help-10

[kindred oasis]

No, it's filled up to 0.25ft from the top, so from the floor up to 1.5-0.25=1.25ft

And it asks you to find the volume of the water, not how much water you need to fill up the pool

[eternal dome]

so my eq would be V=pi 6^2 x 1.25

[kindred oasis]

Yes

,calc pi66*1.25/27

[eternal dome]

and then is that the answer

[kindred oasis]

Maybe you can convert it to yard³ as you did before, but yeah, that's the answer

[eternal dome]

ok so just divide by 27 again

[warm shale]

Result:

5.235987755983

[kindred oasis]

Yes

[eternal dome]

thank you.

[kindred oasis]

You're welcome

[eternal dome]

.close

[obtuse pebble]

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[mighty pelican]

Is the part I painted yellow correct?

[mighty pelican]

the entire problem is

[obtuse pebble]

@mighty pelican Has your question been resolved?

[obtuse pebble]

@mighty pelican Has your question been resolved?

[obtuse pebble]

@mighty pelican Has your question been resolved?

[obtuse pebble]

@mighty pelican Has your question been resolved?

[mighty pelican]

I asked about the part where I found dz/dt @rotund canyon

Oh

Yeah you're right it's the long way

Yess 🥂 😎

I'll be careful:)

@rotund canyon

can you check my solution

[obtuse pebble]

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[heavy summit]

just wanted to confirm, it's 20 right?

[royal basin]

yes

[obtuse pebble]

@heavy summit Has your question been resolved?

[proven zephyr]

yes

[obtuse pebble]

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[heavy summit]

tytyty

[obtuse pebble]

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[stable hawk]

I need help solving an equation math problem

[timid silo]

just ask

[stable hawk]

Im not sure how to solve this problem

like I need a break-down step by step of it

[proven zephyr]

you have the equation

t is usually time

just insert the time

[stable hawk]

Oh, so the time would be 35?

oh wait

[floral canopy]

It says 25.

[stable hawk]

so 25 is the t

[floral canopy]

Yes.

[stable hawk]

So I would need to do like 2 x 25, and 2 x 35?

so...

[floral canopy]

Why would you do that?

[stable hawk]

N = 500

x 2

what would be the exponent

exponent of 2

[floral canopy]

What is t?

[stable hawk]

25?

[floral canopy]

Correct.

So plug it in.

[mighty pelican]

the derivative of sx and this comes from triangle similarity rule

[stable hawk]

so do I need to divide 35 divide by 25

[floral canopy]

Divide 35 by 25.

[stable hawk]

ok

[floral canopy]

This will be your exponent.

[stable hawk]

it says 1.4

So 2 to the power of 1.4?

[floral canopy]

How

How did you get to that?

25/35 is not 1.4

[stable hawk]

oh its

0.71?

[floral canopy]

Yes.

But its not precise

I would write it as 5/7.

[stable hawk]

Okay got it

so 2 to the power of 0.71..

I forgot how to solve the next step

Would you multiple 2 by 0.71?

[floral canopy]

Don't write it as 0.71

its 5/7.

[stable hawk]

oh ok

ok

so, N = 500 x that fraction or do we need to solve that fraction first

[floral canopy]

wait no I'm sorry

I was confused for a second

[stable hawk]

oh

thats fine

i think i shud um

ping a helper?

[floral canopy]

You're allowed to use a calculator, right?

[stable hawk]

of course

this is just a homework assignment

[floral canopy]

Ok then just put everything in

2^(5/7) *500

This will be the answer

[stable hawk]

oh

820

thanks.

[floral canopy]

Yea

[stable hawk]

this question is almost the same

but its saying how many wolves were present at the beginning

so Idk how to solve the t

what would u do to solve the t

[floral canopy]

the beginning.

[stable hawk]

is it 35?

like

35 over 35

[floral canopy]

How many years have passed?

[stable hawk]

none

[floral canopy]

so what is t?

[stable hawk]

0?

[floral canopy]

Yes.

[stable hawk]

ok so 0 over 35

that would be 0 for the exponent right

[floral canopy]

So you get 500*(2^0)

that's correct.

[stable hawk]

ok but if we have a 0 as an exponent

that 2 will become 1?

[floral canopy]

Yep

[stable hawk]

ok

so the answer will be 500

[floral canopy]

that's right.

[stable hawk]

500 wolves were present

alr thx

i got like some few more questions if u dont mind sticking around

so about this graph..

Im not sure how to solve for the x

@floral canopy are u there

<@&286206848099549185> Please help me with this graph and teach me how to solve x

thanks

[timid silo]

x-intercept is when y=0

[stable hawk]

oh

so..

I see that its starting from 0 and going to 4

[timid silo]

(0,0), (4, 0)

[stable hawk]

is x = 4 and x = 0

the answer?

[timid silo]

yes

[stable hawk]

cool, thanks

wow its easier than I thought

anyways I have like 2 or 3 more questions if u dont mind helping me

im not too sure how to solve that by using 10 and 13

[stable hawk]

do u mind helping me on this one

[kindred oasis]

Pythagoras theorem?

[timid silo]

u know pythogreas

[stable hawk]

yes

[timid silo]

use it

[stable hawk]

kinda

like not too familiar with it

[heavy thicket]

Do you need the formula

[stable hawk]

yes pls

:)

[kindred oasis]

Look at this picture
https://images.app.goo.gl/q3YgX2Q2Wo1y26f96

[heavy thicket]

a^2 +b^2=c^2

Where a and b are the legs and c is the hypotenuse

[stable hawk]

oh I just searched up pythagorem theorum and it just gives me like a calculator for it

i got the answer, thanks

[heavy thicket]

Do you know how to simplify radicals or is your teacher ok with not simplifying them

[stable hawk]

yeah she wants the simplified form

[heavy thicket]

Alright so what’s the unsimplified radical

[stable hawk]

oh my mistake

i meant not simplifying em

[heavy thicket]

So she doesn’t want you to simplify

Ok so what’s the radical

[stable hawk]

16.4 cm

[heavy thicket]

Hmmm

She wants you to use approximate decimal form?

[stable hawk]

ye she just wants me to write how much cm is on the other side of the triangle

[heavy thicket]

How many decimal places

[stable hawk]

1

[heavy thicket]

Then you’re fine

[stable hawk]

i have one more

[stable hawk]

How do I solve this

I missed a lesson on it

[heavy thicket]

Pythagorean theorem again

[stable hawk]

is this the same as the previous question

oh

lol ok

[heavy thicket]

Where c is hypotenuse

[nocturne minnow]

Nope

[stable hawk]

so its 6.3?

[nocturne minnow]

You need to use trig functions

[stable hawk]

oh..

[nocturne minnow]

SOH CAH TOA

[stable hawk]

Im not too familiar with it

[nocturne minnow]

Because it's asking for the angle

[stable hawk]

so what would my first step be in solving it

[heavy thicket]

Oh shit i didn’t read the problem

I thought it asked for side length

[nocturne minnow]

https://youtube.com/watch?v=I3jyBUyjg48

Watch some videos on that

Here's one

[stable hawk]

k ty

[heavy thicket]

@stable hawk sin cos or tan which one to use

[stable hawk]

cos?

[heavy thicket]

Try again

[stable hawk]

sin

[heavy thicket]

Ok

So sin(angle b) is?

[stable hawk]

angle b doesnt have a number

[heavy thicket]

Don’t worry we’ll get there

Sin(b) is?

[stable hawk]

um

Im not sure sorry

[heavy thicket]

sin=opposite/hypotenuse

[stable hawk]

hyp would be 7

so it would be 7?

[heavy thicket]

You got the hypotenuse

What is the opposite

[stable hawk]

3

[heavy thicket]

So sin(b) is?

[stable hawk]

2.3?

[heavy thicket]

Use fractions

[stable hawk]

I did 7 over 3

[heavy thicket]

Flip the numbers

And that is sin

[stable hawk]

0.42

[heavy thicket]

Use fractions

[stable hawk]

wait no

-.43

0.43

[heavy thicket]

Because the decimal will repeat

Just use 3/7 as sin(b)

[stable hawk]

yeah i got the answer

its 0.43 degree

[heavy thicket]

1 more step

Do you know inverse trigonometric functions

[stable hawk]

nope

[heavy thicket]

Listen closely

$sin(\theta)=\frac{a}{b}$

[warm shale]

% Openglobe %

[heavy thicket]

$\theta=sin^-1 (\frac{a}{b})$

[warm shale]

% Openglobe %

[heavy thicket]

Negative 1 is the exponent

Do you understand?

[stable hawk]

oh

I think I do

[heavy thicket]

So you calculate sin^-1(3/7)

[stable hawk]

i dont believe I need to do that extra step, 0.43 is the answer I need

thanks for helping tho

[heavy thicket]

That angle is too small

It’s not an extra step

[stable hawk]

alr

[heavy thicket]

It’s a required step

[stable hawk]

so if i calculate this

sin^-1(3/7)

[heavy thicket]

Yes

[stable hawk]

i would get the answer?

[heavy thicket]

Yes

[stable hawk]

25.38

thats my answer

[stable hawk]

thats correct right

[heavy thicket]

Try again

Actually

[stable hawk]

this is the last question I need to do

and im finished for my homework assignment

im not too sure what it means by factored form

[heavy thicket]

Yes

Checked it in calc

[stable hawk]

ye

[heavy thicket]

What is the GCF

Of the two terms

[stable hawk]

um

12?

[heavy thicket]

Ok let’s break this apart

What is GCF of 6 and 12

[stable hawk]

2?

[heavy thicket]

That’s a factor but not the greatest one

[stable hawk]

oh..

its 6

right?

[heavy thicket]

So factor a 6 out of the parenthesis

Now what’s the GCF of t^2 and t

[stable hawk]

hmm

t -2?

[heavy thicket]

Ok

Not the answer

[stable hawk]

im not sure

[heavy thicket]

Do you know t*t=t^2

[stable hawk]

i do]

ik that

[heavy thicket]

So what’s the GCF of t^2 and t

[stable hawk]

(tx2) (tx2)

wait no

t^2 - 2

[heavy thicket]

Why the -2

[stable hawk]

im not too sure

[heavy thicket]

t can be split into t and 1
t^2 can be split into t and t

[stable hawk]

ok

[heavy thicket]

Therefore the GCF is?

[stable hawk]

t

[heavy thicket]

Ok

So put the t beside the 6

[stable hawk]

ok

6t

[heavy thicket]

So it would be 6t()

[stable hawk]

what would be inside the bracket

[heavy thicket]

6t^2/ 6t is?

[heavy thicket]

This is the first number inside the bracket

[stable hawk]

um

[stable hawk]

i cant calculate this

[heavy thicket]

First, the 6’s are canceled

t^2/t is?

[stable hawk]

um

(t)(t)?

[heavy thicket]

Just one t

[stable hawk]

oh just one

ok

[heavy thicket]

So 6t(t-the last term)

So the last term is 12t/6t

Which is?

[stable hawk]

6t (t-2)

right?

@heavy thicket

[heavy thicket]

Yes

That’s the factored form

[stable hawk]

pog

yes

.close

[obtuse pebble]

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[obtuse pebble]

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• Show your work, and if possible, explain where you are stuck.
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[jagged arrow]

again

[jagged arrow]

so

[proven zephyr]

wait.. is this an exam?

[jagged arrow]

no

[harsh remnant]

Assessment?

[jagged arrow]

its a practive quiz

practice

its not worth marks

[proven zephyr]

ok

the three parts on the bottom is equal btw

[harsh remnant]

Why is this hard

[jagged arrow]

im young

[harsh remnant]

Just rectangle formula and addition

What's the formula of a rectangle?

[jagged arrow]

lxw

what do i add

[obtuse pebble]

@jagged arrow Has your question been resolved?

Channel closed

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• Show your work, and if possible, explain where you are stuck.
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[hasty crag]

$\vec{x} = \sum_{j = 1}^{n} \langle \vec{e_{j}}, \vec{x} \rangle \vec{e_{j}} = \sum_{j= 1}^{n} \left( \sum_{i = 1}^{j} e_i x_{i} \right)\vec{e_{j}}$

[hasty crag]

am I doing this right? I'm not sure if this is how u expand these fuck you latec

finally omg

[rugged kite]

What's your question?

[hasty crag]

if this expansion is correct

[upbeat plinth]

should be e_i*x_i

[warm shale]

texaspb :rainbow:

[hasty crag]

sorry im triggered by latex

is it correct now

[upbeat plinth]

yes

[hasty crag]

lovely, thank you

uuh

then I do the sum that's inside first right

[upbeat plinth]

wait

[hasty crag]

okay

[upbeat plinth]

how many components do these vectors have

[hasty crag]

I'm just gonna post the entire question

[upbeat plinth]

are they vectors in R^n

[hasty crag]

Show that the bilinear form $\langle \vec{x},\vec{y} \rangle = \sum_{j=1}^{n} x_{j}y_{j}$ in $V = \bR^{n}$ is a scalar product in V. If $S ={\vec{e}{1}, \dots, \vec{e}{n}}$ is the standard basis of V and $\vec{x} \in V$, show that we can write: \

$$\vec{x} = \sum_{j=1}^{n} \langle \vec{e}{j}, \vec{x} \rangle \vec{e}{j}$$

[royal basin]

did you miss a = right after the S

[hasty crag]

yes thanks ann

[warm shale]

texaspb :rainbow:

[hasty crag]

@upbeat plinth this is the entire thing

so n components

[upbeat plinth]

ok so the inside sum should be up to n

[hasty crag]

ooh ok

gotcha

[upbeat plinth]

and u should write e_{ji} for the ith component of e_j

[hasty crag]

$\vec{x} = \sum_{j = 1}^{n} \langle \vec{e_{j}}, \vec{x} \rangle \vec{e_{j}} = \sum_{j= 1}^{n} \left( \sum_{i = 1}^{n} e_{ji} x_{i} \right)\vec{e_{j}}$

[warm shale]

texaspb :rainbow:

[hasty crag]

how does this work tho i'm confused

[upbeat plinth]

its just using the def of dot product

[hasty crag]

I don't get the e_ji indexing

[upbeat plinth]

e_j is the jth basis vector

each vector has n components

so e_{ji} is a way to write the ith component of the jth basis vector

[hasty crag]

alright

I think I got it, if there's any questions left I'll post in the linagl channel

thank you so much as always

[upbeat plinth]

np

[hasty crag]

.close

[obtuse pebble]

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[obtuse pebble]

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[fluid meadow]

How can we be certain pq is tangent to that circle

[obtuse pebble]

@fluid meadow Has your question been resolved?

[fluid meadow]

<@&286206848099549185>

[obtuse pebble]

@fluid meadow Has your question been resolved?

[obtuse pebble]

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[smoky panther]

am i correct?

[heavy thicket]

,w calculate cos(pi)

[warm shale]

Results provided by WolframAlpha

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

Requested by % Openglobe %#3085

[heavy thicket]

You got it

[smoky panther]

yayy thanks

.close

[obtuse pebble]

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[smoky panther]

how do i solve this?

[warm canopy]

try graphing sin(x)

[smoky panther]

i did

just now

[warm canopy]

okay, and roughly plot x=4

[smoky panther]

okay yup

[warm canopy]

what on a graph represents instantaneous rate of change

[smoky panther]

slope of tangent line

[warm canopy]

right so you're gonna need to eyeball the slope of the tangent line to sin(x) at x=4

luckily the answers to choose from are very different

[smoky panther]

so then wont it be -0.65?

[warm canopy]

i would agree with you

[smoky panther]

great, thanks for helpin

[warm canopy]

youre welcome!

[smoky panther]

.close

[obtuse pebble]

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[obtuse pebble]

Please don't occupy multiple help channels.

[jaunty stirrup]

PleZe

Help

[knotty crow]

so you have the right triangle with one leg equals 4.5 and hypotenuse equals 5, can you find the another leg?

[jaunty stirrup]

What

Which other leg

I already found that leg

It’s 2.2

[knotty crow]

ah you're asking about b)

right?

or what

[jaunty stirrup]

Yea

B)

[knotty crow]

treat it as a rhombus (every square is a rhombus) and use the formula with the diagonals or find the side length of a base

in order to do that use Pythag or just property of the triangle 45, 45 and 90

[jaunty stirrup]

Whatt

I don’t understand

[knotty crow]

what's your diagonal

[jaunty stirrup]

5

[knotty crow]

nope, result from a)

[jaunty stirrup]

Oh

4.4

[knotty crow]

let's say, but it's approximate

[jaunty stirrup]

Yea

[knotty crow]

and in fact idk how you got that, because it should be

,calc sqrt(19)

[warm shale]

Result:

4.3588989435407

[jaunty stirrup]

Yea

That’s 4.4

[knotty crow]

ok

[jaunty stirrup]

It said to round to nearest tenth

At the top of the page

[knotty crow]

so now let's say the side length of the base is equal to x, ok?

[jaunty stirrup]

Ok

[knotty crow]

diagonal is 4.4

can you find x?

[jaunty stirrup]

No

Don’t I need two lengths

For pitagora

[knotty crow]

not really, you have only one variable

only x

[jaunty stirrup]

Wait

Can u set up the equation for me

I don’t understand

[knotty crow]

ok

so the opp and adj is equal to x

basically

x^2 + x^2 = (4.4)^2

can you solve it?

[jaunty stirrup]

Oh

Yes

Wait

What is x^2 + x^2

[knotty crow]

2x^2

[jaunty stirrup]

Oh

[knotty crow]

like an apple + apple = two apples

[jaunty stirrup]

Ok I did it

Thank you

.close

[obtuse pebble]

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[timid silo]

what gives, why are they dropping the divided by twos but not the + 1?

oh i see, they're adding 0.5cn + 0.5cn

[mint bronze]

Would you mind taking an image where the whole sentence above is in it, I'm just wondering what it says

[mint bronze]

Ah thanks

Yes, you are right

[timid silo]

that'd be the first thing i'm correct on today

[mint bronze]

But it is not exactly adding halves but more so that say if n were 13, then it would be adding 6c and 7c

Which gives the 13

[timid silo]

right right

what about this one:
show that the sltn of T(n) = T(n-1) + n is O(n^2)

.close

[obtuse pebble]

Channel closed

Closed by @frank abyss

Use .reopen if this was a mistake.

Please take a minute to participate in [our survey](#changelog message) if you haven't already!

[mint bronze]

What about it sorry

I was just reading it

[timid silo]

i need to understand how to do it

[mint bronze]

You might want to reopen

[mint bronze]

I mean type

.reopn

[timid silo]

.reopen

[obtuse pebble]

✅

[mint bronze]

Thanks

[mint bronze]

Ok, so I'll talk through the inequalities going on here

First we assume that

,,T(n-1) \leq c(n-1)^2

[warm shale]

Social Capital Gainer

[mint bronze]

And we wish to show that this implies that

,,T(n) \leq cn^2

[warm shale]

Social Capital Gainer

[mint bronze]

Start by adding n to both sides of

,,T(n-1) \leq c(n-1)^2

[warm shale]

Social Capital Gainer

[mint bronze]

,,T(n-1) +n \leq c(n-1)^2 +n

[warm shale]

Social Capital Gainer

[mint bronze]

We see that the left hand side is equal to T(n) so

,,T(n)\leq c(n-1)^2 +n

[warm shale]

Social Capital Gainer

[mint bronze]

Expanding the quadratic

,,T(n)\leq cn^2 -2cn +c +n

[warm shale]

Social Capital Gainer

[mint bronze]

Grouping terms into powers of n

,,T(n)\leq cn^2 +(1-2c)n +c

[warm shale]

Social Capital Gainer

[timid silo]

you're going much too fast here

[mint bronze]

Ok sorry

What would you like me to elaborate on?

[timid silo]

why do they square the n in c(n-1) term and not the next n

[mint bronze]

What is the next n?

The n in

,,T(n)\leq c(n-1)^2 +n

[warm shale]

Social Capital Gainer

[mint bronze]

That one?

[timid silo]

yes. i would like you to assume i know nothing and walk me through step by step, if you have the time

i've been studying this stuff for awhile now but it's not clicking

[mint bronze]

Alright

Would you like me to start at tthe beginning again

[timid silo]

i get why the assumption becomes c*n^2

given the definition of O and whatnot

yes sure

so we have T(n) = T(n-1) + n

and our inductive hypothesis is that this will be in O(n^2)

let's set aide the "c is taken to be max(somestuff)" bc i don't understand that

[mint bronze]

Ok I will address the induction parts

[timid silo]

why do you not only square the n inside the first term

why do you square the entire (n-1)

[mint bronze]

Basically we have some statement dependent on integers n which we need to prove is true for all n. That statement is T(n) <= cn^2

I will explain why

Usually in an inductive argument, we assume some step, say n=k, then show that if we assume this, this implies that the next case for n=k+1 must also be true

In this question however

They are assuming the n-1 case

[timid silo]

i don't think so

the recurrence is just given by T(n) = T(n-1)+n that's just it mathematically before any induction or hypothesis

the assumption is that T(n) <= cn^2

[mint bronze]

That is the recurrence relation, but given this recurrence relation it is easier to assume n-1 and prove n rather than assume n and prove n+1. They both will give the same result, but assuming n-1 lets us apply the recurrence relation straight away, so that is what they do

They finish by showing that T(n) <= cn^2

[timid silo]

this is how i'm seeing it.. is this incorrect

[mint bronze]

I disagree with what tlu have labelled as the cn^2 expression, that is not the cn^2 expression, that is something we will show is less than or equal to the cn^2 expression

Hence showing that T(n)<=cn^2

[timid silo]

by transitive property?

[mint bronze]

Yes, I suppose

So lets do that

Assume that

,,T(n-1) \leq c(n-1)^2

[warm shale]

Social Capital Gainer

[mint bronze]

Now add n to both sides

[timid silo]

wait

how did we get that expression

[mint bronze]

We assumed the n-1 case

Of the expression we want to prove which is

T(n)<=cn^2

Change n with n-1

[timid silo]

why is it (n-1)^2, why isn't it c((n^2)-1))

[mint bronze]

Because n is the variable

Say that n=7

Then

T(7)<=c7^2

Also

T(6)<=c6^2

6 is the n-1, 7 is the n

[timid silo]

i don't understand

[mint bronze]

Another way to think about it is like this

Take this as an example, the function f(n)=n^2

Evaluate f(3) for me

[timid silo]

9

[mint bronze]

Evaluate f(w) for me, where w is some number

[timid silo]

w^2

[mint bronze]

Evaluate f(w-1)

[timid silo]

ah ok

[mint bronze]

(w-1)^2

[timid silo]

bc that variable can be any expression

[mint bronze]

Yes exactly

[timid silo]

ok

[mint bronze]

So we set n to n-1 in

T(n)<=cn^2

To get

T(n-1)<=c(n-1)^2

It's kinda clunky

If i were proving this I would say

Set n=k-1

T(n)<=cn^2

T(k-1)<=c(k-1)^2

So that it flows a bit better

Does that make sense?

[timid silo]

i mean sort of except that we're replacing n-1 in for n it terms of T on the left and replacing T with c on the right

anyway i guess we now have T(n-1) = c(n-1)^2 + n

[mint bronze]

We're not replacing T with c, not sure what you mean

Anyway

We have this

[timid silo]

T (n) = T (n − 1) + n ≤ c(n − 1)^2 + n

to get from the first expression, to the one on the right hand side of the equality, we replace T with c

[mint bronze]

I think you are confusing the brackets here a bit. The brackets in T(n-1) belong to the function T

Like saying f(3)=9

The brackets in c(n-1)^2 are normal brackets like 3(x+y)=3x+3y

You cant expand T(n-1) into Tn -T

[timid silo]

oh i thought we were plugging in stuff into the left hand side to get the right, but we're plugging it into the T(n) <= cn^2 statement

[mint bronze]

Yes, n is a variable and we are plugging stuff into it

[timid silo]

nvm i'm confused ok

i was confused*

[mint bronze]

Ok, but does that make sense now

[timid silo]

so we haven't actually done anything yet except taken the original expression and set it less than or equal to cn^2 where n = n-1

[mint bronze]

Yes that's what we've done so far

[timid silo]

but just to be clear

[mint bronze]

We assume the n-1 case and want to use it to prove the n case

[timid silo]

T(n) = T(n-1) + n is the original recurrence

so there is no n-1 added yet anywhere

no substitutions

[mint bronze]

Um, yeah not yet

[timid silo]

all we did was take it as is and set equal to c(n-1)^2

[mint bronze]

Yeah

[timid silo]

well less than or equal to

[mint bronze]

So lets carry on

From

,,T(n-1) \leq c(n-1)^2

[warm shale]

Social Capital Gainer

[mint bronze]

We are just about to get that

[timid silo]

right they are adding +n to both sides I guess.. why

wait no here it's only added to the cn^2 side

[mint bronze]

It is so that we can form T(n-1) +n, which we recognise is equal to T(n)

That will let us get the left hand side of T(n)<=cn^2 which is what we want. We need to use what we have assumed to get the next case, the n case

[timid silo]

it looks like all they did so far was square n-1 on the right hand side and then add a (+ n)

[mint bronze]

Yes, they started with this and added n to both sides

To get

[timid silo]

no you are misunderstanding

there has been no +n added to the left side

[mint bronze]

,,T(n-1) +n \leq c(n-1)^2 +n

[timid silo]

that is its native form

[warm shale]

Social Capital Gainer

[timid silo]

T(n-1) + n is the original recurrence. there has been no +n added

[mint bronze]

No, we did that, going from left to right in that string of inequalities does not reflect the logical order in which it was obtained

[timid silo]

otherwise it'd be in the form T(n-1)+2n

[mint bronze]

So lets follow the train of thought as I am presenting it and it will hopefully become clear

We go from

,,T(n-1) \leq c(n-1)^2

[warm shale]

Social Capital Gainer

[mint bronze]

To

,,T(n-1) +n \leq c(n-1)^2 +n

[warm shale]

Social Capital Gainer

[mint bronze]

By adding n to both sides

Is that ok?

[timid silo]

ok

[mint bronze]

Cool

So now looking at the left hand side of this, we recognise that T(n-1)+n is exactly just equal to T(n), so we can replace the entire left hand side with T(n)

,,T(n) \leq c(n-1)^2 +n

[warm shale]

Social Capital Gainer

[mint bronze]

This is because T(n-1) +n is exactly the same as T(n) so I can interchange them

[timid silo]

right

so we have T(n) <= c(n-1)^2 + n

[mint bronze]

Yes this is how they got

They tacked on T(n)= onto the left of

What we made

[timid silo]

ok so what step should i do next on paper

(n-1)^2?

[mint bronze]

So from

,,T(n) \leq c(n-1)^2 +n

[warm shale]

Social Capital Gainer

[mint bronze]

You want to expand the quadratic on the right

[timid silo]

n^2 - n - 1

oh i forgot c

[mint bronze]

Ye, and a 2

[timid silo]

c*(n^2 - n - 1)

[mint bronze]

Its not just-n

[timid silo]

ah shit i added the +n in there

[mint bronze]

Ah

[timid silo]

c*(n^2 - 2n - 1) + n

[mint bronze]

And it is also +1

Since it's squared

[timid silo]

oh right

so altogether sofar we have T(n) <= c(n^2 - 2n + 1)+n?

[mint bronze]

Yep

[timid silo]

ok so now what

[mint bronze]

Now expand that bracket too so that there is a c on every term that needs one

[timid silo]

T(n) <= cn^2 - 2cn + c + n

[mint bronze]

Yes

Now you want to factorise n from -2cn and n

[timid silo]

actually in the form its in it takes the form of the answer, the cn^2 bit

[mint bronze]

Yes exactly we have all the pieces we need, and some extra ones too which we have to deal with

[timid silo]

ah shucks

[mint bronze]

Lets write what we have in full

,,T(n) \leq cn^2 +(1-2c)n+c

[warm shale]

Social Capital Gainer

[mint bronze]

This is where we are

Note that I disagree with the screenshot of the solution you posted

I don't believe what they have written is completely correct

Could you post that screen shot again

[mint bronze]

Yeah so they say that

[timid silo]

i mean the book says "show that the solution is O(n^2)"

[mint bronze]

,, c(n-1)^2+n=cn^2+(1-2c)n+1

[warm shale]

Social Capital Gainer

[mint bronze]

That is right, but this step is wrong

As we just showed

It should be +c and not +1

[timid silo]

ok

[timid silo]

well this is where i am

[mint bronze]

Yep, so factor n from -2cn and +n

[timid silo]

n(cn - 2c) + c + n

[mint bronze]

Specifically only from the terms I stated

Not from cn^2

[timid silo]

n(2c + 1)

[mint bronze]

-2c but yeah

Ok so we have

,,T(n) \leq cn^2 +(1-2c)n+c

[warm shale]

Social Capital Gainer

[mint bronze]

Is that ok ?

[timid silo]

you lost me

i rewrote cn^2 - 2cn + c + n as cn^2 - 2cn + n +c

[mint bronze]

Ye, now factor out n from the second and third terms only

[timid silo]

i get -n(2c+1)

[mint bronze]

Not quite

That will give you a -n but we have +n

[timid silo]

how do you factor cn^2 - 2cn + n + c for just n what happened to the -

[mint bronze]

So -2cn+n = n(-2c+1)

Or (1-2c)n which is what I wrote

[timid silo]

ok ok

sorry i'm a bit friend mentally rn

fried*

[mint bronze]

No worries at all

So we are at

,,T(n) \leq cn^2 +(1-2c)n+c

[warm shale]

Social Capital Gainer

[mint bronze]

Yea?

[timid silo]

yeah i have to write it a different way but yes

[mint bronze]

Ok, I wrote it that way so it is as close as possible to what your answers have

[timid silo]

oh i'm not even looking at that

[mint bronze]

Ah ok

Alright

Now it is time to use

,,c=\max(1,T(1))

[warm shale]

Social Capital Gainer

[mint bronze]

What this means is

the value of c will be set to whichever of 1 or T(1) is biggest

Therefore either

c=1, so 1>T(1)

Or

c=T(1), so T(1)>1

[timid silo]

lol what

oh ok

i mean, it's more helpful to just look at 1 or T(1) and check which is largest, but how do we evaluate T(1)

i don't even know which T we would plug that into or what c would be

[mint bronze]

We don't, we just have to acknowledge that it can either be bigger or less than 1

We don't even need to know what number it is

You will see why

Lets look at the first possibility, that 1>T(1)

Therefore c=1

We can use that in our inequality

,,T(n) \leq cn^2 +(1-2c)n+c

[warm shale]

Social Capital Gainer

[mint bronze]

Put c=1 in here for me

Everywhere except in cn^2

Leave that c as c

[timid silo]

wat

[mint bronze]

It will help us get the result we need

[timid silo]

cn^2 - n + 1

[mint bronze]

Yes

Now because we are dealing with natural numbers, what is the smallest value that n could be?

[timid silo]

1

[mint bronze]

So do you agree that

,,cn^2 -n+1 \leq cn^2

[warm shale]

Social Capital Gainer

[mint bronze]

Because n will make the left hand side smaller or at most equal to the right hand side

We are subtracting a number n that is bigger than or equal to 1

[timid silo]

i cannot look at something with so many variables and inherently know anything. what i said before was a guess. i was lost as soon as we plugged in something for most things but not every term in the expression

[mint bronze]

Sometimes we can be flexible like that, what we did was not wrong, it is just a slightly different way of writing it

Here is an example

h=5

h+h+h+h+h

Can be written as

5+5+5+5+5

Or

h+h+h+5+5

Is that wrong?

[timid silo]

no

[mint bronze]

So neither is what we did earlier

Anyway

[timid silo]

was this bc 1 is the min of natural numbers or because we evaluated something

[mint bronze]

Because 1 is the min of the natural numbers

[timid silo]

ok i think i need to take a break from all this stuff. thank you for your time

[mint bronze]

Ok sure, no worries

Would you like to continue later?

[timid silo]

sure i mean i have other examples we could work.. what time is good for you

[mint bronze]

I dunno, you take a break and dm me I guess

I can talk for a little longer later on to finish this example, but maybe not for other examples

Just in terms of time

[timid silo]

alright lets just pick it up another day

i thought this one was done?

[mint bronze]

Almost yeah

I actually could have done the very last thing we did a little more simply

I dunno

There are several ways to go about it

I think what I would recommend for you is to practice inequalities questions in general, as I feel that is mkstly what we have been discussing

[timid silo]

its a bit frustrating for me bc i have this algorithms course coming up and i am focusing on solutions to the book problems found online but i don't know exactly where i should be focusing. i guess i could look over the example homeworks i have access to and try to find similar ones in the book

[mint bronze]

Ah you are computer science

Ok

[timid silo]

not quite

bioinfo

[mint bronze]

Well I would still try to practice inequalities questions if you can find any

Bio informatics

That's cool

[timid silo]

correct. and unfortunately its a little too much like mathe matics 😂

[mint bronze]

Haha, that's what makes it cool imo

[timid silo]

no totally just not my strong suit. anyhow, thanks again

[mint bronze]

No problem, feel free to ping me a message if you have a specific question I might try find some inequalities questions and send them

If you would like

[timid silo]

ok that's probably a good idea

[mint bronze]

Ok

Well unless there's anything else

[timid silo]

no thanks have a good one

[mint bronze]

You too

[timid silo]

.close

[obtuse pebble]

Channel closed

Closed by @frank abyss

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• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
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[fierce lagoon]

Alright, so let's set up some dummy variables for the regular and the expanded

I'm gonna use r for regular, p for expanded

You can set up a system of equations from that

$$r + p = 650$$
$$25r + 65p = 21250$$
$$r, p\in\bN$$

[warm shale]

Umbraleviathan

[fierce lagoon]

You can't sell half a board, r and p must be natural numbers

Just as a reminder in case you somehow get half a board

Or something

[vast wren]

I have a question unrelated to the original problem

does the set of natural numbers include zero? or is it included in the set of whole numbers?

[fierce lagoon]

It does include 0, sometimes

[vast wren]

I see, thank you

(but it’s defo included in whole numbers, right?)

[fierce lagoon]

$\bN$ is similar to $\bZ^{+}$, sometimes 0

[warm shale]

Umbraleviathan

[fierce lagoon]

@leaden elbow

[vast wren]

all right, thanks

sorry for the interruption

[fierce lagoon]

I mean it's not much of an interruption

Still I'd rather you make a separate help channel

[leaden elbow]

yes

[leaden elbow]

yea i did thus

this*

but its not giving me the right answer

[fierce lagoon]

You sure?

[vast wren]

I found the answer

you just need to solve the system of eqs

[fierce lagoon]

I got natural number answers

[leaden elbow]

k im following my teachers steps and let me shoiw u

im confused where she got 43.5 from

[fierce lagoon]

This is a different question

[leaden elbow]

yes ik

u still follow the same steps dont u?

[fierce lagoon]

That's because 0.50 • 87 = 43.5

She distributed the 0.50

And yeah you follow the same

I did it similarly

[leaden elbow]

ooooh ok

im duuuuumb damn

so it would be 65 • 650?

[fierce lagoon]

What are you substituting

p = 650-r?

[leaden elbow]

uh ...

oh

[fierce lagoon]

65(650-r) = 65(650) - 65r

[leaden elbow]

are we substituting it the same bc u said you are doing it similarly

[fierce lagoon]

I am substituting but not the same way

I substituted r+p = 650 into 25(r+p)

[leaden elbow]

cause i got 323 and that doesnt seem right

[fierce lagoon]

For what

R?

No

[leaden elbow]

yea

[fierce lagoon]

No that's not right

[leaden elbow]

yea i got mixed up im gonna restart

Stuck here

[vast wren]

it’s not 25+65=21,250

r + p = 650
25r + 65p = 21,250

[leaden elbow]

mb i was jus following what my teacher was showing

[vast wren]

we can say

r = 650 - p

so

25(650 - p) + 65p = 21,250

25*650 - 25p + 65p = 21,250

40p + 25 * 650 = 21,250

from here you can find p

[vast wren]

then find r

[leaden elbow]

p=125? r=525?

[vast wren]

✅

[leaden elbow]

THANK UUUU

.close

[obtuse pebble]

Channel closed

Closed by @leaden elbow

Use .reopen if this was a mistake.

Please take a minute to participate in [our survey](#changelog message) if you haven't already!

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[timid silo]

can someone help please?

[vast wren]

y = 5x - 3
y = -6x + 8

we can say that 5x - 3 is equal to -6x + 8, can’t we?

since they’re both equal to y

[timid silo]

yeah

so x is 2?

[vast wren]

5(2) - 3 = 10 - 3 = 7
-6(2) + 8 = -12 + 8 = -4

no

since 7 is not equal to -4

[vast wren]

here

5x - 3 = -6x + 8

[timid silo]

do i put that in a calculator?

[vast wren]

no

you solve it

[timid silo]

is it 1?

[vast wren]

yes

how did you find it?

[timid silo]

i put it in a calculator

[bitter folio]

Lol

[uneven palm]

do you know how to solve equations?

[timid silo]

not really

[uneven palm]

alright so do you see how you can add 6x to both sides of this equation?

it cancels the -6x + 6x on the right side

[timid silo]

im on this now

[vast wren]

we want to collect the like terms on one side of the equation

#❓how-to-get-help

please don’t use this thread

[timid silo]

im really new to algebra, i dont really understand it

[uneven palm]

hmm

perhaps you should find a youtube video on how to solve basic equations

if you’re not already learning it

[vast wren]

go to Khan Academy

and create an account

[uneven palm]

it’d odd that you’re doing systems of equations without knowing how to solve a single variable equation first lol

[vast wren]

then start taking the algebra courses

[timid silo]

i would really just like to understand this first

[uneven palm]

yeah and you need a basic knowledge of equations to do this though

like 5x - 3 = -6x + 8

[timid silo]

could you help with this?

im leaning more towards A

[uneven palm]

how come?

[timid silo]

wait actually i think B i think the y step goes after that

[uneven palm]

no, you do want to set the equations equal to each other

so it is indeed A

[timid silo]

oh

[uneven palm]

eliminating x^2.. I dunno how you’d even do that

[timid silo]

okay so this one would be A now

[uneven palm]

no

you need a better grasp of algebra.. I can’t really explain everything here lol

but in short, solving a quadratic (with x^2) requires you to set it equal to 0

dividing both sides by x^2 just makes things more complicated

use Khan Academy like headshotcem suggested

[timid silo]

is it free?

[uneven palm]

yeah

[timid silo]

okay

[vast wren]

https://www.khanacademy.org/math @timid silo

[timid silo]

im not sure if its B or not because theres no Y only x