help-23

[floral osprey]

think about where the 0 can go

[lean otter]

nowhere

[floral osprey]

why

[lean otter]

because are equalitys

[floral osprey]

there is a place for the 0 look harder

[lean otter]

no because

0/x=0

and 0*x=0

[floral osprey]

some of the numerators are 2 digit numbers

[lean otter]

yep

and

????

need to be different digits in every case

[floral osprey]

mhm

[lean otter]

i can say a fraction is equal to 0

if isnt 0/x

[floral osprey]

the 0 can’t be in the denominator

[lean otter]

yep

[floral osprey]

and it can’t be the first digit

what does that leave open

[lean otter]

nothing

[floral osprey]

incorrect

[lean otter]

why

[floral osprey]

look again

[lean otter]

y*x=0

with only integers

exists only when y or x is equal to 0

and 0 cant be the final result

[floral osprey]

indeed

but there is still a place for it

[lean otter]

cant be the divisor

[floral osprey]

alr imma just tell you
it’s in the 1s place of the numerator

[lean otter]

why

[floral osprey]

because a0 is not 0

[lean otter]

a_0

?????

[floral osprey]

where a is the 10s place

[lean otter]

but

and taking that need to be different numbers in every case

[lean otter]

may not exist

[floral osprey]

what

I told you theres an answer

that should help

[lean otter]

can you take screenshot and explain why it exist??????

[dusty fiber]

Its not multiplication

[lean otter]

it is

[dusty fiber]

No

it isnt

[lean otter]

damn

i just realized

😟

wasnt what i was thinking

i was thinking too much mathematically

apparently it means a number

like 02 is 2

😫

[floral osprey]

well 02 wouldn’t work

[lean otter]

i know

[floral osprey]

alr

[lean otter]

But that refutes what I said

so

theres a solution

[lean otter]

thanks

if you can solve the problem may be good

but i know you will not

[floral osprey]

well im not telling you the answer

!nosols

[safe radish]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[lean otter]

thanks

.close

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[dusky jay]

Just needing some guidance with part i and ii

[green bane]

!status

[safe radish]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[dusky jay]

1

Sorry should of stayed more quickly

I’ve find the function from doing the integral idk if that’s right

[safe radish]

@dusky jay Has your question been resolved?

[safe radish]

@dusky jay Has your question been resolved?

[safe radish]

@dusky jay Has your question been resolved?

[dusky jay]

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[oak storm]

Sorta stuck

[rotund wolf]

what have you tried

[oak storm]

I seperated the fractions with numerators A B and C

[still charm]

nope

wrong

there's your issue

[rotund wolf]

well you can do that

there is a way to have only constants on the numerator

[oak storm]

so A/2-x + B/3+2x +C / (3+2x)²

??

[still charm]

you should have A, B, then Cx + D

[rotund wolf]

looks good

[oak storm]

so like Cx + D/(3+2x)² ?

[rotund wolf]

there will be no Cx + D

https://www.youtube.com/watch?v=QfQxVAZKQpw

when you break it up like that, you get only constants on the top

[oak storm]

I'll take a look, thank you

[rotund wolf]

nw

[oak storm]

bro

i realised

I wrote 6x²+8x+8 on the left side

[rotund wolf]

oof

but at least the initial partial fraction decomp was right

[oak storm]

yup

thanks

[rotund wolf]

nw

[oak storm]

i thought i was actually doing something wrong

[rotund wolf]

lol

[oak storm]

but thank you anyways

[rotund wolf]

all g

[oak storm]

have a nice day

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[formal cove]

How would I find the modulous for this at s=4?

i did it but I'm getting a different answer

then myprof

[final halo]

modulous?

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[potent solstice]

I am pretty sure I am having an incorrect y calculation, I am being asked to calculate this question, for my first iteration I am getting 5.E-14 for y_1 but I am getting -1.6569 via my algorithm?

[plucky elk]

Where's your question

[potent solstice]

Sorry, the original question is this.

[potent solstice]

^

[plucky elk]

What is YOUR question

That's the problem you're working on

[potent solstice]

Am I calculating the correct value for y, if not can you show me the correct value if not 5.E-14? If it is the case my algorithm is incorrect could you point it out for me?

y_theta = 1/2; % denotes value for y_0
y = 0; 
x = 1/sqrt(2); % denotes value for x_0

%% 1
iterator = 0;
while (abs((1/pi)-y) >= 10^(-15))
    if (iterator == 0)
        y_1 = y_theta;
    else
        y_1 = y;
    end
    x = (1-sqrt(1-x^2))/(1+sqrt(1-x^2))
    y = y_1*(1 + x)^2-(2^(iterator+1))*(x+1)
    iterator = iterator + 1;
    if (iterator == 1)
        break
    end
end

[safe radish]

@potent solstice Has your question been resolved?

[safe radish]

@potent solstice Has your question been resolved?

[plucky elk]

`y = y_1*(1 + x)^2-(2^(iterator+1))*(x+1)`

why do you have x + 1 at then end on this line?

[safe radish]

@potent solstice Has your question been resolved?

[plucky elk]

@potent solstice are you still working on this question?

[potent solstice]

Yes, I have removed the + 1 from the x term at the end, sorry I must have just been sloppy, it still did not change the answer to what I am calculating, what else do you think it could be?

[safe radish]

@potent solstice Has your question been resolved?

[vocal portal]

https://www.wolframalpha.com/input?i=(1%2F2)(1%2B0.171573)^2+-+2(0.171573)

[potent solstice]

Isn't the exponent for 2 ^ 2?

For the zeroth iteration this is the correct answer, but does that mean the condition always True after 1?

[potent solstice]

I'm just assuming that the y_0 is y_0 and not y_theta.

Otherwise it makes no sense to me haha

[potent solstice]

I think you just helped me solve this thank you.

[potent solstice]

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[south leaf]

How to solve this syllogism problem?

[buoyant shadow]

this is an exercise you just think

[south leaf]

There are two possible venn diagrams.

[buoyant shadow]

it's like if it said "do 6 pushups" and you asked "how do you do 6 without giving up"

smh

sorry for the outburst

[south leaf]

It's okay. Can you help me out?

[buoyant shadow]

okay

[south leaf]

According to the first venn diagram, conlusion I is wrong and conclusion II is right.
And according to the second venn diagram, both conclusions are right.

[buoyant shadow]

there's probably like 5 possible diagrams

[south leaf]

Really?

[buoyant shadow]

well they could be completely nested inside each other

so like II doesn;t follow

[south leaf]

Aaha yes!

[buoyant shadow]

i think this is also valid

"some" doesn't actually preclude "all"

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@south leaf Has your question been resolved?

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[void agate]

It is given f(x) = 3x - 5

Find the value of t, given that f(t) = 10.

Woulb it be x is now t?

[safe radish]

Please don't occupy multiple help channels.

[thin bridge]

to get f(whatever)
replace all x in the original equation with whatever

[void agate]

Oh I see

So it wuld be 3(10) - 5?

[thin bridge]

no

equations don't become expressions
and also not what you want

[void agate]

3t - 5 = 10

?

[thin bridge]

yes

[void agate]

Ohhh I see

[thin bridge]

you were told info about f(t)
so it'd make sense to replace all x with t
to apply that

[void agate]

I see

[thin bridge]

do not conflate $f(a) = c$ with $f(c)$ \
I've seen you do that multiple times now

[flat frigate]

ℝamonov

[void agate]

How about for this one

f(x) = 4x - 7, find

The value of p when f(p) = p

Are they the same concept?

[thin bridge]

yes

same concept

[void agate]

So it would be p = 4p - 7 right?

[thin bridge]

yes

[void agate]

Oohhh man

Thank you so much

Ive been trying to figure out hkw to do thid

Thanks for the help and your time

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[gritty pier]

$x^2+y^2-xy+x+y$

[flat frigate]

rainy

[gritty pier]

Find the absolute min and max of the following function over x<=0 ; y<=0 ; x+y>=-3

Okay, so I determined the critical point -1, -1, and that gives -1 as smallest value

But what about x<=0 ; y<=0 ; x+y>=-3

Dont rly get how I should calculate this

[safe radish]

@gritty pier Has your question been resolved?

[crimson field]

Have you graphed the region described by the three inequalities?

This theorem basically says either the extrema occur on the boundary curve or in the interior but if they do not occur on the boundary they occur at the critical points. You need determine the possible values of f(x, y) along the entire boundary curve.

@gritty pier

You need to determine the possible values of f(x, y) on these boundary curves.

[gritty pier]

Nope I havent, but its not needed i think, we learned it without graphing anything

[crimson field]

You can sketch it and you should do.

[gritty pier]

Hold on

[crimson field]

I only use Desmos to graph it so you have a neat illustration of what I'm suggesting.

[gritty pier]

Could you explain how you interpreted the given domain, and sketched it?

How did you go from x<=0 ; y<=0 ; x+y>=-3 to the graph ^

[crimson field]

For x <= 0 we have the entire half plane to the left of the red line;
for y <= 0 we have the entire half plane below the blue line; and
x + y > = -3 or equivalently y >= -3 - x we have all the points above the green line.
The only region which satisfy all three inequalities at the same time is the region bounded by the three lines.

To illustrate, Desmos has shaded for x + y >= -3 only.

We just have to restrict it further according to the remaining two conditions.

[gritty pier]

Thx

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[white swallow]

I got half of the answer?

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@white swallow Has your question been resolved?

[white swallow]

Nvm I think I got it

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[grizzled shoal]

We are looking for local and global extremes. The script says "f:D->R, x is an inner point and f is differential in x. Then x is a local extreme if f'(x)=0.".
Then it says "the condition that x is an inner point is significant because f:[0,1]->R, x->x has local extremes at 0 and 1 but f'(0)=f'(1)=1.".

[grizzled shoal]

I just need help understanding this a bit

It sounds like it is needed that x is an inner point

But they also say that the points on the "edge" of the interval are local extremes

[final halo]

The point here is that global extremes happen either at the boundary, or at points where f'(x)=0

[grizzled shoal]

So they mean that it is a sufficient statement but not a neccesary one?

That makes sense

The wording just felt weird

Ty

I got ir 👍

[final halo]

Sufficient but not necessary yep

[grizzled shoal]

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[midnight patio]

I'm not sure where I went wrong here?

[granite idol]

just thinking aloud here. should it be 9-choose-2, and 7-choose-3, and 4-choose-4?

[midnight patio]

Actually you need the 10

[buoyant shadow]

i don;t know what you;re both doing

[granite idol]

right, I didn't mean just those terms

[buoyant shadow]

10c1 × 9c2 × 7c3

[midnight patio]

yeah

Also do you know what I did wrong here?

[buoyant shadow]

i don;t know how to solve it 🤔

[granite idol]

this one is tougher...

[buoyant shadow]

i have to go

[merry pelican]

I haven't done anything similar to this before, but my first intuition would be to split it up into two cases: numbers less than 300000 and numbers greater than or equal to 300000. Reasoning being that the upper bound for some digits (after the first) changes depening on if the first digit is 3 or less than 3.

[midnight patio]

that was what I was ding

[merry pelican]

For numbers greater than or equal to 300000, how many different numbers would you say there can be?

[midnight patio]

wait but I finding it equakl to the number

[midnight patio]

there are 7*10^3

[merry pelican]

What is your reasoning for that

[midnight patio]

So I consider how many numbers there are above or equal to 3

and then I multiply it with how many numbers I can pick from 0 to 10 from the other digits

[merry pelican]

We can try going over them one digit at a time. Assuming the first digit is 3, since we're looking at cases of the number greater than or equal to 300000, how many ways to pick the second digit?

[midnight patio]

10

[midnight patio]

So I did that, but not sure why it not work again

@merry pelican what did i do wrong?

[granite idol]

what if you count all the 1xxxxx, 2xxxxx then figure out the 3xxxxx. not sure if that's any easier

[midnight patio]

can you please see what I did wrong?

I don't think there is anything wrong with this?

[granite idol]

those first two should be 10 x 9^5 I think

[midnight patio]

what first two?

[granite idol]

oops, actually all numbers starting w/ 1 should be 9^5, right? then the ones starting w/ 2 should be 9^5 as well. so that's 2 x 9^5

[midnight patio]

This si apparantly the solution

but I don't quite understand it

zohj I get it now thanks

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[crimson steppe]

i picked c but they say like the answer it is wrong

[green bane]

ok isolate y on 1 side

[crimson steppe]

what

yah

i checked that

the slope is negative

so either c or a remkains

@green bane

[granite idol]

each grid line is 2 units, right?

actually it doesn't matter

unless you want to plug in the numbers

[modest karma]

can you input the points?

that's the easiest method if it is single correct

[crimson steppe]

hmm

like the value of x

[modest karma]

the points

of graphs in the equation

to see if it satisfies

[crimson steppe]

both are satisfied

[granite idol]

what is the point shown on graph C

[crimson steppe]

0,1

[granite idol]

no, it is (0, 2)

and what is the other point?

[crimson steppe]

no it is 0,1

where it is 0,2

[granite idol]

look at the scale, the axes go to 10 and there are 5 lines

each line means you go up/down/left/right by 2

[granite idol]

yes, that is (0, 2)

[crimson steppe]

my goodness

so tricky question

a is correct

[granite idol]

yes

[crimson steppe]

i have to check both values

[modest karma]

its not actually

its quite easy and direct

just find out the points from the graphs and check by putting in the eqn

[crimson steppe]

2 values in order to get perfect

this question too

i picked b

a the slope is positive

the perpendicular line slope should be m1*m2=-1

[green bane]

look at the y intercept

[crimson steppe]

so it must be negative the slope of l

and y intercept is positive therefore it is

b

someone reply

[granite idol]

what is the slope of line b

[crimson steppe]

negative

[granite idol]

are you sure? get it into y = mx + b form

[crimson steppe]

yes

sure

see the

figure

it on positive x axis

means

its slope positive

[safe radish]

@crimson steppe Has your question been resolved?

[safe radish]

@crimson steppe Has your question been resolved?

[safe radish]

@crimson steppe Has your question been resolved?

[safe radish]

@crimson steppe Has your question been resolved?

[cold kettle]

hey sanji

answer is C, right?

@crimson steppe

[cold kettle]

this questipn

[safe radish]

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[quiet citrus]

Trying to solve using elimination method

[green bane]

is that the one where you get rid of 1 varbile

[quiet citrus]

yes

[green bane]

what have you tried

[quiet citrus]

no idea how to get rid of the x

[green bane]

hint

get rid of y

[quiet citrus]

i got 9 for x

[green bane]

so just

solve

it

[random star]

well you get the y from one of the equations and substitute it in. that way you'll have an equation with a single variable (x)

[quiet citrus]

ok i got the right answer

.close

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[lean otter]

i am confused about the general process of finding the sum of a series

[neat kiln]

What about it ?

[lean otter]

if you are asked to find the sum of an alternating series, will it always be using the estimation theorem?

[safe radish]

@lean otter Has your question been resolved?

[safe radish]

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[keen birch]

help

[keen birch]

do i find the primitive of $x^2lnx$

[obtuse plover]

U tryna differentiate?

[flat frigate]

RMS

[obtuse plover]

Use product rule

What does primitive mean

[keen birch]

antiderivative

[keen birch]

stephen brother... cmon man

[obtuse plover]

?

[obtuse plover]

How is this image relevant if u wanna find antiderivative of x^2 lnx

Is it related?

[lean otter]

u can integrate the right side

or differentiate the left side

but i thing integrating is easier

[obtuse plover]

Ohh lol i was like what

[lean otter]

just integrate by parts

[keen birch]

me too

[lean otter]

and its done

[keen birch]

aight imma try that

[obtuse plover]

U rlly think ibp easier than 2 step product rule?

Maybe just straight tabular but it’s defo more work thru ibp

[keen birch]

rq question, how do i find the antiderivative of the right hand side if i wanted to do so (i know that finding the derivative of the left hand side is easier but im just asking hypothetically if i wanted to)

[obtuse plover]

He alrdy said this

Or u can set it up thru tabular which is basically the same

[lean otter]

integrate

[lean otter]

integrate by parts

[keen birch]

by parts?

[obtuse plover]

U shd probably know antiderivative of lnx by heart

[lean otter]

did u learn integration

[keen birch]

i know integration

[lean otter]

so there is a rule

for these functions

its called integration by parts

[keen birch]

what is it called ill just search it up

nevermind

[lean otter]

so we have 2xlnx+x

so we need to integrate 2xlnx by parts

and x is an easy integral

[keen birch]

of course, it is just: $x^2/2$

[flat frigate]

RMS

[keen birch]

but the 2xlnx is the difficult part

[lean otter]

yes this one by parts

we assume u=ln(X) and dv=2x

and apply the rule

[keen birch]

rule?

hold on lemme search it up

[lean otter]

u.v-Integrate[v.du]

[keen birch]

u mean this

[lean otter]

غثس

yes

this is the formula

use it assuming u=lnx and dv=2x

[keen birch]

u mean نعم

[lean otter]

yes

[keen birch]

the problem is i do not know how to do this

( the second part) of the second side btw

[lean otter]

well look

now out integral is

this

[keen birch]

this is the issue

[lean otter]

when we assume u=lnx and dv=2x we get this

[granite idol]

your original question just calls for using the product rule of differentiation...

[lean otter]

yea but he asking how to integrate it

[keen birch]

this isnt about the orginial question

[lean otter]

so now u have to apply the formula

we have u.v

we get v from integrating dv

dv is 2x so v=x^2.dx

now x^2lnx

the first part

second part is

• integral of x^2 which is v .du

[granite idol]

Technically dv = 2x dx

[lean otter]

du=1/x.dx

[lean otter]

yeah right

mb

or u can write dx/x

so

its the -integral of x^2.dx/x

therefore -integral of x

and u get (-x^2)/2

[keen birch]

so essentially it is: $2xlnx - x^2lnx$

[flat frigate]

RMS

[lean otter]

nono there is no ln on other side

how do u right that stuff lemee show

[keen birch]

yes there is so what is v?

[lean otter]

lemme write on a paper and send it

[flat frigate]

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

[lean otter]

1 sec

check it

[keen birch]

thank you brother 👏

[safe radish]

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[normal cedar]

hey, can someone explain to me why sqrt(68) is less simple than 2sqrt(17)?

[devout shale]

because 68 has a factor of "4" which has a perfect square root of 2

68=4*17

so $\sqrt{68}=\sqrt{4*17}=\sqrt{4}\sqrt{17}=2\sqrt{17}$

[flat frigate]

AustinU

[normal cedar]

i don't really understand

why does this make it simpler?

[hallow moss]

the smaller the number in sqrt = simpler

[devout shale]

because the number inside the square root is smaller

2 is exact

when we approximate square root of 68

[normal cedar]

ohh

[devout shale]

we get a less exact answer than approximating the square root of 17 and multiplying by 2

since we were able to pull out the exact factor of 2 that we know is in there

we don't muck that up by leaving it in and approximating

[normal cedar]

alrighty

i understand

thank you both

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[fresh tree]

anyone know this?

[safe radish]

@fresh tree Has your question been resolved?

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[fresh tree]

3

[tardy mango]

Send what you've done so far (in general, it's good practice to send whatever you've done so far so that it offers a jumping off point - worst comes to worst, you start from square one)

[fresh tree]

Printing Area = (w - 2 × 2) × (h - 2 × 4)

(w - 4) × (h - 8) = 72
wh - 8w - 4h + 32 = 72

wh = 4w + 8h - 40

Total Area = w × (4w + 8h - 40) / (h - 8)

d(Total Area) / dw = (4h - 32) / (h - 8) - 4w / (h - 8)^2 = 0

4h - 32 - 4hw / (h - 8) = 0

4hw = (h - 8) × 32

w = 8

8h - 32 = 9h - 40

h = 8

[tardy mango]

Total Area = w × (4w + 8h - 40) / (h - 8)

I'm not sure this is correct

b/c (4w + 8h - 40) / (h - 8) should be height

you have that 4w+8h-40=wh

but wh/(h-8) isn't h

[fresh tree]

Total Area = (w + 4) × (h + 8) = (w + 4) × (80 / (w - 4))
d(Total Area) / dw = 80 / (w - 4)^2 - (w + 4) × 80 / (w - 4)^2 = 0

but i still get 8

[tardy mango]

But it seems like you defined w and h as the dimensions of the original paper before taking away the margins

[fresh tree]

oh Width = x + 4 = 24 in
Height = y + 8 = 12 in

nope

i have 1 more chance

[safe radish]

@fresh tree Has your question been resolved?

[safe radish]

@fresh tree Has your question been resolved?

[safe radish]

@fresh tree Has your question been resolved?

[safe radish]

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[cobalt falcon]

Hello Im doing a related rates problem for calc 1 and im having trouble identifying what information is givin to me?

[cobalt falcon]

Sorry this is better so you don’t have to tile your head

[safe radish]

@cobalt falcon Has your question been resolved?

[cobalt falcon]

<@&286206848099549185> Sorry no rush just letting yall know i still need help

[safe radish]

@cobalt falcon Has your question been resolved?

[safe radish]

@cobalt falcon Has your question been resolved?

[safe radish]

@cobalt falcon Has your question been resolved?

[floral osprey]

is x the angle

find an equation involving the angle and y
(think trig)
then take deriv with respect to t
and rearrange to solve

[safe radish]

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[true estuary]

why would the standard dev in this question be 4?

[quasi bison]

where does the graph of y = e^(-x^2/2) have inflection points?

[true estuary]

x=31

[quasi bison]

read the question i ask you

[true estuary]

there is a point of inflection at x = 31

[quasi bison]

read the question i ask you

i'm not asking where your bell curve has inflection points.

[true estuary]

(0,1)?

[quasi bison]

what kind of points am i asking about?

[true estuary]

inflection points

oh

would it be when

the curve touches or approaches the x axis

?

[quasi bison]

[places two slices of bread to your ears]

[gordon ramsay voice] what are you?

[true estuary]

im so baffled rn

[misty bay]

try to answer the questions without using big words and somewhat related words to make a salad

[quasi bison]

https://tenor.com/view/gordon-ramsay-idiot-sandwich-angry-mad-what-are-you-gif-4169547

@true estuary wouldve been better to admit you dont know what an inflection point is.

an inflection point is a point where y'' = 0

for the graph of y = e^(-x^2/2), of which any bell curve can be obtained via horizontal translations and horizontal/vertical scalings, the inflection points happen at x = ±1.

for the general normal distribution curve they happen at 1 standard deviation from the mean in both directions

[safe radish]

@true estuary Has your question been resolved?

[true estuary]

the second derivative, that tells you whether a point is max min or point of inf

so if u let the second derivative equal to 0, u could solve for x and that would find u the points of inflection?

[quasi bison]

this is just barely comprehensible.

[true estuary]

.close

[safe radish]

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[gray remnant]

Hi, I have a bearings question?

[green bane]

[gray remnant]

I could swear that I'm doing everything right, but my maths homework platform (Sparx) is telling me I'm wrong. It's in video form, is that OK?

[green bane]

hmmm ok..?

[lean otter]

again

[gray remnant]

oh, the video is too long to send, ill send pics

one sec sorry

Sorry for messy working out

But I've followed every step of the video and still it's telling me I'm wrong

[lean otter]

.heic? Seriously?

[gray remnant]

?

[lean otter]

Just a weird format

Windows is taking forever to open the frickin' file explorer ajsdklajdkasjd

[gray remnant]

oh sorry, I just ss it

I guess macs are annoying like that

[lean otter]

plz work

sh*t

[gray remnant]

ah my bad

ill find another way hang on

btw, it asked for the angle rounded to the nearest whole integer

so its not a rounding problem

[lean otter]

,w calc atan(52/33)

[flat frigate]

Results provided by WolframAlpha

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

Requested by Labyrinth#3979

[lean otter]

Ah shoot it was just ,calc

,calc atan(52/33)*180/pi

[flat frigate]

Result:

57.600159826081

[lean otter]

I...?

@gray remnant Sorry for the late reply, how'd you get 64°?

[green bane]

rad?

[gray remnant]

is this for me?

[green bane]

yea

[gray remnant]

cuz dude idk maths language

whats rad

[green bane]

radian?

or degrees your calculator

[lean otter]

This can only be degrees

Radians wouldn't go up to 64

[gray remnant]

fr

[green bane]

ok so

thats the reason it is wrong

[gray remnant]

i dont understand

[green bane]

not the same

[gray remnant]

oh

er

[green bane]

i wonder why

[gray remnant]

mate idk i drop my calculator quite a lot

[lean otter]

Because the second one's in degrees?

[green bane]

o

[lean otter]

I'm not CS expert but I can tell you this is either does nothing, breaks the screen, or breaks the calculator entirely

[green bane]

nvm

[lean otter]

No way for it to start giving slightly wrong answers

[gray remnant]

i would hope so

Dw guys. Thank you for your help anyway, Sparx is a god awful website so, maybe, they've just formatted the question wrong or something of the sorts. I've emailed my maths teacher (not that she'll be any help) but it means I probably wont get a detention for not completing it.

[lean otter]

"they've just formatted the question wrong or something" because your calculator is giving wrong answers?

,calc atan(52/33)*200/pi

[flat frigate]

Result:

64.000177584534

[lean otter]

Oh no way

Your calculator is in grads

[gray remnant]

yooo

wtf is that

i mean my calc is spanish

it could be that

[lean otter]

It being Spanish is unrelated

[gray remnant]

they specialise in making this difficult

haha ok

[lean otter]

The units for angles are the same worldwide

[gray remnant]

right

so how can i convert my calculator back to degrees

[lean otter]

Try to find the settings

[gray remnant]

ok

[lean otter]

It looks like a Casio, is it?

[gray remnant]

yeah

[lean otter]

Have you found the settings? It should be straightforward

[gray remnant]

I think so

'Unidad angular' means angle units so it hsould be in here

ok I have 3 options, an o, a g, and an r

[lean otter]

I'm guessing r means radians and g grads

[gray remnant]

i think so

[lean otter]

why

[gray remnant]

wait no

yeah

that

i was gonna say that

[lean otter]

Does it say what mode it is on right now?

WHY

[gray remnant]

hahaha

[lean otter]

Okay at least this makes sense

[gray remnant]

ok

[lean otter]

So the o is probably referencing the ° symbol

[lean otter]

Ah yes

[gray remnant]

ooh smart

[lean otter]

It's not an o

[gray remnant]

ah

ok my bad

ok lemme try now

tysm for your help

[lean otter]

@gray remnant Pro tip: .close

[gray remnant]

thanks

[lean otter]

(as in type that into the channel)

[gray remnant]

.close

[safe radish]

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[cunning moon]

can someone explain what P(2x-1) is, i forgot what that was and how one can solve it

[thin bridge]

P(2x-1) is P(2x-1)

the above equation is an equation showing you the relation between multiply things

to get the desired result
compare the form of

P(2x-1) - P(x)
to
P(3) - P(2)

and deduce that it would be helpful if there exists and that you find
the value of x where
2x-1 = 3 AND x = 2

and substitute that into the original equation

[cunning moon]

got it thanks

[peak estuary]

P(2x-1) means that you sub in 2x-1 everywhere.

if you had an expression for P

which you dont have here

[cunning moon]

ye, that confused me a lil

.close

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[night pendant]

"in a class of 25, the average mass of the school bags was 4,6 kg.

John came in late. After measuring his bag's mass, the class average bag-weight increased to 4,7 kg.

How heavy was john's bag?"

[steep pewter]

lets assume after john came all 25 are present

[night pendant]

yes

[steep pewter]

the weight of 24 bags was 4.6 times 24

right?

[night pendant]

Maybe...?

well it says "average" weight for the bags but i dont see how else you could calculate this

[steep pewter]

you know the mean formula right/

?

[night pendant]

I was absent today

i dont

[steep pewter]

suumation/number = average

so total weight divided by number of bags = average weight of a bag

[night pendant]

ah

[steep pewter]

now you can also find the weight of those 25 bags

same method

i leave john's bag up to you

[safe radish]

@night pendant Has your question been resolved?

[safe radish]

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[magic meteor]

can someone help with this

[peak estuary]

!status

[safe radish]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[magic meteor]

2

i've shown iu is a homomorphism for part a, not sure how to show bijectiveness

lost with b and d

i have c done

[peak estuary]

for bijective you could try to find the inverse map

what do you have to show in b

[magic meteor]

uhh

identity, closure

inverse

identity is obviously i_e

is closure just like

$i_u \circ i_v ( x) = i_{uv}(x)$

[flat frigate]

ball

[magic meteor]

wait ok nvm that makes sense

inverse too

still lost on d though

[peak estuary]

Z is cyclic

where can you send generators

[magic meteor]

wdym send

[peak estuary]

if phi is some automorphism, what can phi(g) be for some generator

"where can phi send g"

[magic meteor]

to g ?

so identity is an automorphism

uhh

[peak estuary]

one of them, yes

[magic meteor]

0 messes a lot of stuff up right

[peak estuary]

what are the generators of Z

[magic meteor]

1

and -1 ?

[peak estuary]

yes

[magic meteor]

so identity and n-> -n

[peak estuary]

yes

[magic meteor]

how do i notate that though

[peak estuary]

if you know what phi(1) is then you know all of phi because it's a homomorphism

[magic meteor]

$\text{Aut}(\mathbb{Z}) = {??}$

[flat frigate]

ball

[peak estuary]

for example you could write $\operatorname{Aut}(\bZ) = {x\mapsto x, x\mapsto -x}$

[flat frigate]

Denascite

[peak estuary]

or you give each of these a name

id and -id would be classic names

[magic meteor]

ok ok

would Inn be the same thing ?

[peak estuary]

well first, what is U(Z)

[magic meteor]

1

[peak estuary]

if anything it should be a set

and you are missing something

[magic meteor]

uhh

-1? but how

[peak estuary]

is there an integer that when multiplied with -1 gives 1?

[magic meteor]

oh

lol

thats how it works

then how does Inn work

[peak estuary]

well what is the map x->uxu^-1 for u=1 ?

what about for u=-1?

[magic meteor]

both identity

oh so they're the same

so Inn = {x mapsto x}

[peak estuary]

yes

[magic meteor]

ok tysm man

[safe radish]

@magic meteor Has your question been resolved?

[safe radish]

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[keen mica]

Im confused about 17 and am just stuck

[keen mica]

Calculus 1 is where this is from

[brave wolf]

whats the upper bound of second integral?

[keen mica]

2

And the other is 3 above the x

[brave wolf]

ohh

so the other is $\int_{0}^{x^{3}}3x^{2}f\left(\frac{x^{8}}{4}\right)dx$

[flat frigate]

MathIsAlwaysRight

[keen mica]

No 2 and x^3

Int for both Is 2 and 0

[split terrace]

like this?: $\int_{0}^{2}3x^{2}f\left(\frac{x^{3}}{4}\right)dx$

[flat frigate]

Winus

[keen mica]

Yep

[brave wolf]

I dont quite get what you've done in your work. But consider u-substitution. U=x^3/4

[keen mica]

Ok I was honestly just making stuff up in the work section I got really turned around

[brave wolf]

so when you do the substitution, what do you get?

[keen mica]

3x^2/4

[brave wolf]

Do you know how to do substitution?

[brave wolf]

That's du/dx, right?

[keen mica]

Yes

[brave wolf]

okay, now you will need to express dx using that

du/dx=3x^2/4
dx/du=4/3x^2
and find dx

[keen mica]

Is it 24x^2/4

[brave wolf]

what is this?

[keen mica]

Nvm I emailed teacher

.close

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[high wharf]

is anyone free

[safe radish]

Please don't occupy multiple help channels.

[high wharf]

@loud loom whats process here

[safe radish]

@high wharf Has your question been resolved?

[high wharf]

<@&286206848099549185>

[safe radish]

@high wharf Has your question been resolved?

[safe radish]

@high wharf Has your question been resolved?

[safe radish]

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[paper phoenix]

can someone explain this

[green bane]

which part

[paper phoenix]

in the botton section he put the sum of the formula applied to the sequience to find the general term

[green bane]

yes...?

[paper phoenix]

2+(n-1) = 2n-1? no?

[green bane]

no

[paper phoenix]

i feel like its basic stuff iam forgeting

[green bane]

2n-2

[paper phoenix]

wouldnt it be -1 because of the formula?

constant in the formula

[green bane]

the 2 distributes

[paper phoenix]

ah i see, iam gonna practice that a bit, should be fine, thanks

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[austere ravine]

Can someone help me with this inequality?

[fluid token]

!status

[safe radish]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[austere ravine]

4

got an answer but not rlly sure

[fluid token]

!show

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Show your work, and if possible, explain where you are stuck.

[austere ravine]

apologies for the messy work

I basically took √2x-1 as y

and applied y throughout the inequality

and towards the end I used the wavy curve method

I got x as (1, 5) in the end

but turns out values from 1/2 to 1 work as well

so not sure what I am missing

any help would be appreciated

(also, I can provide a better pic, if this one isn't readable)

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[tall hare]

double angle problem

not sure what i did wrong

im also not sure how to figure out if cos should be negative or positive

i tried both ways and both are wrong though, so it's not that

i see my problem

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[tall hare]

.reopen

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✅

[tall hare]

actually still wanna know how to know if cos would be negative or positive in this case

should be either 4sqrt5 or -4sqrt5 i believe

<@&286206848099549185>

the answer is negative, just need to know why cos is positive

<@&286206848099549185>

<@&286206848099549185>

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[tall hare]

.clsoe

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[proven burrow]

im sorta confused on this question ^

in terms of the setup, I couldn't figure out which specific part was wrong and so I pursued calculating it by my own methods. I got 8640, which is 4x the result for the given method

I lined up the steps I made, and realised out the issue is somewhere in 2) and 3), and those steps correspond to the point where i calculate all the possible combinations of the dice. as the dice are distinguishable, this is 5! = 120

i suspect that it has something to do with the pairs in step 3 but i dont know what specifically to add to fix it :/

<@&286206848099549185>

<@&286206848099549185> :((

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[proven burrow]

<@&286206848099549185> smh

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[wind shoal]

did i do a mistake here or no

?

[obtuse plover]

U good

👍

[wind shoal]

👍

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[jovial flicker]

Hello, I need help with this geometry question. AD is a height/altitude of triangle ABC, and it is expanded until the circle, and the intersection point is E. H is the orthocenter of the triangle. I need help proving that HD=DE

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.reopen

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✅

[jovial flicker]

<@&286206848099549185>

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[stray axle]

how can i find the interval of convergence here? is it just -1<2x+4<1 which is -5/3<x<3/2

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@stray axle Has your question been resolved?

[obsidian oracle]

you can find the radius of convergence by finding $\frac{1}{R} = \lim_{n\rightarrow \infty} \frac{|a_{n+1}|}{|a_n|}$, here you didn't use $a_n$ but $a_nx^n$

[flat frigate]

rafilou2003

[obsidian oracle]

or $a_n(x+1)^n$ to be more precise

[flat frigate]

rafilou2003

[stray axle]

wait wdym i used Anx^n. didnt i use (x+1)^n

also noticed a mistake, interval should be -3/2<x<-1/2

[obsidian oracle]

yes

[stray axle]

alright, also it converges to 0 as well right? because just divide by N and you get 0/2

[obsidian oracle]

?

[stray axle]

not the interval but the limit of (2n ln(n+1))/((2n+2)ln(n+2))

[obsidian oracle]

that limit is 1

[stray axle]

why?

[obsidian oracle]

2n equivalent to 2n+2

ln(n+1) equivalent to ln(n+2)

[stray axle]

huh

infinity sure is weird, how would i write that if it was asking what it converges to?

just do a comparision test or limit comparison or can you just say by observation

[obsidian oracle]

for n :

$\frac{2n}{2n+2} = \frac{1}{1+\frac{1}{n}}$

[flat frigate]

rafilou2003

[stray axle]

alright yea

makes sense

[obsidian oracle]

for ln :

$ln(n+1) = ln(n) + ln(1+\frac{1}{n})$

[flat frigate]

rafilou2003

[obsidian oracle]

so $\frac{ln(n+1)}{ln(n)} = 1 + \frac{ln(1+\frac{1}{n})}{ln(n)}$

[flat frigate]

rafilou2003

[stray axle]

huh, i didnt know that one

thats a cool identity

thank you!

[obsidian oracle]

no problem

[stray axle]

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[faint seal]

\begin{align}
y&=f(x)\nonumber\
f^{-1}(y)&=x\nonumber\
\frac{d}{dx}(f^{-1}(y))&=1\nonumber\
\frac{d}{dy}(f^{-1}(y))\cdot y'&=1\nonumber\
\frac{d}{dy}(f^{-1}(y))&=\frac1{y'}\nonumber\
\end{align}
Say, y = sin(x) then if I'm not wrong doesn't this say that
$$\frac{d}{dx}(\sin^{-1}(y)) = \frac1{cos(x)}$$

[flat frigate]

kheerii

[faint seal]

also the original statement that I derived itself doesn't make sense to me, since f^-1(y) should be x

so I've just proven that dx/dy = 1/(dy/dx)

I'm trying to prove/formulate a solution for the derivative of an inverse function

[obtuse jackal]

(f^-1)' = 1 / (f' o f^-1)

dy/dx = 1 / (dx/dy)

[faint seal]

ye

but that's not what I wanted to prove

maybe I can take x = f(f^-1(x)) and differentiate that

[obtuse jackal]

cos x = sqrt(1-sin²(x)) = sqrt(1-y²)

then you get the usual formula

[faint seal]

ahh you're right

but again, isn't sin^-1(y) just x

also that should be a d/dy on the bottom line mb

okay it makes sense to me now, thanks

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[obtuse jackal]

sometimes you just need to write it out

[faint seal]

yep

thanks

[obtuse jackal]

I say that a lot

14 times in 6 months