help-23

[quick crater]

what is it asking?

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@loud cedar Has your question been resolved?

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[near moon]

I have part a done but I need help with part b

I do not know how to even start this question

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[near moon]

<@&286206848099549185>

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[abstract rune]

completely lost.

how does 10x + 8 and 5x-4 get to 12x +16??

how do i get units out of that??

[arctic raven]

given is that $LM=10x+8$, $MN=5x-4$, and $LN=12x+16$

[flat frigate]

Flip

[arctic raven]

the fact that L, M, and N are all points on a line restricts what x can be

so in effect you're solving for x in order to determine LN

[valid olive]

LN = LM + MN

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[lean otter]

Is there way to know in which quadrant sin^-1 is lying
I want to find cos using Pythagoras theorem but it gives two values positive and negative

[arctic raven]

arcsin only gives an angle between -pi/2 and pi/2

[valid olive]

you can use the identity tan (a +b)

that will fix that positive and negative value problem of yours

[lean otter]

Oh, Thanks 🙏

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[hollow olive]

Where does the minus one come from

[trim swan]

Are you talking about this?

The second term in the binomial was -sin^2(x), they just separated (-1) from sin^2(x)

@hollow olive

[hollow olive]

Oh that's why then

Thanks

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[somber cape]

I'm confued on the part starting with "Conversly,"

[somber cape]

not understanding what they are trying to say

also what is

not sure what symbol that is

I think ph(omega)

what does that notation mean?

[steep lily]

I suspect that's a curly C

$\mathcal C$

[flat frigate]

Edward II

[steep lily]

Ok default latex font has a boring version but still reasonably sure it's a C from context

$C^n(X)$ is common notation for the set of functions with continuous $n$ th derivative on X

(So C^0 is used for continuous ones, C^1 for continuously differentiable)

[flat frigate]

Edward II

[steep lily]

$\mathscr C$

[flat frigate]

Edward II

[steep lily]

Oh there we go

[arctic raven]

$\mathscr{C}$

ahhh too late

[steep lily]

As for what they're trying to say, they're saying these conditions are sufficient to know a function is complex differentiable from real and imaginary parts separately

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[somber cape]

ah ok I see thank you

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[dense wadi]

after fliiping row 1 and 2

is adding row 3 to row 4 a good idea to get a 0 in the first column of row 4

[sick scarab]

yeah, you can

you can get the 0 in the first column of row 3 as well

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[jade kernel]

Need help with part a and b

[devout shale]

Is this a test

[lean otter]

singaporean math practice paper

idk it looks like it

Anyways

[jade kernel]

No

[jade kernel]

How you know?

[lean otter]

i did almost every schools math paper!

Anyways

ANYWAYS!! Focus

1.2x is your new 100%

[jade kernel]

Your from Singapore bro?

[lean otter]

india

anyways ANYWAYS

[jade kernel]

Yeah

[lean otter]

Sorry so, in 2022

[jade kernel]

So do I have to find 85% of 1.2x?

[lean otter]

sure

[jade kernel]

Huh?

[lean otter]

1.02x

so just find 1.02 x 100%

Which would just be 2% increase from 2020

[jade kernel]

So it’ll be (1.02-x)/x X 100%?

[lean otter]

Take this for example,
In 2020, the company value is $100.
2021, 20% increase.
$120.
2022, 15% decrease from 2021.
$102.
Difference = $102-$100 = $2
therefore you gained $2.

this is a way of visualizing it.

[lean otter]

(Final-Initial) * 100%

[jade kernel]

Is it 2%?

[lean otter]

A 2% increase.

If they ask you to show what's the overall % would be 102%

Yup ur qn asked % increase

[valid olive]

I can't understand why is it 2%

[lean otter]

Why

[valid olive]

120x/100 is the value for 2021 however it reduced 15% again so (120x/100)-(15x/100) = 105x/100. so total percentage increased since 2020 is 1.05%

[lean otter]

Can you really use 15x?

[jade kernel]

Where you get 105x?

[lean otter]

Because it did say it's
15% of 2021.

[valid olive]

Oh

[lean otter]

Meaning that 2021 would be your new 100%.

i get your idea,

What you did was

You took the original as the 100%

[valid olive]

Yeah

[lean otter]

And in 2021, it's just 120%

But the question wanted the decrease from 2021.

If i'm not wrong it means that

How much did the company lose from the valuation of 2021.

[valid olive]

It's 2020 btw

[lean otter]

For 2022.

[valid olive]

Oh so 15 percent was decreased from 2021

Now I get it

[lean otter]

Yes

HAHAHAHA

Sorry i made it confusing.

[valid olive]

No you made it clear

Thanks

[jade kernel]

I got another math qn to ask. Can I ask?

[lean otter]

Are you abel to understand?

ofc

[valid olive]

Yes yes

[lean otter]

:> I Meant for the OP

[valid olive]

Just read the question again and found out

[lean otter]

HAHAHA

[lean otter]

isok This question is v common and its irritating

[jade kernel]

I don’t understand what they’re tryna ask here

[lean otter]

c or d?

[jade kernel]

For part d

[valid olive]

Total time is 5 years

[lean otter]

Sorry i didnt read the qn properly

Is there an initial

Wastage? I see the qn is part d

[valid olive]

And 20% food wastage is being reduced every year

[lean otter]

So notice for this question its similar to the above,

[valid olive]

I think it's based on the pie chart

Total 1000 households

[lean otter]

You can't take the initial and minus a hundred percent, you need to find the weight after each year.

[valid olive]

But what is the initial weight

[lean otter]

^ Precisely

goodluck flex ill leave this qn to you! im gonna be offline alreadys

[safe radish]

@jade kernel Has your question been resolved?

[valid olive]

The amount of food wasted per person in 2019 is 30 Kg.

Food wastage decreased in a year = 1 - 20% = 4/5
Food wastage decreased in 5 years = (4/5)^5 = 1024/3125

Amount of food = 30 x decrease
= 30 (1024/625)
= 6144/625 = 9.8 kg approx.

So David is correct

[jade kernel]

Oh wait sorry maybe y’all might need part a and b information

[valid olive]

This question is a case of exponential decay so you should use this formula in this case

[jade kernel]

Actually the answer key says that David is not correct

Hold on I’ll show you

[valid olive]

If you see the question carefully it is indirectly saying 30 kg is initial weight. that's how I got that answer

[jade kernel]

The top one

[valid olive]

Initial weight 🫠

[jade kernel]

That’s 30kg right?

[valid olive]

I thought so but the answer says something diff

Obv 30 can't reduce to 47

Is there any chance that the answer might be wrong

[exotic cypress]

below 30kg in 2024, not starts at 30kg 🫠

[jade kernel]

Actually the pie chart represent the total kg of food being wasted right?

[exotic cypress]

where is part a and part b

[jade kernel]

Wait

[exotic cypress]

ok there it is

wait no thats the other qu

[jade kernel]

That’s part a and b

[valid olive]

Oh

[exotic cypress]

ok so it's 130kg start and each year u reduce the average amount by 20%

[valid olive]

Should have sent it earlier

[jade kernel]

Oh my bad

[valid olive]

Np

[jade kernel]

So we have to find 20%?

[exotic cypress]

so it goes 130* 0.8 = 104* 0.8 = ... for 5 years

[exotic cypress]

just times the amount by 80%

[jade kernel]

Can you explain?

[exotic cypress]

david wants people to reduce food wastage by 20% each year right?

so u just -20% from the average yearly waste which means u get 80%

and since he is proposing this over 5 years u repeat this process for 4 more times

using that year's average wastage value

[jade kernel]

Is 0.8 80%?

[exotic cypress]

yea

1 is 100%

0.8 is 80%

[valid olive]

Well it is being reduced by 20%

Every year

For 5 years

[valid olive]

@jade kernel your answer

[jade kernel]

👏

🙏

Thanks

I understand now

[valid olive]

Welcome

[jade kernel]

You know this working kinda reminds me of the compound interest formula

[exotic cypress]

it is pretty much it

[valid olive]

All of this percentage thing

[exotic cypress]

like literally same thing

[jade kernel]

So 130 is like the initial value 5 is the number of years

[valid olive]

Like how much the money decreases by time

Math is beautiful isn't it

Interconnecting things from here and there

And making perfect sense

[jade kernel]

Definitely

[jade kernel]

I think we just have to understand the problem tho

[valid olive]

Yes

[jade kernel]

Thanks

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[cyan forge]

I dont know where to begin in question 1)a)

[nocturne phoenix]

See 1/infinty means really close to 0

And in limit we find close values right?

Same for 1/0

[fathom jewel]

no

[nocturne phoenix]

WDYM

[fathom jewel]

f(x) = 1/x

[nocturne phoenix]

Oh i got 1/0 is not close to zero.

[fathom jewel]

from the right side it's +inf from the left -inf

[nocturne phoenix]

I didnt want to say that earlier I wanna say that concept was same

[cyan forge]

yeah

im just confuse how I should frame the answer

like what should I say lol

[nocturne phoenix]

It's just common sense thing. So I think you can explain it in your language anyway

[cyan forge]

ah okay

how would you say it?

[nocturne phoenix]

See If I were to write ans I would write it this way
Yes its true, since greater the divisor less the quotient.
Divisor is infinty that's why Lim x tends to infinity for the function 1/x=0.

[cyan forge]

ah okay thank you so much!

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[jade kernel]

How to do part c(i)?

[tranquil ice]

You have points for A, B, and C and you know ACBD is a rhombus. Try to draw it

[jade kernel]

But 3,1 is quite low tho

[cold sphinx]

a rhombus has all equal sides so u know 3 points need the D point so use distance formula for CD AND BD OR AD and get the 2 equation to solve for D(x,y)

[jade kernel]

How am I gonna find two unknown?

[tranquil ice]

I'd say just draw the rhombus, there's only one way to do it

[cold sphinx]

u know A,B,C coordinates and want D(x,y) 2 variables form 2 eqn and get x and y

[jade kernel]

How do I draw from here

[valid olive]

only one point D is unknown

[tranquil ice]

Line from C to B

[jade kernel]

X and y coordinates

[tranquil ice]

The line AB is not a part of it

[jade kernel]

That makes a triangle

[cold sphinx]

2 variable make 2 equations

and get x&y

[tranquil ice]

Read it carefully. ACBD is the rhombus. So A to C, C to B, B to D and D to A

[cold sphinx]

btw the slope of 1st ques is -1/2 u have written 1/2

y=mx+c

[valid olive]

try CD = BD

[jade kernel]

Oh I think I got it hold on

[tranquil ice]

If you find the point by drawing it, you should also see how the equations work out. It's gonna help your intuition later

[jade kernel]

I got 9,13 as the points

What’s the formula for area of rhombus

[tranquil ice]

Do you know how to calculate the area of a triangle?

Remember how you said that the line AB make a triangle of the bottom half of the rhombus? You could use that to calc the area

[jade kernel]

1/2 x base x height

[tranquil ice]

Yeah

So what would be the are of a rhombus then?

[jade kernel]

1/2 x AC x CB?

[tranquil ice]

Well if AC is the base and CB is the height, then that would be the formula for the area of a triangle as you said

But if you draw the AB line in the picture, you see that the rhombus consists of two triangles

[jade kernel]

So I have to times 2?

[tranquil ice]

That's right! But now to avoid extra work, let's think about it generally

So if you have two identical triangles and you want to know their total area:
The formula for the area of a triangle is 1/2 * b * h. But as you correctly stated, for two of them, you have to multiply by two so you get:
2 * 1/2 * b * h

2 * 1/2 is just 1 so the whole thing reduces to b * h.

And so you've derived the formula for the area of a rhombus by noticing that you can cut in into two triangles and by knowing how to calculate the area of a triangle. Well done!

[jade kernel]

Got it

Thanks for helping bro

[tranquil ice]

np

[jade kernel]

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[forest rapids]

Do i have to times it in?

[cold sphinx]

p=4

take log both sides

and then solve

[forest rapids]

How come p is 4

[cold sphinx]

you know rules of log?

[forest rapids]

Uhhh

No

[cold sphinx]

then its tough for u ig

[forest rapids]

But how come the p is 4

[cold sphinx]

u should know logarithm to solve

wait lemme check again

[forest rapids]

I studied all of them but im dam weak

[cold sphinx]

ya ans should be 4

see the below part after line

log a^b=b log(a)

this rule is apllied everywhere in the answer try to understand

[forest rapids]

So how do you know you need to use log

And not other formula

@cold sphinx

[cold sphinx]

it comes with practice + if we see something raised to the power then it can simplified using log

as i had to find p which is something raised to the power p so brought it down means log will help to get rid of powers

[forest rapids]

Oh wow

[cold sphinx]

yep super useful

[forest rapids]

Do you have any coming exams

[cold sphinx]

yeah clg exams but i dont study maths now

i dont have maths as a subject this sem

[forest rapids]

Ohh

Ok well thx for the help

Im just looking for a math buddy

Rn

[cold sphinx]

u are preparing for which exam?

[forest rapids]

Secondary school uhhh

Idk how to explain

Second to last year

Final

And next year will be my big exam

[cold sphinx]

ohh ok all the best

[forest rapids]

Thx

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[forest rapids]

.reopen

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✅

[forest rapids]

You can do that?

[cold sphinx]

5/6=(6/5)^-1

reciprocal bro

so i just wrote this took down -1 so log 6/5 cancels

[forest rapids]

Woah

[cold sphinx]

i just wrote 5/6 as (6/5)^-1

[forest rapids]

And you take the negative one to the other side?

[cold sphinx]

ya we can take it down cuz log a^b=b log(a)

[forest rapids]

Ohhhh

Ok thx

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[forest prairie]

Doing a statistics investigation for my class right now. I was wondering if there is an easy way to convert a z score to confidence level. I have found a website to do this (https://evolytics.com/resources/calculators/abtesting-zscore-to-confidence/) but I need to do it with a large amount of data, and so it doesn't make sense to keep putting z scores into a random online calculator manually, without any idea how it is generating its results.

[forest prairie]

It seems to work with the common z-score values like 1.96, 1.645, etc. so i do trust it

it's just very inefficient, moving values from excel, to website, then word document,over and over again

there doesn't seem to be any API for linking to the calculator through some quick spaghetti code either

[grizzled fossil]

It's implemented in any stat software. For example in R type
pnorm(z_score) and it does that.

[forest prairie]

any manual calculation steps? may need to include it in my investigation

otherwise all good

[grizzled fossil]

Calculate the integral by hand

[forest prairie]

ah, how fun

no thanks, lol

thanks for your help though! <3

.close

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[gleaming ridge]

I need help with this radical.

[cedar void]

!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 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

@gleaming ridge Has your question been resolved?

[gilded fog]

no i am not able to get the answer

it is like: x+root( -x^2 -4x +12) = -1

this is what i got using (a+b)^2 formula

[dapper star]

you won't get a equation under the root

[safe radish]

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[dapper star]

.reopen

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[uncut star]

How do I complete the square of x^2 - 9x + 7?

[lean otter]

you need sth like x^2 - 2ax + a^2, which becomes (x-a)^2.

[fickle monolith]

do 2(9/2)x

[lean otter]

so 2a = 9. go from there

[uncut star]

that makes sense to me

[fringe oxide]

quadratic polynomial

[uncut star]

that's how I started

then I would have x^2 -(9/2)^2 +7 + (9/2)^2

whoops

how do I continue or did I already go wrong?

wait

[thin bridge]

missing the 9x

[uncut star]

i did go wrong

yes

i see that now

[fringe oxide]

u have to solve x^2 - 9x + 7 = 0 right ?

[uncut star]

I need to take my paper out one sec

[lean sparrow]

theres a set formula for it

if theres a general eqn ax^2+bx+c

[fringe oxide]

man i am doing it go find another serv pls

[lean sparrow]

you add and subtract b^2/4a

[fringe oxide]

don't give him the solution delete ur text

[lean sparrow]

its not the solution its a formula

[fringe oxide]

yes it is

u have to ask him for it

[uncut star]

x^2 - 9x +7

then

[fringe oxide]

so what do u k about quadratic polynom ?

how to solve ?

[uncut star]

(x^2- 9/2 * 2x + (9/2)^2) + 7 - (9/2)^2

Is this right?

Then I go (x - 9/2)^2 + 7 - (9/2)^2

[fringe oxide]

what do u need to solve

is x^2 - 9x + 7?

[uncut star]

I need to factorise so I can find roots

[fringe oxide]

u have to solve it by the factorise form right ?

[lean sparrow]

he need completing the square meth

ig

[uncut star]

yes

that's the name

[fringe oxide]

or do u have to find the roots and after factorise ?

[uncut star]

I need to find the roots after I factorise but I think I can do that on my own

I am having trouble with this expression

[fringe oxide]

then u k how to solve the roots come back when u find them

[lean sparrow]

did u get this?

[uncut star]

No I do not, I haven't seen that before

If there is a general formula to solve these I would like to know it

I have a good memory but suck at logic

[lean sparrow]

see you know the general eqn?

[uncut star]

I don't know the general equation

[lean sparrow]

whats the gen eqn of a quadratic

ax^2+bx+c

[uncut star]

oh

[lean sparrow]

thats the general eqn

[uncut star]

yes I am familiar with the general equation

not the b^2/4a

[lean sparrow]

so now compare the values with ur eqn

and tell whats a b c

[uncut star]

a = 1, b = -9, c = 7

[lean sparrow]

yes correct

now in this eqn

[uncut star]

Yep

[lean sparrow]

what would be b^2/4a

[uncut star]

It would be 81/4

I can't simplify that expression I don't think

[lean sparrow]

wait

now would adding 0 change anything in this eqn>?

[uncut star]

Just an observation, maybe not relevant

Nah

It wouldn't

If I add and subtract something

Nothing changes

I know this

[lean sparrow]

yess

[uncut star]

If I add or subtract

[lean sparrow]

now add and subtract this

[uncut star]

I can rewrite it as (x-a)^2

[uncut star]

Okay let me jot that down

see if I understand

[lean sparrow]

and also dont add the 7 right away

just add and subtract

[uncut star]

but

I did this already

[uncut star]

or is this wrong

I thought I already did this

[thin bridge]

you did all of this already

[lean sparrow]

oh yes

[uncut star]

Okay, I don't know how to continue from there

[lean sparrow]

so now u just equate it to 0

[uncut star]

Okay

[thin bridge]

no need

[lean sparrow]

he wants the roots right?

[thin bridge]

the goal is factorisation first

[uncut star]

Yeah we haven't factorised everything yet

I can do it to a certain point but then I don't understand how to continue

[thin bridge]

from
(x - 9/2)^2 + 7 - (9/2)^2
combine the bolded part into a single fraction

[uncut star]

okay

I will do that

(x-9/2)^2 + 53/4

I think

[thin bridge]

check your signs

[uncut star]

right

sorry i see now

(x-9/2)^2 - 53/4

Sorry I am terribly sloppy

[thin bridge]

yes

and then factor that as a difference of two squares

[uncut star]

sorry english isnt my native language

what is the difference of two squares

is that the conjugate rule?

[thin bridge]

p^2 - q^2 = ?

yeh, its related to conjugates

[uncut star]

(p-q)(p+q)

[gilded fog]

right

letmme try that

[uncut star]

So I assume that 53/4 is the result of a square?

[thin bridge]

Please don't answer questions intended for the op.

[uncut star]

(sqrt(53/4))^2

[thin bridge]

yes

[uncut star]

(x-9/2)^2 - (sqrt(53/4))^2 = ((x-9/2)-(sqrt(53/4))((x-9/2)+(sqrt(53/4))

probably have some unnessecary paranthesis but I didn't feel comfortable writing that without them

[thin bridge]

doesn't hurt to have them

[uncut star]

Now I try to find roots?

[gilded fog]

not possible coz its + not -

try to use a different formula

[uncut star]

I think that formula works

there is a -

[thin bridge]

simplify those a bit first

[uncut star]

Okay I will try to simplify

[gilded fog]

but there is a + sign between p and q

[elfin latch]

Can anyone teach me integration

[thin bridge]

first with simplifying sqrt(53/4)
and then combining the fractions

[uncut star]

(x-9/2)^2 - (sqrt(53/4))^2

anyway I will try to simplify now

[gilded fog]

oh its a different question

[thin bridge]

to make it easier to identify the roots, you'd want to get the factorisation in the form
(x-m)(x-n)

[gilded fog]

sorry misunderstanding

[uncut star]

I am sorry but I am having trouble simplifying this

[thin bridge]

which part?

[uncut star]

The sqrt symbol

sqrt(53/4)

how do I get that to jive with (x-9/2)

Do I square it?

[thin bridge]

no

[uncut star]

nah that seems illegal

yeah

[thin bridge]

first try simplifying the sqrt(53/4)

[gilded fog]

guys i got: x^2-9x-77/2

[thin bridge]

do you know any properties related to square roots?

[gilded fog]

is that right?

[uncut star]

It is unfortunately not correct

Properties of square root, could you remind me of them

[thin bridge]

$\sqrt{\frac mn} = , ?$

[uncut star]

If I do know them I have forgotten them

[flat frigate]

ℝαμOmeganato5

[gilded fog]

r there any MCQs or something

cancel root and square

i guess

[uncut star]

sqrt(m)/sqrt(n)?

[thin bridge]

yes

[uncut star]

okay

I don't see how it helps me yet

oh wait

[thin bridge]

apply that to what you have

[uncut star]

I can turn the denominator

wait ok let me try again

woah

is it just the same as sqrt(53)/2?

WAIT

WRONG

I meant that

[thin bridge]

yes to the edited result

[uncut star]

yes

okay good i will continue

i think i can continue now

(2x - 9 - sqrt(53))/2

[thin bridge]

don't worry about combining the terms with x

leave the x alone by itself

[uncut star]

alright

[thin bridge]

you'd want to get the factorisation in the form
(x-m)(x-n)

[uncut star]

x - (9 - sqrt(53))/2

[thin bridge]

and your other factor?

[uncut star]

oh yeah

i have TWO XD

I forgot

[gilded fog]

u can't coz x is not common for everything

[uncut star]

(x - (9 - sqrt(53))/2)(x + (9 - sqrt(53))/2)

Is this right?

I think it is correct

So then the roots would be

x_1 = (9 - sqrt(53))/2 ?

no wait

[thin bridge]

messed up a few signs

[uncut star]

i think that's correct for the lefthand expression

i edited again

right hand would be x_2 = (-9 + sqrt(53))/2 ?

[thin bridge]

no

[uncut star]

damn

ok let me try again

i forgot

the /2

?

[thin bridge]

can you type out your final factorisation

[uncut star]

or still a problem with the signs

(x - (9 - sqrt(53))/2)(x + (9 - sqrt(53))/2)

if there is a way to continue from there I have no clue where to go

(x - (9 - sqrt(53))/2)

this part was for x_1 = (9 - sqrt(53))/2

[thin bridge]

still a problem with signs
are you able to show your work for the second factor

[uncut star]

sorry I tried doing them in my head so I have no work to show: I can explain my thought process; I want each expression to equal 0. Those are the roots. If x = (9 - sqrt(53))/2 then

[thin bridge]

do the work on paper

[gilded fog]

guys please resend the question once

[uncut star]

Ok I will try on paper

on paper i get the same thing

[thin bridge]

can you show your work on paper

[gilded fog]

send the question guys

[uncut star]

Wait

Is this the part that has gone wrong?

(x - (9 - sqrt(53))/2)(x + (9 - sqrt(53))/2)

Or is it me trying to find the roots

[thin bridge]

yes, i've mentioned multiple times that the factorisation itself is wrong

[uncut star]

ah wow

ok

I will get back to you

I really need to go gym now

I have it scheduled in 9 minutes

Sorry for cutting it short and ty for trying to help

I will try to factorise is correctly when I am back

[thin bridge]

in the previous step, you had
$$\br{x \blue{- \frac92 - \frac{\sqrt{53}}{2}}}\br{x \red{- \frac92 + \frac{\sqrt{53}}{2}}}$$
factor out -1 from the red and blue parts

[flat frigate]

ℝαμOmeganato5

[uncut star]

ohhh

yeah I did not see that

and I would not have figured out how to do that on my own

Is it wrong to just combine the terms with like denominators?

[thin bridge]

depends on how you're doing it

which is why i requested that you show your work

[uncut star]

I just did (9-sqrt53)/2

[thin bridge]

yeh, so that'd be wrong

[uncut star]

ok

that's where I went wrong then

among many other places

I will try this again when I am back like I said

Maybe start over with a different expression

[thin bridge]

by doing that, you're treating the expression as though it were:

[uncut star]

ah

there is a difference?

[thin bridge]

yes

[uncut star]

I don't see the difference that would explain my error

Bad thing I don't see a difference

but I really have to go now brb

will be 40 or something

[thin bridge]

1 - 1 - 1
isn't 1

[uncut star]

true

[thin bridge]

is essentially what you did

you're calculating it like
1 - (1 - 1)

[uncut star]

yeah

[thin bridge]

subtraction isn't associative

you could however slightly modify it to
1 + (-1 - 1)
to get 1 + (- 2) = 1 - 2 = -1
which would be valid

$$\br{x + \blue{\br{- \frac92 - \frac{\sqrt{53}}{2}}}}\br{x +\red{\br{- \frac92 + \frac{\sqrt{53}}{2}}}}$$

[flat frigate]

ℝαμOmeganato5

[thin bridge]

and simplify those fractions

similar to how you'd do
-1/7 - 2/7

or
-1/7 + 2/7

[safe radish]

@uncut star Has your question been resolved?

[safe radish]

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[hushed laurel]

3-22. Consider the two-link planar robot
with l1 = l2 = 5 in. and oriented in the
elbow-down position as shown in Fig.
P3.22. If the tip of the robot is located at
the point P(x, y) = (4.83, −8.36), determine the values of 𝜃1 and 𝜃2., i cant figure out how to get a answer that matches with the point P

[hushed laurel]

i love reverse kinematics lol

the steps i used was i connected the point p with the origin and found the length by using pythag theorem and got 9.65 inches and labeled that as d

then i solved for 𝜃2 using law of cosine

then used 𝜃2 to solve using law of tan to get 𝜃1 and got -134.38= 𝜃1 and 𝜃2 =150 and its wrong

<@&286206848099549185>

[heady topaz]

i dont think you can use pytheg teorem for that

[hushed laurel]

i had to solve for length D tho

so you would use a diff version of it by using the x and y cords of P instead for the leights of the lines

to get the hypotenuse

[safe radish]

@hushed laurel Has your question been resolved?

[safe radish]

@hushed laurel Has your question been resolved?

[lean otter]

well I think you could do something like that

use cosine law and pythagorean theorem on d

then tangent on $\theta_1 + \frac {\theta_2} 2$

[flat frigate]

./Kotarou

[lean otter]

that's the idea, I won't make any calculations cuz I'll be busy by now

[safe radish]

@hushed laurel Has your question been resolved?

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[quartz ibex]

How would I get "g"?

[plucky elk]

solve for g algebraically

x = 1 / 2 * b implies b = 2x

[quartz ibex]

where did you get b?

15.25 = 1/2 * 2(15.25)?

[plucky elk]

b = g t^2

do the same for dividing by t^2 on both sides

[quartz ibex]

ah okay

give me a second

30.5 = g (1.67^2)
t= 1.67^2 = 2.79
2.79/2.79
30.5/2.79
b = 10.93
b=g so
G = 10.93?

[plucky elk]

looks like a lot of rounding errors, but sure the steps are right

[quartz ibex]

oh

Rounding errors

could you correct me ?

[plucky elk]

,calc 1.67^2

[flat frigate]

Result:

2.7889

[quartz ibex]

oh

would that be 2.8 then?

i calculated to nearest hundreth

or should i not round

[plucky elk]

you should, in general, keep all the digits when doing algebraic manipulations and only round at the end

[quartz ibex]

so round it to 2.789?

im a bit rusty with my algebraic manipulations so sorry if i frustrate you a bit

30.5 = g (1.67^2)
t= 1.67^2 = 2.7889
2.7889/2.7889
30.5/2.7889
b = 10.936 --> 10.94
G = 10.94?

[plucky elk]

you also rounded t itself

don't

[quartz ibex]

OH

30.5 = g (1.668^2)
t= 1.67^2 = 2.782224
2.782224/2.782224
30.5/2.782224
b = 10.9624530591--> 10.96
G = 10.96

wait

2.782224 would be 2.78 no?

[plucky elk]

,calc 15.25 * 2 / 1.668^2

[flat frigate]

Result:

10.962453059135

[plucky elk]

10.96 sounds right

[quartz ibex]

okayy thank you

could i get help with this aswell?

oh

my average time was wrong

it was 1.686

<@&286206848099549185> Could I get someone to check my answers

😭

bruh

[glossy cypress]

one second

i'm checking

[quartz ibex]

thank you sm 🙏

[glossy cypress]

i got like -9.37% so you're fine

[quartz ibex]

alrr

ty

.close

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[idle rover]

I want to know what is the slope if the equation is y=18 and x=0

[sly needle]

is it a straight line or a slanted line

[red delta]

slope of both

[sly needle]

if all you have is y=18 then its most likely a striaght like

[idle rover]

slanted line

no

[red delta]

the first has 0 slope

[idle rover]

straight

[red delta]

the second has infinite

can i explain how

[idle rover]

yea sure go ahead

[red delta]

ok so slope is the ratio of change in y to change in x

for the first y does not change at all

regardless of x

y=18 always, slope = $18-18/x_2-x_1 = 0$

[idle rover]

so its a horinzontal line

[flat frigate]

David

[red delta]

yes

the second line is vertical

[idle rover]

But it doesnt have a slope

[red delta]

the slope is infinite or undefined

[idle rover]

so in a cartesian plan, how would it look like?

[red delta]

,w plot x=0

[flat frigate]

Results provided by WolframAlpha

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

Requested by ishigamisenku_

[red delta]

it will be the line for the y axis

[idle rover]

got it thanks, also another question, if the original abscisse and the b is 0 (y=mx+b), what would be the equation?

[red delta]

original abscissa and b

?

[idle rover]

its cause i have a french version

and idk what is a and b in english

[red delta]

so if m=0 and b=0?

[idle rover]

well we need to find the slope to indentify the equation

because we already have b, and the equation formula is, y=mx+b

[safe radish]

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.

[idle rover]

wait nevermind i found the answer

thank you for helping me out

.close

[safe radish]

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[safe radish]

@magic tendon Has your question been resolved?

[plucky elk]

use binomial theorem twice

[magic tendon]

I don't understand
I was told to replace the variables with 1 to find the sum of coefficients in binomials
But i don't understand trinomial

[plucky elk]

(a + b + c)^2 = ((a+b) + c)^2

[short topaz]

so note that

if we had a polynomial like

ax^2 + bx + c

then to find the sum of the coefficients

plug x=1 to get

a + b + c

[plucky elk]

oh that's smarter.

[short topaz]

that was my idea too lol but then they gave their own hint

[safe radish]

@magic tendon Has your question been resolved?

[safe radish]

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[signal token]

Factor (5x+1)^3 + 8, using the a^3 + b^3 rule

[signal token]

I understand how to do it but i keep getting (5x+3)(25x^3 + 10x) which is apparently the wrong answer pls help

[fathom adder]

a^3 + b^3 = (a+b)(a^2 + ab + b^2)

[signal token]

When i use that way i get the wrong answer

[fathom adder]

So (5x+1 +2)(25x^2 - 2(5x+1) + 4)

Oh

Ok -ab

[signal token]

Yea thats what i got so far then it goes weong

[fathom adder]

(5x+3)(25x^2 - 10x + 2)

[fathom adder]

How do you get that 25x^3 ?

[signal token]

Wait nvm i got (25x^2 + 10x)*

But still its wrong idk how

[fathom adder]

+2 at the end

[signal token]

Ohh

Thats it?

[fathom adder]

Should be

[signal token]

Alright tyvm🙏🙏

.close

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[fathom adder]

You're welcome

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[lavish wharf]

can someone explain how they found the second derivative

[lavish wharf]

like what happened in the first line

[severe pond]

!xy

[safe radish]

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.

[severe pond]

what did you get for dh/dt

[junior smelt]

chain rule

You have $\dv{P}{t} = \dv{P}{h}\dv{h}{t}$, here replace $P = \dv{h}{t}$

[flat frigate]

@junior smelt

[lavish wharf]

$$ \frac{2}{\pi (3\cos^2h + 4)^2} $$

[flat frigate]

seonhee

[lavish wharf]

wait sorry i dont follow

[severe pond]

that’s the chain rule

that’s what they were doing in the first line

[lavish wharf]

wait so they just differentiated dh/dt

oh that makes sense actually i just didnt differentiated properly

.close

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[lavish wharf]

.reopen

[safe radish]

✅

[lavish wharf]

im still not getting the same answer??

[severe pond]

was your dh/dt right

from the answer key

[lavish wharf]

yeah

$$ \frac{2}{\pi (3\cos^2h + 4)^2} $$

is it something implicit because of the variables?

[flat frigate]

seonhee

[lavish wharf]

because i used the chain rule before as well

and got this

@severe pond

[severe pond]

hello

youre differentiating dt

h is a function of time

thus you need a dh/dt

when differentiating a term that has h

such as the cos^2

$\frac{2}{\pi} (3\cos^2(h) + 4)^{-2}$

[flat frigate]

knief

[severe pond]

would you like me to differentiate it

@lavish wharf

[lavish wharf]

sorry im back

[lavish wharf]

yeah could you show me the steps?

dont really need to show me the algebra but i think it has smth to do with the first line in the answer key

if its an application of the chain rule are they saying that differentiating d/dt gives 1?

[severe pond]

what do you mean

[lavish wharf]

well i dont know how d/dt(dh/dt) = dh/dt x d/dh(dh/dt)

if they're using chain rule

[severe pond]

you have $\frac{dh}{dt} \cdot \frac{d}{dt}(\frac{dh}{dt})$

[flat frigate]

knief

[severe pond]

you already know dh/dt

[lavish wharf]

why?

[lavish wharf]

is this from the double derivative

[severe pond]

yes

[lavish wharf]

so $$ \frac{d}{dt}(\frac{dh}{dt})=\frac{dh}{dt} \cdot \frac{d}{dt}(\frac{dh}{dt}) $$

[severe pond]

another way to think of this is implicit differentiation

[flat frigate]

seonhee

[severe pond]

let us differentiate the dh/dt expression

to see why we get this result

[severe pond]

from here

lets differentiate

to get

$\frac{-4}{\pi} (3\cos^2(h) + 4)^{-3} (6\cos(h))(-\sin(h))(h’)$

[flat frigate]

knief

[lavish wharf]

ohh i see what i did wrong

missed the h' at the end becuase implicit differentitation

[severe pond]

forgot the h’

[lavish wharf]

and you sub in h'

[severe pond]

see how the h’ is precisely dh/dt though

[lavish wharf]

yeah

[severe pond]

from here

[lavish wharf]

ohh i see

tysm

.close

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[zenith crystal]

Hey, i just need to get three questions in a row correct. This has nothing to do with the class im taking and I just need to get it done. If someone could help me get the three questions right, it would be much appreciated. Thanks!

[soft prism]

Well, do you know dot products?

[zenith crystal]

not even a little lol

[soft prism]

https://youtube.com/playlist?list=PLAEgL_CQRjwTF3iqXIUKZvnaLHnZav0bv&si=85l3JKIL9WCjWwo5

Just spare about 30 mins of your day, and you'll "get it done" properly

[zenith crystal]

lol i just need 3 right to move on and not do it again (this is for a chem class) but i appreciate you pushing me to go the right direction

[safe radish]

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@zenith crystal Has your question been resolved?

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[modern flower]

wanted to check my answer to this question

[safe radish]

Please don't occupy multiple help channels.

[eternal carbon]

your result is correct

[modern flower]

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[granite eagle]

I’m confused about question 2. How would I be able to compute these if I dont know the dimensions of A?

[junior smelt]

While you may not know the number of columns of A, you know it has 4 rows, and should hopefully remember that elementary matrices are square and invertible, and how to multiply by them to carry out row operations...

[granite eagle]

Oh okay
I was assuming that to be the case but I wasn’t 100% since I always see “A is a square/n x n matrix”

Thanks

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[umbral solar]

guys i tried to calc this with row operatons

but very false

can someone spot mistake?

even after last operation to add bottom to top row multiplied by /-(6/5)

[obsidian oracle]

what happened to the bottom left coefficient on the right matrix

when did it become 0

[umbral solar]

when top row multiplied by 6 and added to bottom

1*(6) + (-6)

[obsidian oracle]

ok

but why did it become 0 again afterwards

what happened here

[umbral solar]

oh that 1

damn!

i try fix

[obsidian oracle]

also you multiplied by 5/71

6/71 is the coefficient on bottom left

bottom right should be -5/71

[umbral solar]

hmmm

yes very much

how to not make these stupid mistakes?

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[cosmic epoch]

when negating something like this

[cosmic epoch]

why does it turn into a greater than or equal to sign?

I don't understand it

[peak estuary]

well if its not smaller then its either greater or equal

[cosmic epoch]

but why

like

is there an example of why this occurs?

[peak estuary]

what do you think the < should turn into

[cosmic epoch]

I thought it would just turn into a greater than sign

But ig

tehnw e're leaving out the 1 value we're looking for

which would be the value in the middle

of both

ok nvm

ur right it makes sense

tysm

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[turbid marsh]

apparently the answer is 729pi/2, but i cant figure out how to get that

im in another tab, ping me in your reply

[safe radish]

@turbid marsh Has your question been resolved?

[turbid marsh]

<@&286206848099549185>

[pure agate]

What have you tried?

[pure agate]

.

[turbid marsh]

Shell method, disk/washer

[pure agate]

Shell Method would be the preferred method.

What integral did you make?

[turbid marsh]

1 moment

@pure agate

[pure agate]

You forgot r(y) for the radius.

[turbid marsh]

like this?

[pure agate]

The interval is incorrect.

[turbid marsh]

then what should it be? If I recall correctly the intervals are the x intercepts

[pure agate]

You are integrating with respect to y so you need to integrate along the y-axis.

[turbid marsh]

Then would it be 9 instead of 12? My professor didnt really explain this part so im pretty confused

[pure agate]

Yes, you will integrate from 0 to 9.

https://www.geogebra.org/classic/qezvnj24

[turbid marsh]

ah i see

thank you for your help

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[pure agate]

yw

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[broken cloud]

how do i find the correct graph for g

[red delta]

The graph f(x) is y=5 right?

[light plover]

yes

[red delta]

So the value of y is constant

[light plover]

yes

therefore g(x)=5

[red delta]

Meaning it is unaffected by the value of x

[light plover]

yes

it won't be affected by x

[red delta]

@broken cloud

[broken cloud]

ok

[light plover]

so the answer would be the last option

[safe radish]

@broken cloud Has your question been resolved?

[light plover]

@broken cloud

[arctic raven]

f is a line segment, defined on [-3, 3]. so g is, at best, defined on [-6, 6]

we can see that the length of the curve f is significant by the varying lengths in the available options

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[primal sluice]

The question is: Show that the equation has at least one solution.

[primal sluice]

I don't understand what the question is asking me to do

[junior smelt]

Hint: you may want to show that sin(x) + x - 1 = 0 has at least one solution, which is equivalent (you just need to justify why one exists, assumedly you don't need to find it)

[primal sluice]

how do i prove that