help-17

[dim sinew]

somewhat

more than cubes

[swift bane]

So we have

3(16x^6 - 25)

[dim sinew]

so a^3 = 16x^6

[swift bane]

Uhh no

We can't take a number

raise it to 3rd power

and get 16

[dim sinew]

wait

then I don't know

[swift bane]

4 squared is 16 right

[dim sinew]

yes

[swift bane]

and x^3 squared is x^6

[dim sinew]

yeas

yes

[swift bane]

So we can plug it in the difference of squares formula

16x^6 - 25

You can try plugging it in

find what a and b is first

[dim sinew]

I don't know how to find a or b if we have that 3 multiplying the equation

[swift bane]

Don't let it confuse you

We're just representing $48x^6 - 75$ in a different way

[twin meteor]

asm

[dim sinew]

right

[swift bane]

We can just do the work inside of the brackets

[dim sinew]

but then a^3 and a would change because we divided the number by 3?

like

it cant equal the same thing once we divide it by 3, we have to change the power

[swift bane]

Uh no

we factored it out

If u distribute that 3 you'll get the original expression

It doesn't affect the powers

[dim sinew]

oh

so then we are still looking for a^3

[swift bane]

Not anymore

It's difference of squares

$16x^6 - 25$

[twin meteor]

asm

[dim sinew]

isnt that in the parenthesis

[swift bane]

Yea but

I took it out

[dim sinew]

okay i get that

[swift bane]

So that you can see the a^2 - b^2 better

[dim sinew]

okay let me paint a scenario because its the only way i see it

[swift bane]

sure

[dim sinew]

If I see ** 48x^6 - 75** , and **16x^6 - 25 ** I'm going to think they need the same type of solving because they generally look the same

but you're saying that it changed

[swift bane]

Changed as in

we simplified it

3 * 16 = 48 ;
25 * 3 = 75

[dim sinew]

I know but

[swift bane]

Because I don't see where to go with the original expression

They're not cubes so

[dim sinew]

that means simplifying the expression changes something

in a

[swift bane]

Well

Makes it easier to see

That's the purpose of simplifying

[dim sinew]

okay ill just try to roll with that

so now that we have 16x^6 - 25 after we know that we simplified

[swift bane]

We simplified it to the point where it can be put into a formula

16 and 25 are perfect squares

[dim sinew]

oh you didn't point that out

[swift bane]

Yea

[dim sinew]

i was just confused why we were suddenly using difference of squares

i get that now

[swift bane]

We simplified it down in order to get the perfect squares

48x^6 - 75

Don't see where to go with that

[dim sinew]

right

simplifying makes sense

we are definitley supposed to do that

since we got lucky with that factoring find

[swift bane]

Yeah, I don't think they would give you an expression where it is unfactorable

[dim sinew]

so now I have to find the expression that can be multiplied by itself to equal 16x^6 - 25 ?

[swift bane]

Uh

$a^2 - b^2$

[twin meteor]

asm

[swift bane]

where $a = 4^2 \cdot x^3$

[twin meteor]

asm

[swift bane]

Yo u agree with that right

[dim sinew]

4^2 * x^3 = 4x^6 so yes

[swift bane]

Same goes for 25

b = 5

now you just use the formula

Don't forget that 3 outside the brackets tho

[dim sinew]

yeah

[swift bane]

sure

[dim sinew]

how would we get a negative number by squaring

-25

[swift bane]

We ignore that part 😁

that's how the formula goes

$a^2 - b^2 = (a + b)(a - b)$

[twin meteor]

asm

[dim sinew]

wait

lets talk about b

[swift bane]

Mhm

[dim sinew]

its just 5

[swift bane]

b^2 = 25; b = 5

yeah we ignore the minus

Power of the formula

[dim sinew]

okay

?

[swift bane]

Remove 4^2 btw

[dim sinew]

oh right

[swift bane]

a^2 = 4^2
not
a = 4^2

[swift bane]

Yea u gotta put

4

[dim sinew]

wait no ur right

[swift bane]

$a^2 - b^2 = (a + b)(a - b)$

$(4x^3)^2 - 5^2 = ....$

[twin meteor]

asm

[dim sinew]

so how come we mention 4^2 if in the equation and formula its never used

[swift bane]

What we're essentially trying to do is show that 16x^6 and 25 can be put in the formula
a^2 - b^2

a^2 is indeed equal to 16x^6

[dim sinew]

man this is tough

[swift bane]

We're just trying to find a so that it looks "intuitive" I guess

You don't have to find a and b

[dim sinew]

thanks for the help tho

[swift bane]

16 - 25 = (4 + 5)(4 - 5)

You can just straight up square root it

[dim sinew]

-9 = -9

yes

[swift bane]

Yea

(4 + 5)(4 - 5) is another way of representing -9

[dim sinew]

bruh wtf is this

[swift bane]

5^3 = 125 right

[dim sinew]

do i gotta make it a = 3x^3

[swift bane]

Yeah

3 * 3 = 9

U gotta put it in cubes formula

[dim sinew]

b = 125 not -125 right

[swift bane]

No it's never that

[dim sinew]

its never what

[swift bane]

we're dealing with a formula so

[dim sinew]

b^3

[swift bane]

We just don't care

b is just 125

rather than - 125

[dim sinew]

doesnt b^3 = 125

[swift bane]

yea

[dim sinew]

ok ok

alright if i get this right in formula first try im probably good to close it

or somewhat first try

is there a different formula for tihs one

difference of perfect cubes

[swift bane]

There's only difference of squares

U have these for cubes

[dim sinew]

does it matter which one

[swift bane]

I GUESS it is called difference of cubes

Not really sure how you call them in english

[swift bane]

Yeah of course

[dim sinew]

what language is your native?

[swift bane]

Georgian

[dim sinew]

how do i know

[swift bane]

So are you asking

Which one to use?

[dim sinew]

Yeah for my problem

[swift bane]

a^2 - b^2 or a^3 - b^3

Well

[dim sinew]

both work

[swift bane]

is 27 a square of something

[dim sinew]

no

[swift bane]

So it doesn't work

[dim sinew]

so cube

[swift bane]

I think you could say something like

$(sqrt(27) + sqrt(125))(sqrt(27 - sqrt(125)))$

[twin meteor]

asm

[dim sinew]

well yeah that'd be the same i guess

[swift bane]

yeah

[dim sinew]

just ugly numbers if you tried to simplify

[swift bane]

We don't see any squares

only cubes

[dim sinew]

okay

okay so

i have perfect cubes with a - in my expression

which means

[swift bane]

What do you mean

[dim sinew]

like whats the nexts tep

[swift bane]

its this one right

[dim sinew]

yes

[swift bane]

Well now u just plug them in the formula

[dim sinew]

well you showed 2 of them

[swift bane]

So your problem has a -

we need the difference one

[dim sinew]

yea

[swift bane]

you can't apply +

[dim sinew]

so whats the formula called for this one? or if u could post it

[dim sinew]

the bottom

[swift bane]

yea

[dim sinew]

okay

you know ive never even heard of your language lol

[swift bane]

Yeah, americans ironically call it a state

[dim sinew]

😭

[swift bane]

I mean my country, not my language

[dim sinew]

mhm

wait so im stuck on this little part again

the 9x

to what power

[swift bane]

a^2 right

you have your a right here

[dim sinew]

when 3x^3 with the ^2

yeah just the power im stuck on

[dim sinew]

it's 5 or 6

[swift bane]

$3^2 \cdot (x^3)^2$

They ge tmultiplied

[twin meteor]

asm

[swift bane]

They never get added

[dim sinew]

so 9x^6

[swift bane]

yea

this is the only time they get added

you can save these properties later on

[dim sinew]

alright sweet i got it

[swift bane]

Are you doing these from khan academy?

[dim sinew]

No from my teacher

[swift bane]

Oh

[dim sinew]

10th grade algebra II

[swift bane]

If you're really interested into math, I would recommend getting a book

Like an algebra book

[dim sinew]

potentially

i feel like these type of problems im doing have lots of area for mistake when you're first learning them

in comparison to other units we've learned

[swift bane]

Self-study a lot, you might not learn a lot from school (talking from personal experience)

[dim sinew]

man i gotta shower in 4 hours for school

[dim sinew]

i feel that

[swift bane]

Yeah it's natural

What you wanna do is

Self-study and experiment on your own

Messing around is a good way of learning

And of course, refer to the properties

[dim sinew]

I heard a way to get good specifically in math is doing problems quickly

[swift bane]

so that you don't make up your own math

[swift bane]

Well

maybe

Doing problems quickly just trains you at doing problems quickly

You gotta learn how to think about it

[dim sinew]

yeah im sure

alright im probably gonna close this and headout, thanks for your help. maybe ill be back here another day lol

[swift bane]

np 👍

[dim sinew]

.close

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[halcyon oracle]

hi if i wanted to create a transition matrix to model a M/M/3 queue what would it look like?

[halcyon oracle]

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[blissful sequoia]

i decided to assume that bd = df and ec = ef in the beginning but now after finishing the proof how should i prove that bd = df and ec = ef? also my proof kinda feels wrong

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Please don't occupy multiple help channels.

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@blissful sequoia Has your question been resolved?

[blissful sequoia]

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[vital scroll]

Help please, how do I solve this?

[atomic jasper]

!status

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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

[vital scroll]

1

[atomic jasper]

ah i see where you went wrong

[atomic jasper]

that -1 in the exponent is over the t, not the 125

$8t+125t^{-1}=8t+\frac{125}{t}$

[twin meteor]

Bonk

[vital scroll]

so where would it go when i factorise ?

if i need to factorise-

[atomic jasper]

probably easiest to multiply by t

[vocal sleet]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[atomic jasper]

$8t+\frac{125}{t}=t\left(8+\frac{125}{t^2}\right)$

[twin meteor]

Bonk

[vital scroll]

I did tho-

[vital scroll]

OHH okay

[vital scroll]

Why t squared tho if there's already a t outside the bracket

[atomic jasper]

im sure you know $a^ba^c=a^{b+c}$

[twin meteor]

Bonk

[twin horizon]

$t \cdot \frac{1}{t^2} = \frac {1}{t}$

[atomic jasper]

so $t^1t^{-2}=t^{1-2}=t^{-1}$

[twin meteor]

Bonk

[vital scroll]

yes

[vital scroll]

OHH

[atomic jasper]

that is whats happening here

[twin meteor]

David

[vital scroll]

wait but do we want t^-1 or t^-2

Cos I think it's t^-2 after being differentiated

[atomic jasper]

well, we had t^-1 before, and we took another t out, so we get t^-2

$8t+\frac{125}{t}=t\cdot 8+t\cdot\frac{125}{t^2}=t\left(8+\frac{125}{t^2}\right)$

[twin meteor]

Bonk

[atomic jasper]

make sense?

[vital scroll]

yes

brb

.close

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[atomic jasper]

oh.

alright

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[tawny nacelle]

I don't understand how to solve part 4 of this question

[tawny nacelle]

the solution I found in part 3 for initial condition $x(0) = x_0$ is $x = (t + x_0^{\frac{2}{3}})^{\frac{3}{2}}$

[twin meteor]

fallenstars

[tawny nacelle]

in the case where $x(0) = 0$, this reduces to $x = t^{\frac{3}{2}}$, right?

[twin meteor]

fallenstars

[tawny nacelle]

I'm confused as to how I'd find more solutions with this initial condition

I tried multiplying t^3/2 by a constant k, but then x no longer solves the ODE

I'll take any suggestions

[quiet echo]

wait

is it alright if i solve the ODE

[tawny nacelle]

go ahead

[quiet echo]

dx/dt=3/2x^(1/3), x^(-1/3)dx=3/2 dt, 3/2x^(2/3)=3/2t+C, x^(2/3)=t+2C/3, x=(t+2C/3)^(3/2)

odd

what does the sketch look like?

[atomic jasper]

,w x'=3/2*x^(1/3)

[quiet echo]

i feel like there can be some intuition places from there

[twin meteor]

Results provided by WolframAlpha

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

Requested by .bonk__

[quiet echo]

[tawny nacelle]

slope field looks like this

I'm not actually sure how to read the initial condition information using this though

[atomic jasper]

the line that goes through (0,0)

is IC

[tawny nacelle]

why is that the case?

[oak magnet]

Desmos art

[atomic jasper]

because x(0) = 0

[atomic jasper]

at t=0, x=0

[quiet echo]

yeah my intuition has been activated

something tells me that the majority of these slopes pass through

[tawny nacelle]

I can tell that all the slopes pass through (0, 0)

but not sure how to find them

[tawny nacelle]

not all

but the ones above the x axis

[tawny nacelle]

also, I still don't follow

[quiet echo]

that suffices as infinitely many

right?

[tawny nacelle]

I should find em though

[atomic jasper]

x(t_0)=x_0 is IC

at t_0, x(t_0) is x_0

given is that t_0=x_0=0

[tawny nacelle]

but the axes are not t and x

[atomic jasper]

but im kinda confused how there are infinitely many solutions

[tawny nacelle]

that's why I'm confused

the axes are x and x'

[atomic jasper]

mby it has something to do with the exponent?

[tawny nacelle]

I don't even understand the initial condition thing yet

[tawny nacelle]

but if the exponent is changed, the function is surely not a solution

[atomic jasper]

yeah, thats what i figured out

there are at least two solutions

x(t)=0 and x(t)=t^(3/2)

right?

[tawny nacelle]

x = 0 and x = t^(3/2) work

yeah

[atomic jasper]

if you find another nontrivial solution you can construct infinitely many

[versed bane]

not true

the ODE isn't linear

[atomic jasper]

okay nvm

[atomic jasper]

im trying to think of how you can find a third solution

[versed bane]

how are you constructing a slope field for the diffeq if the axes aren't x and t

[atomic jasper]

x and x'

[versed bane]

i know

that that's what the axes are

but what does the slope mean

[tawny nacelle]

my professor told us that the a curve that follows the slope lines is a solution, but he didn't say what the slopes at each pt mean

[versed bane]

like when you're representing the slope like that, the visualization implicitly assumes that the horizontal axis is the indep. variable

[tawny nacelle]

so I'm confused

math libre says this

but x seems to be taking the role of t here?

[versed bane]

yes but then you have to represent your axes right, you want dx'/dx to be the slopes if you're representing the slope field like that in the image that you sent

idk

[tawny nacelle]

now I'm really confused, because I had assumed that it was dx/dt

on the y axis

[versed bane]

what are the slopes here

like if one axis is x' then that's dx/dt

and your slopes should be constant on one axis

is the horizontal axis x'

?

[tawny nacelle]

I feel pretty sure that the horizontal axis is x

[versed bane]

actually this doesn't make sense anyways, because the space of (x,x') that is allowed in this diffeq looks like x' = 3/2 x^(1/3)

so what is the slope field

like you plotted this, what did you write down to be the slope at each point?

[tawny nacelle]

I actually used an online plotter

but I tried it by hand first

[versed bane]

what did you tell it to render the slopes as

[versed bane]

see but that isn't right

that's x' = 3/2 t^(1/3)

[tawny nacelle]

I see

okay that makes a lot more sense

[versed bane]

if you're plotting x' and x, then you have some curve which represents the allowed states

it'll look like y = 3/2 x^(1/3)

[tawny nacelle]

like this, yeah?

uhh, I had to use x for t

[versed bane]

what is the slope grid

[tawny nacelle]

I have no idea

[versed bane]

let me pull up desmos 1 sec

[tawny nacelle]

okie

[versed bane]

how do i make a slope grid

[tawny nacelle]

you're asking me?

here's the one I'm using https://www.desmos.com/calculator/p7vd3cdmei

[versed bane]

okay so

[tawny nacelle]

so the confusion stemmed from the fact that x was implicitly a function of t earlier, but the calc assumed it was independent

[versed bane]

i bet that you can, for each r in R, construct a solution such that x''(0) = r

[tawny nacelle]

wait, why do we want to consider the 2nd derivative?

[versed bane]

here's a proper slope grid

[tawny nacelle]

okay looks like mine now

[versed bane]

well, i wonder if it'll help

i mean, let's just try to compute derivatives until we get some degree of freedolm

[tawny nacelle]

can you send me what you have actually?

like, how you generated the slope field

[versed bane]

g(x,y) = 3/2 y^(1/3)

[tawny nacelle]

I don't think I understand

[versed bane]

idk, i'm just throwing out ideas tbh

[atomic jasper]

how tf do you find the 3rd solution

it might give a hint

[tawny nacelle]

heck if I know

[versed bane]

if you have x = 0 for t < 0 and x = t^3/2 for t geq 0 then that's a solution right

the only thing that you'd have to prove is differentiability at 0

call that solution x = f(t)

then x = f(t-t_0) is also going to be a solution for all positive t_0

i think that's the intention

because the differential equation is invariant under horizontal shift

so the only remaining condition is that x'(0) = 0

but that holds for the new solutions

the solution isn't analytic

[tawny nacelle]

wait, it does?

[versed bane]

of course

[tawny nacelle]

x(0) = f(0 - t_0) = f(-t_0) = 0 for any t_0 in R?

[versed bane]

t_0 has to be positive

[tawny nacelle]

okay sure, but even then

[versed bane]

then that's good

x will be identically 0 in a neighborhood of 0 then

so x'(0) = 0

and at t=t_0 you switch to the other solution

[tawny nacelle]

the condition isn't x'(0) = 0, it's x(0) = 0

[versed bane]

yeah

okay

[tawny nacelle]

but this works for that too

maybe

[versed bane]

then x(0) is also 0

you need x'(0) to be 0 from the diffeq

so on the left end you get the diffeq holding because x is 0 in a neighborhood of t

and on the right you get the diffeq holding from separation of variables

and then at t_0 you need to prove that the righthand limit as h goes to 0 of (f(t+h) - f(t) )/ h is 0

in order to guarantee that 1. f is differentiable at t = t_0 and 2. the differential equation holds at t = t_0

[tawny nacelle]

icic

I think I got it then

thank you so much for the help: smay, exiled, Flappie!

.solved

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[vast shale]

can someone be my permanent friend and help me with math problems

[twin horizon]

Not how it works
#❓how-to-get-help

[vast shale]

if that's not outside the rules

[dense cedar]

Bro what 🙂

[vast shale]

sorry

[coral chasm]

You can just send your help here and eventually someone will help you

[vast shale]

ok sorry

I am new here

[coral chasm]

Its okay

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@opal trout Has your question been resolved?

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[balmy furnace]

Hey, I'm just curious what's wrong with my Excel formula here? I'm trying to analyze a large dataset and I don't see what the issue is with what I have.
=COUNTIFS(B2:B12429, U2, (C2:C12429, E2:E12429, G2:G12429, I2:I12429, K2:K12429), "DesiredOutput")
The logic here is that I want it to count the number of lines where the cell value in the B range matches the cell value in U2 AND where any of the corresponding cells in the other ranges are "DesiredOutput".

So if B2 is "Yes" and "U2" is "Yes" and then any of C2, E2, G2, I2, or K2 are "DesiredOutput" it should add to the count.

[vocal sleet]

@balmy furnace Has your question been resolved?

[flat whale]

Excel isn't really popular here. Wolfram and desmos are

Try asking in an excel discord server

There's one called excel

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[violet bloom]

Hello, can someone check my work for me. I boxed the answers and the way I got them is all written on the respective questions. Thank you I really appreciate it!

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@violet bloom Has your question been resolved?

[quiet echo]

that’s right

[violet bloom]

thank you!

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[twin meteor]

ten

need some help with how to do this question, new to laplace.

Q: Find i1(t), i2(t), i3(t) for:

$$ \begin{align*} L_1 \frac{di_1}{dt} + R_1 i_2 &= E_1(t) \\ L_2 \frac{di_3}{dt} - R_1 i_2 + R_2 i_3 &= 0 \\ i_1 - i_2 - i_3 &= 0 \end{align*} $$

given the values L1 = 0.005, L2 = 0.01, R1 = 10, R2 = 20, E(t) = 50

I have done this

$$
\begin{align*}
  0.005 s I_1(s) + 10 I_2(s) &= \frac{50}{s} \\
  0.01 s I_3(s) - 10 I_2(s) + 20 I_3(s) &= 0 \\
  I_1(s) - I_2(s) - I_3(s) &= 0
\end{align*}
$$

no clue where to go next
```Compilation error:```! Package amsmath Error: Erroneous nesting of equation structures;
(amsmath)                trying to recover with `aligned'.

See the amsmath package documentation for explanation.
Type  H <return>  for immediate help.
 ...                                              
                                                  
l.53 ... &= 0 \\ i_1 - i_2 - i_3 &= 0 \end{align*}
                                                   $$
Try typing  <return>  to proceed.
If that doesn't work, type  X <return>  to quit.```

[flat whale]

you have a system of equations

Solve for I1 or I2 and plug into second equation.

repeat until you just get one function

yes and then you have a 3rd equation to solve for one of the functions

[twin meteor]

ten

[ionic willow]

im not sure what id do after that then

.close

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[ionic willow]

.reopen

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✅

[flat whale]

you should have only one variable I1, I2, or I3 by the time you substitute into your 3rd equation

https://en.neurochispas.com/algebra/systems-of-equations-3x3-examples-with-answers/

once you get I3, you use the inverse laplace transform to get i3

[twin meteor]

ten

[flat whale]

yea probably partial fractions

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[normal heron]

Hi i managed to solve a and b pretty easily but c i cant seem to grt can someone tell me what am doing wrong

[normal heron]

Q14 btw

[quiet echo]

You can use the sine rule on angle ABD and DBC, find their measurements, add them together, then use cosine rule on the segment AC.

[normal heron]

I did this but the answer i got was wrong

The book says the answer is 5.15

[quiet echo]

Huh.

[hollow kite]

Use cosine rule

for the angles

$(BD)^2 = (BA)^2 + (AD)^2 - 2 * BA * AD * cosA$

[twin meteor]

alex <3

[normal heron]

I did i just put it straight jnto calc

[quiet echo]

I...don't think ADC would be 106.7 degrees.

Check your work again.

[normal heron]

Yessir

[hollow kite]

I got 136.332 degrees for DAB though.

Idk I used an online calculator and different triangle that was similar so it may be wrong, was too lazy to do it on my own

But hold on

[normal heron]

Am i like

Stupid or am i just

[hollow kite]

you're using the sine rule

you can't use sine rule if you don't know any angles in the triangle

[normal heron]

Let me try with cosine rule and ill be back

[hollow kite]

in the question only sides are given

[normal heron]

But we do know angles no?

[hollow kite]

no

look at your question

[normal heron]

The first part of the question

[hollow kite]

you only have the sides

[normal heron]

Is to figure out angles DAB and BCD

[hollow kite]

yes

[normal heron]

Why cant i use those in sine ruie?

[hollow kite]

not the same triangle

[normal heron]

Okok

Ty

Ill try it with cosine rule then

[hollow kite]

Angle DAB is in the DAB triangle, with no known angles, only known sides -> Cosine rule
Angle DCB is in the DCB triangle, not in the same DAB triangle, there's again no known angles in the triangle, only sides -> Once again cosine rule

Since angle DAB is the angle at point A.

and angle DCB is the angle at point C.

[normal heron]

Ahh

Okay tyvm

[hollow kite]

Np

Let me know when you're finished w the calculations

[normal heron]

@hollow kite i got it as 5.15

Thank you very much

So when am dealing with multiple triangles i should use cosine rule?

[hollow kite]

you can use the cosine rule when you are given two sides and an angle (but in the form of SAS) or when you are given three sides and want to find an angle. However, for the sine rule, you can use it when you have either two angles and a side (in the form of ASA) or two sides and a non-included angle (SSA)

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[marsh nexus]

how do i rewrite this

as powers

i odnt know how to do the 6/rootx

[sharp lynx]

recall $\frac 1a = a^{-1}$

[twin meteor]

Steakanator

[sharp lynx]

or perhaps more generally $\frac 1{a^b} = a^{-b}$

[twin meteor]

Steakanator

[marsh nexus]

so would it be x^-6/2?

[sharp lynx]

not quite

perhaps it would help if you (mentally) rewrote it as $6 \cdot \frac 1{\sqrt x}$

[twin meteor]

Steakanator

[marsh nexus]

so if it was 1/rootx

its x^-1/2?

[sharp lynx]

https://tenor.com/view/cat-nod-catnod-yes-vibe-gif-21529329

[marsh nexus]

so 6x^-1/2?

[sharp lynx]

https://tenor.com/view/cat-nod-catnod-yes-vibe-gif-21529329

[placid cradle]

👍

[marsh nexus]

thank you

so would i write the whole thing as x^1/2 - 6x^-1/2 - 1 = 0

[sharp lynx]

that works

[marsh nexus]

as it tells me to do this

[sharp lynx]

fair enough

in that case there's a little more work to be done since you don't have a quadratic yet

[marsh nexus]

then i just factorise right?

wait no

sorry im not sure what to do

[pale perch]

how could you get rid of the pesky x^(-1/2)

[marsh nexus]

oh do i times by x^1/2

[sharp lynx]

https://tenor.com/view/cat-nod-catnod-yes-vibe-gif-21529329

[marsh nexus]

does this seem right

[pale perch]

one of your 'solutions' was x^1/2=-2, do you think this one is valid?

[marsh nexus]

is it incorrect

because i squared both sides to get rid of the power

[pale perch]

what values can sqrt(x) be?

[marsh nexus]

• • ?

[pale perch]

wdym by that

[marsh nexus]

like +-sqrtx

[pale perch]

im speaking solely about sqrt(x)

[marsh nexus]

oh -2

and 3

[pale perch]

i mean generally

what values can sqrt(x) be

,w graph y=sqrt(x)

[twin meteor]

Results provided by WolframAlpha

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

Requested by aldrnari_

[marsh nexus]

0 to 1

[pale perch]

no thats just a snippet of it

sqrt(x)>=0, so sqrt(x)=-2 isnt a solution

[marsh nexus]

ah okay

so it cant be -2

what do i do to change it

[pale perch]

you dont

its just not a solution

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[fading wasp]

Could someone tell me if my proof is sound here? The theroem im trying to prove is : Let n>=2 be an integer. Then the largest integer that is guaranteed to divide n^5 - n is d = 30.

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Please post images (such as PNGs or JPGs) of the question rather than other filetypes such as PDFs which have to be downloaded. Non-image downloads can potentially contain viruses or other security risks.

[vocal sleet]

@fading wasp Has your question been resolved?

[quiet echo]

it works out well enough

[fading wasp]

do you see anything that i may be overlooking? @quiet echo

[quiet echo]

not rly

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[shrewd stone]

on what intervals is the function cube root x increasing?

[shrewd stone]

becase i thought it was always increasing but does x=0 not count as increasing?

[glad trail]

Strictly speaking, it's increasing everywhere but x=0

[tawny nacelle]

what does it mean for a function to be increasing at a single point?

any notion of monotonicity (increasing or decreasing) needs to consider pairs of points

[shrewd stone]

yeah but cant a function be increasing within a range

[coral chasm]

What do you mean by that

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[eager drift]

You can prove that 30 divides $n^5-n$ much more simply by factorising:

$n^5-n = n\left(n^4-1\right) = n\left(n^2+1\right)\left(n^2-1\right) = n(n-1)(n+1)\left(n^2+1\right)$

$n(n-1)(n+1)$ must include an even number and a multiple of 3, and so is divisible by 6.

You now only need to establish that when there is NOT a multiple of 5 amongst $n$, $n-1$, and $n+1$, that $n^2+1 \equiv 0$ (mod 5).

[vocal sleet]

Please don't occupy multiple help channels.

[twin meteor]

Jay

[eager drift]

This is advice for @rron on an earlier question.

Sadly, the bot is grumpy, so...

.close

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[dim ivy]

$$\text{Calculate the period of a wave with frequency }$$
$$2 * 10^5 Hz$$

[twin meteor]

Dev

[dim ivy]

How do i calculate this

thats all there is to the question

[solar cape]

The period is the reciprocal of the frequency

[dim ivy]

so $\frac{1}{2 * 10^5Hz}$ ?

[twin meteor]

Dev

[lilac pebble]

hz is 1/s

1/second

[dim ivy]

oh

[solar cape]

$\frac{1}{2 * 10^5}s$

[twin meteor]

denzio321

[dim ivy]

ah ok

[solar cape]

s being seconds

[dim ivy]

next?

[solar cape]

Yea that's it

[dim ivy]

oh....

Thats the period?

[lilac pebble]

you can tidy the expression if you wish

[solar cape]

The period is the time it takes for one full osscilation to occur

[dim ivy]

Right.....

[lilac pebble]

$\frac{1}{2\cdot 10^5}=5\cdot 10^{-6}$

[twin meteor]

Axe

[solar cape]

What's with the elipses 😭

[solar cape]

Do you have any doubts?

[dim ivy]

nope

i thought there was more to it

and forgot which formula to use

And that bascially is it

Thanks!

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[pliant violet]

Hi can someone do this and explain it

[grizzled vault]

can you not find the minimum and the maximum sum and then count the number of possible sums in between the two?

they’ll differ by a multiple of 3

[mental egret]

Problem reduces to finding number of different sums that can be formed using a triplet from the set {1, 2, ..., 675} (Why?)

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[fading wasp]

Ahhh thats so much simpler. Thanks for the advice. I appreciate it.

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[woeful igloo]

dot product is a projection of one vector on the other

So, in this case, u.w means projection of w on u (or the other way around)

And so, you can see, because they are anti parallel, the dot product is not gonna be positive

(This has nothing to do with whether the vectors are end to start or whatever since you can always move them parallely to arrange them however you want)

[vocal sleet]

@vast shale Has your question been resolved?

[woeful igloo]

yes

You can see it to be antiparallel

I took projection of w on u, so its way easier to see within the given diagram

But, since the dot product is commutative, u.w is same as w.u, so you can do the projection of w on u or u on w, whichever is convenient for you

This is projetion of u on w

u on v is trivial, you can do it easily

[woeful igloo]

Arrows point away from each other in this case, so the dot product would be less than 0

Yep

Well, simpler way is, you dropped a perpendicular from the arrowhead, so you put the arrowhead on the projection

[woeful igloo]

Here, I dropped the perpendicular from the tail end, so I put the tail there

Can you show the other ways to put the arrows?

I am not sure what you mean

If you show the other ways I would tell you why they are wrong or arent

Ok, so you drop the perpendicular (black line) from the 'head' of u

so, the projection must also have the 'head' at the base of the perpendicular

so the arrow goes on the lower part, so this is correct

This one is wrong coz the arrow is only at one end of the perpendicular

Well, you also gotta consider the tail of u

Consider this situation.

Generally, you draw the perpendicular from both ends of u

But in your problem one end of u is already on v, so you dont need the second perpendicular

[woeful igloo]

If you still consider drawing both the perpendiculars, you would notice that the perpendicular from arrow end of u gets the arrow on projection and the non-arrow end doesnt get the arrow on projection

u1 and u2 are same, but you have u1 in your figure, so the second perpendicular is not considered relevant. Its "obvious" how the second perpendicular would be

Uhh, the direction is along the arrow?

I am sorry, i dont get what you mean

Projection of u has its direction, and v has its own direction

If they are same, the dot product is positive, if they are not, the dot product is negative

See, I drop a perpendicular from the arrow end of u

So, the end of the projection there gets the arrow

Like this

And if I draw the other perpendicular, from the non arrow end, because it was from non-arrow end, it wont get the arrow

So I end up with this

Did they not cover that in highschool? I see you got undergrad maths role. Thats why I thought youd know

np

So, same idea here. Non-arrow end gets no arrow

The arrow end gets the arrow, non arrow end gets no arrow

yea

You know basic trig right?

You make a right triangle so that you can make "components" which are trig ratios

u, red projection and blue makea right triangle

so the projection of u on v (red vector) is cos theta times u

thats why the dot product u.v is |u| |v| cos theta

npnp

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[gusty heron]

In here why to find J or H = A + B - C?

[gusty heron]

I mean to count J or H, I could just do (4 - 1) + (13 -1) right

Wich remove the double card

Why must 4 + 13 - 1

[edgy gulch]

you removed the jack and hearts card twice

[gusty heron]

I mean both the 4 and 13 have 1 J and H right, so I just - 1 both

Eh

[austere goblet]

P(J or H) = P(J) + P(H) - P(J and H)

[gusty heron]

Yeah right

[edgy gulch]

the 4 jacks already contain one of the 13 hearts

so you just remove one of the double counts

[gusty heron]

So 4 + 13 - 1

[gusty heron]

Can you give me one extra example?

[edgy gulch]

if you count the jacks, you get jack spade, jack clubs , jack hearts and jack diamond

if you count the 13 hearts, you get 1 heart 2 heart ... until king hearts

but inside the hearts you count again the jack hearts

[gusty heron]

Yeah that mean I counting twice the jack and heart card

On the both set

[edgy gulch]

yes

so you remove the double counted card

[gusty heron]

Ok....

I count twice

Ohhh

I see

Yeah that makes sense

Thanks

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[vast shale]

I need help with identifying who this is

Its a Prominent mathematician

Do we known who he is?

Also

I put this without the 1 13/28 mixed number

I just put

41/28

would the answer still of been correct?

My answer

I put the 41/28 and I didn't convert into a mixed number

Would I still of been correct with my answer?

[outer warren]

yes

depending on what grade you're in, your teacher may want you to convert to mixed numbers
but as you get higher, improper fractions are usually preferred

[vast shale]

And who is this?

[oak magnet]

You can search it up on google

With google image

[vast shale]

Shall we go to DM?

Me and you?

[oak magnet]

No

[vast shale]

Btw im not in a Grade, I am in England. In the UK. I am doing a Mathematics Level 2 Functional Skills Maths course and this is the same as GCSE Level, so I was using this website for revision

I will be doing GCSE maths next year

Adult GCSE Maths

How to pass this course?

Can u tell me how?

@oak magnet

What do i do?

[vast shale]

What do i do?

[outer warren]

ask your teacher whether they want mixed or improper fractions

[vast shale]

I mean how to make sure I passes the course?

Functional skills level 2 Maths

[outer warren]

practice

[vast shale]

Are their proper resources for revision? How do I find them?

[outer warren]

just look online

past papers are good

[vast shale]

I did a Past paper and I only got 13 out of 64 Marks, when my teacher marked it. He said the pass mark would be around 36

What do I do?

[outer warren]

review the questions you got wrong/had difficulty with

[oak magnet]

Ramonov said all

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@vast shale Has your question been resolved?

[vocal sleet]

@vast shale Has your question been resolved?

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[lone temple]

Anyone can help me with part c

[vast shale]

What about it

[lone temple]

Ap cal

[vast shale]

oh part c ?

[lone temple]

I dont get part c

Yea

I'm getting 128 but the answer is 128/3

[strong grove]

show working?

[lone temple]

8 to 4 is the same

X²-16x+64

The other half thats blocked

[strong grove]

shouldnt the integral be more like this?

[lone temple]

Oh lord bless you I have been stuck on this question for a while

[strong grove]

u understand tho? or is it ok

[lone temple]

Wait

Why x^2

[strong grove]

this the area to find

[lone temple]

Ohhh I see

[strong grove]

split it like this and u get 2 integrals

[exotic ravine]

bro just integrate x^2 from 0 to 8 and the line from 4 to 8
this would be easier to calculate

[open kraken]

How to find Differential Equation

[lone temple]

Hmm, I get it now

[strong grove]

"easier" to calculate?

both are same things

[lone temple]

Yea

[exotic ravine]

nah i mean if ur using pen and paper what i said is quicker i believe

the result is the same obv

[strong grove]

ah ok , yeah u can say that

both ways work , but they are using a calculator xD

[exotic ravine]

cuz x^3/ 3 would be in both integrals

[exotic ravine]

yeah shit i didnt see 😭

[lone temple]

Yessir

[open kraken]

How to find differential Equation by Separable method

[strong grove]

also by "same things" i mean like really same things

[lone temple]

School is bit easy on us

[exotic ravine]

apple tesfaye please cant wait no more for da album my goat 🙏

[strong grove]

u can split integral of x^2 from 0 to 8 into integral of x^2 from 0 to 4 and integral of x^2 from 4 to 8

[exotic ravine]

yeah they are lol

[lone temple]

12 more days brother

[strong grove]

combine that integral of x^2 from 4 to 8 with the integral of tangent line and boom

[exotic ravine]

exactly

[open kraken]

Bro read my Question as well

[strong grove]

!help

[vocal sleet]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[exotic ravine]

they done got my goat to do calc b4 the album 😭

[lone temple]

Yessir dropped of school at 17 it’s hard life out here uk what I mean

Atleast I’m getting my APs done

[exotic ravine]

shit sad af

[lone temple]

B4 the album

[vocal sleet]

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[left finch]

.close

[vocal sleet]

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[frosty prawn]

Hello. What is the easiest method for creating Negative or positively definite matrices? It seems like a really long process.

[vocal sleet]

Please don't occupy multiple help channels.

[frosty prawn]

Figured it out

.close

[vocal sleet]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[rugged jetty]

what's the best way to approximate this integral numerically (where n is a real number except negative integers)

[smoky iron]

doesn't the result change on changing the value of n?

[mild flower]

yes, the result will be in terms of n

[rugged jetty]

yes, the approximation formula has to be infact in terms of n

[rugged jetty]

excactly

[mild flower]

because S² + C² = 1 it seems like the denom should simplify a lot

[rugged jetty]

if n=1/2

what if it's not, how can i get the best numerical approximation from it in terms of n?

[mild flower]

oh a real number I misread

[rugged jetty]

so...?

[hard atlas]

well for each fixed n you can numerically integrate it

giving you some data point

I suppose you could try to interpolate those results

[rugged jetty]

yes even my calculatore can

but i wanted kinda of a closed form approximation of it

[dense cedar]

You can substitute tanx = z, that gives you int 0 to infinity ()
Idk if this helps ! But maybe after that binomial expansion or some sort of approximation

[rugged jetty]

lemme try it

it now looks something like this

[quiet echo]

oh god

[vocal sleet]

@rugged jetty Has your question been resolved?

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[onyx spade]

how do i do this

[onyx spade]

using IBP

[quiet echo]

you're not gonna believe it but integrate 3

[onyx spade]

oh

[steady plover]

Integrate 3 and differentiate sin inverse

[midnight cloak]

yeah in inverse trig integrals always try to differentiate the inverse trig function in IBP

[quiet echo]

i've already said this 👍

[steady plover]

mb

[onyx spade]

but whats antiderivative of arcsin

@quiet echo @steady plover

[quiet echo]

integrate 3

differentiate arcsin

this problem teaches you the antiderivative

[steady plover]

you don’t need it

[quiet echo]

so thats a circular question to ask

[steady plover]

I hope you know the derivative

[vocal sleet]

@onyx spade Has your question been resolved?

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[snow bluff]

need help with finding the riemann sum for this

[atomic jasper]

!status

[vocal sleet]

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

[hallow crescent]

seems like you already inputted a value, do you want to check the answers?

[vocal sleet]

@snow bluff Has your question been resolved?

[snow bluff]

no it’s not right

I need help finding the Riemann sum for both the left-end point and right-end point but I didn't get the right answer

[atomic jasper]

!show

[vocal sleet]

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

[snow bluff]

give me a second

nvm

[vocal sleet]

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[slow gale]

Why is this not point of inflection if concavity clearly changes

[vocal sleet]

@slow gale Has your question been resolved?

[vocal sleet]

@slow gale Has your question been resolved?

[flat whale]

Who says it's not

[quiet echo]

[slow gale]

😂

The website

It’s tweaking right

[flat whale]

Show what the website says

[slow gale]

It was on my friends laptop

Oh well

Guess it was tweakin

Thanks for confirming guys

[slow gale]

Btw love that emoji

.close

[vocal sleet]

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[quiet echo]

its from the other math server 😄

[slow gale]

[vocal sleet]

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[marsh nexus]

ive done a)

[mint badge]

dyk how to complete the square

[marsh nexus]

yes

[mint badge]

so what part

do u not know how to do

wait it says otherwise

u don't even need to

just differentiate both sides

A' = 1/4 (3 - 2x)

let A' = 0

x = 3/2

idk if u have to prove max but if u do

A'' = -2

[mint badge]

therefore max for all values of x

[marsh nexus]

i meant to write a -1/4

at the end

so i know x is 1.5

[mint badge]

ye

[marsh nexus]

but is the max value

-12.25?

[mint badge]

no

[marsh nexus]

or do i have to multiply tat by -1/4

[mint badge]

u didn't do that right

[marsh nexus]

oh which bit

[mint badge]

ill show u wait

@marsh nexus

[marsh nexus]

so i shouldve multiplied -12.25 by -0.25

[mint badge]

yes

[marsh nexus]

ah okay

and would 49/16 be my max value