way

if it's right, he/she approves it

[high lily]

anyway,
pythagorean trig identities / conjugates / difference of two squares are closely linked

you could consider applying pythagoras to the sin^2(x) on the numerator

in simple terms, the conjugate of a+b is a-b

[gilded viper]

oh

so

[high lily]

you could consider conjugates and multiply the numerator and denominator by the conjugate of the denominator and ultimately apply pythagoras

[hot mist]

i'll present my way, u can apply it or not
assume the identity is correct with all values
=> sin^2(a) = (1-cos(a))(1+cos(a))
=> sin^2(a) = 1 - cos^2 (a)
=> sin^2 (a) + cos^2(a) =1 (true for all values a)
therefore, the identity is true for all values of a

[short spruce]

what

[hot mist]

is it wrong anywhere?

[short spruce]

yeah you're not proving the identity at all

[hot mist]

if u assume it's correct and from that u come up with something that is obvious

it's correct

[short spruce]

why are you making assumptions when proving trig identities

[spiral maple]

You cant start with the thing you're proving

[spiral maple]

if you do that backwards, then you'd have a valid proof.

[hot mist]

there

that's why i said "u can apply it or not"

[high lily]

technically you "can" with the appropriate notation
using stuff like iff

[hot mist]

why can't i use "assume"tho?

[spiral maple]

Proof:
Assume true.
QED

[hot mist]

we don't do "QED" in my country

sorry :<

[spiral maple]

Ok... and?

[hot mist]

we just prove it(with the correct conditions)

or we have smth like QED

but I don't understand what it is

[spiral maple]

"It has been demonstrated"
The proof is done.

[hot mist]

in my country we can assume an identity we are proving is correct, then from that we start to move on until we get smth obvious with all values(some of which are restricted in certain values) then we confirm the identity is correct(with values at which the identity makes sense)

[hot mist]

nice tks for that 🙂

i'm just in grade 9 so my knowledge about trigonometry and English statements is not deep

my main language is not English either

[high lily]

depends on the teacher and notation

[hot mist]

or if ur teacher doesn't like my way of assuming

then go backwards

[junior inlet]

the problem is that you're not stating that it can't be incorrect

[hot mist]

in this scenario i'm talking about proving identities to be correct

and i'm assuming they are

but u can assume it's not

and then use contradictions

[high lily]

you could do something that involves similar algebra that'd be less frowned upon like \
$$\text{LHS} = \text{LHS} \cdot \frac{\text{RHS}}{\text{RHS}} = \underbrace{\br{\text{LHS}\cdot\frac{1}{\text{RHS}}}}_{\text{show that this is 1}} \cdot \text{RHS}$$

[warm shale]

ℝamonov

[hot mist]

possibly

he's a pro

but anyway

@gilded viper close this

we are done

[obtuse pebble]

@gilded viper Has your question been resolved?

[obtuse pebble]

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[pallid abyss]

I need help

[mild ocean]

send a question then

[pallid abyss]

gimme a sec

nvm

.close

[obtuse pebble]

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[cloud spear]

What do you do when -4AC is smaller than B^2 in a quadratic equation and the square root becomes negative in a quadratic equation

[cloud spear]

*quadratic formula

-28.32+/-sqrt(-571.7952/43.92)

This is in a distance formula problem

So the anwser cannot be i

How do I find the distance between parametri equations

[obtuse pebble]

@cloud spear Has your question been resolved?

[cloud spear]

Hello?

[short spruce]

if you're meant to find a real answer, you did something wrong

i don't know the context though

[junior inlet]

if the discriminant is smaller than 0, you've probably made a mistake

there would be no solutions

[cloud spear]

What is the distance between Steve and Elsie at time t ? (your formula will be valid for the times t when your formula for Elsie 's coordinates is valid).

The parametric formulas for Steve:

y=-3t+4

For Elise:

x=-3.6(t-1.5)-6

I did the distance formula

Then got a quadratic

[obtuse pebble]

@cloud spear Has your question been resolved?

[obtuse pebble]

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[cloud spear]

Yeah my dumbass lost a negative

[obtuse pebble]

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[untold pike]

I know the answer is C, but idk how

[mild ocean]

use the quadratic formula on the first function

then plug in the two x values in each answer

wait solution?

as in singular?

[obtuse pebble]

@untold pike Has your question been resolved?

[near rock]

I mean you could try expanding the equations

Or completing the square

[obtuse pebble]

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[still seal]

For number 4 I have tried with yp=Ae^x and yp=Axe^x, but A gets cancelled out. How should I do it?

[obtuse pebble]

@still seal Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[humble wind]

i expanded the lhs and it comes out to 4cosacosbsinasinb dk what to do from there

[obtuse pebble]

@humble wind Has your question been resolved?

[lofty quiver]

And have you tried expanding RHS

?

[jolly ginkgo]

Show your working

[humble wind]

can't atm sorry 😦

oh solved it

.close

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[jolly ginkgo]

So can you show your working

I don't get this

You have to find v such that (A-(-2)I)v = 0

v=(1,a)

Yes

First tell me what the matrix A-(-2)I equal to

Uhhh

[jolly ginkgo]

I'm not asking for final answer just this

$\begin{bmatrix} 56&98\-28&-49\end{bmatrix}\begin{bmatrix}1\a\end{bmatrix} = \begin{bmatrix}0\0\end{bmatrix}$

[warm shale]

it's Sam

[shy plank]

scalar multiplication

Then we cancel leading coefficient

[jolly ginkgo]

[shy plank]

top should equal 1 and bottom zero

[jolly ginkgo]

Why?

Just find a like this

[warm shale]

it's Sam

[shy plank]

yes

[jolly ginkgo]

That's what I did

You just have to solve the thing I gave

[jolly ginkgo]

Solve this and tell me the value you get for a

Does that means you don't understand eigenvalues and eigenvectors ?

What are they

You just have to substitute given v in Av=lambda v

Everything is given

It's not even like some tricky question

[warm shale]

it's Sam

[jolly ginkgo]

You know how to multiply matrix

Dude

If you call this trolling

I can clearly see what it is

I can answer you but it's not about me answering you it's about me just hinting you towards the answer

And this hint is almost like answering

I would be really surprised if you won't get answer from this

You need to show me what you did

No I want to see working not just answer

Ok ping any helper from this server and show where you explained and what

Where point it here too

[jolly ginkgo]

Show your work on this leave what you did earlier

It's a simple matrix multiplication which will give answer in a step or two

Multiply matrices

You surely know how to do that if you are doing eigen stuff

[warm shale]

it's Sam

[shy plank]

Why would I multiply twice?

When I already did so.

I am trying to reduce the matrix

but you want 56+98a top row

-28-49a bottom row

[jolly ginkgo]

So what does 56+98a=0 give

[shy plank]

lets see

subtracting 56 on both sides

98a=-56

diviiding to isolate a

a= -56/98

=-4/7

[jolly ginkgo]

So

[shy plank]

I see another error I made.

The problem gave lamba 1 = -2

I started with 5 first initially.

So to solve for b it should be 49+98b top row

[jolly ginkgo]

You got one answer right now I think you can get other on same lines

[jolly ginkgo]

No

[warm shale]

it's Sam

[jolly ginkgo]

Just solve this

[shy plank]

When lamba is 5

We have 49 98 top row

-28 -56 bottom row

-28-56b=0

add 28 to both sides

-56b=28

dividing both sides by -56

b=-28/56

b=-1/2

is that correct?

I think so since I get -1/2 when I compute: 49+98b=0 aswell.

[obtuse pebble]

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[full aspen]

is dot a multiplication?

[warm ledge]

yes

[full aspen]

and you only need to prove that $$2^{k+2} * 5^{2k+1} + 3^{k+2} * 2^{2k+1}$$ is divisble by 19?

[warm shale]

Epifor

[warm ledge]

$$2^{k+3} * 5^{2k+3} + 3^{k+3} * 2^{2k+3}$$ is divisible by 19

[warm shale]

r4z33n

[warm ledge]

given $$2^{k+2} * 5^{2k+1} + 3^{k+2} * 2^{2k+1}$$ is divisible by 19

[warm shale]

r4z33n

[full aspen]

not exactly sure about this but did you try lft?

since 19 is a prime

[warm ledge]

lft? i dont think we did that

[full aspen]

fermats little theorem

[warm ledge]

yeah its not in our syllabus so cant use that

[full aspen]

did you do congruences?

[warm ledge]

nope, for most of it, we just use basic facts or algebra manipulation to take out 19 as the common factor

[full aspen]

and k is a whole number?

[warm ledge]

yeah natural

this is the main question

[full aspen]

see that 2k+1 for 2 and 5 is always odd

[warm ledge]

right

.close

[obtuse pebble]

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[warm ledge]

.close()\

[obtuse pebble]

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[viral forge]

is this correct?

[real orchid]

It is close. You need to think about the domain of f(x).

Nevermind I just looked at it again. You are correct. I'm just tired.

[viral forge]

ohh okay

thanks for your help haha

[real orchid]

No problem.

[viral forge]

.close

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[violet merlin]

Hello. Can someone help me with either confirming my solution or better yet telling me how to make Mathematica confirm it for me?
We have an integral as seen above and the solution I got is -infinity but both Mathematica and Wolfram refuse to give me an answer

[obtuse pebble]

@violet merlin Has your question been resolved?

[violet merlin]

<@&286206848099549185>

[obtuse pebble]

@violet merlin Has your question been resolved?

[final python]

,w integral of 0 to pi/2 of (-4coszx+cosxsinx)/(1-cos^2x-2sinx)dx

[warm shale]

Results provided by WolframAlpha

Click here to refine your query online
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Requested by Master Butler#4230

[final python]

it doesn't converge

[obtuse pebble]

@violet merlin Has your question been resolved?

[violet merlin]

I see. And how do I know whether the result is + or - infinity then?

[obtuse pebble]

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[warm shale]

azeem321

azeem321

[strong vale]

This is apparently incorrect. Where did I go wrong?

[quaint glen]

The solutions do lie on those lines, but they're points on those lines, not the entire lines. You can substitude y=x-1 and y=x+5 into one of the equations now

[strong vale]

Thanks. Without a graphing app or something. how did you conclude that it wasn't the entire lines?

[quaint glen]

If you do some random row operations on a linear system, you could certainly get a new, different line that your solutions lie on. It does not guarantee however, that that entire line solves the system. This situation is analagous. All you've actually done is show that that system holding implies y=x-1 or y=x+5. Not that the system holds for any y and x satisfying y=x+5 or y=x-1

Also, I used a graphing app

[strong vale]

Thanks for the explanation got it now

.close

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[obtuse pebble]

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[stable kernel]

can someone show me how to do this? i have like no idea how this would work

[timid silo]

Start by cubing both sides as this cancels the cube root. Then you can solve the cubic

[stable kernel]

ye but i ended up with this and i'm not sure how to factor it:

[obtuse pebble]

@stable kernel Has your question been resolved?