help-42

[calm coral]

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[halcyon ore]

wohop

[halcyon ore]

sry

how does 4.84 x 10^3 have only 3 sf?

isnt it 4840.0?

[kindred estuary]

In scientific, x 10^n part doesn't count

Only the stuff before

[halcyon ore]

oh ok

[kindred estuary]

So the 4.84

[halcyon ore]

but when im mutliplying

i should use it has 4840x3.6?

[kindred estuary]

No

You can still multiply the two like that

[halcyon ore]

oh ok

oh

i just realized sry

theres a denominator lo

Ok so

I multiply 3.6 x 4.84 first

answer can only have 2 sig figs

so 17?

and then division 17/0.5020?

or wrong approach so far

[kindred estuary]

Apply sig figs at the very end

[halcyon ore]

ohh ok

[swift laurel]

you don't round in the middle of a calculation, only at the end

[halcyon ore]

so

i got 34.70916335

butt only allowed 2 sig figs

so 35 if i rounded

er

what the heck did i do here

it supposed to be 3.4709

but i dogt 34.709

i keep putting

3.6 X 4.84

then dividing w/0.5020

Dah heck

[kindred estuary]

You still need to take in consideration of the x10^3

[halcyon ore]

Oh

oooh

[kindred estuary]

That doesn't count towards sig figs but it still affects the value

[halcyon ore]

Right

Ok thnxs i try again and see

ok

34709.16635

answer 2 sf only

so in this case

do imove the decimal place?

or just would be 35?

[kindred estuary]

Neither

[halcyon ore]

uh oh

[kindred estuary]

34709 is not the same as 35

[halcyon ore]

oh ok

for sure yeah

[kindred estuary]

Why would you move the decimal

[halcyon ore]

maybe i can make it

like scientific notation

34709.163 x 10^-4

and round the rest?

[kindred estuary]

That's not the same value

34709.16635 is not 34709.163 x 10^-4

[halcyon ore]

oh i meant 35 at the end

so 34709.16635 x 10^-4 not valid

[kindred estuary]

34709.163 x 10^-4 is not equal to 34709.16635

[halcyon ore]

oh

i am so lost now

[kindred estuary]

What's 34709.16635 rounded to 2 sf?

[halcyon ore]

would i use the number b/w dec point

like 34

i would say i want to use those two first digits

but since the decimal point is all the way there

[kindred estuary]

Why are you focused on the decimal point?

[halcyon ore]

so only 2 sig figs allowed

ah

[kindred estuary]

But you don't need to focus on the decimal point

[halcyon ore]

i can use non sig digits?

[kindred estuary]

You know you need to round 34709.16635 to 2 sf

Where would 2 sf be?

[halcyon ore]

first ones?

3 and 4

[kindred estuary]

Yes, then you would use the next digit to round appropriately

[halcyon ore]

7 make it round up

[kindred estuary]

Yes

[halcyon ore]

Ok

so 3 and 5

[kindred estuary]

So what would your final answer be?

[halcyon ore]

well

ofc not 35

but

[kindred estuary]

You need to review your sig fig rules

[halcyon ore]

Non zero = sig

Leading zero = not sig

Uh

Interior zero = sig

[kindred estuary]

You started with a value of 34 thousand 7 hundred 9, it does not become just 35

[halcyon ore]

oh

[kindred estuary]

It gets round as 35 thousand

35 000

The ending zeros do not count as sig figs unless a decimal point is there

[halcyon ore]

uh

[kindred estuary]

Where are you confused?

[halcyon ore]

so ending zeros

Where is that rule i just want to make this distinction

so 6

if opposite of 6

Not sig ok

[kindred estuary]

Rule 7

[halcyon ore]

oh

[calm coral]

@halcyon ore Has your question been resolved?

[halcyon ore]

hey

lets say i had like

0.005736000

idk

this would only have 4 sig fig then

those last 4 zeroes

are not before a decimal point

Oh but they are after a non zero

so 7

[calm coral]

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[naive steppe]

im incredibly lost

i struggle w vectors a lot and

what i would do is draw all of those vectors

what im confused with is i learnt that gravity always has a vector that is drawn vertically down so this one

but technically the gravity pulls you to the left because the force/tension is pulling you up or is this only the kinetic friction and not necessarily the gravity?

because for question b what ive seen is that you do: W = m x g x sin(32) x d = 20 x 9.81 x sin(32) x 3.80

but to me this doesnt make sense if gravity is always pulling you vertically down

and also for the normal force youd have an action reaction pair with a force thats facing the opposite way but what would that force even be in this case because normally its the grav force

<@&286206848099549185>

[fierce violet]

Work done on the suitcase by the force F
is given by:

WF=F⋅d⋅cos⁡(θF)

Since the force is parallel to the ramp, θF=0∘θF​=0∘, so cos⁡(0∘)=1. Therefore:
WF=160 N×3.80 m×1=608 J

Work done by the gravitational force

The gravitational force acts vertically downward. The component of the gravitational force along the incline (parallel to the ramp) is:
Fg=m⋅g⋅sin⁡(θ)

where g=9.8 m/s2g=9.8m/s2. Thus,

Fg=20.0 kg×9.8 m/s2×sin⁡(32.0∘)

Fg​=196N×0.5299=103.86N

The work done by the gravitational force is:
Wg=−Fg⋅d

Wg​=−103.86N×3.80m=−394.67J

The negative sign indicates that gravity is doing negative work (opposing the motion of the suitcase).

The normal force is perpendicular to the surface of the incline. Since the displacement is along the ramp and perpendicular forces do no work.

Wnormal=0J

Work done by the friction force

The frictional force is given by:
fk=μk⋅N

The normal force N is the component of the gravitational force perpendicular to the incline:
N=m⋅g⋅cos⁡(θ)
N=20.0 kg×9.8 m/s2×cos⁡(32.0∘)
N=196 N×0.848=166.2 N

So, the frictional force is:
fk=0.300×166.2 N=49.86 N

The work done by the friction force is:
Wfriction=−fk⋅d

Wfriction​=−49.86N×3.80m=−189.47J

The total work done is the sum of the work done by all forces:
Wtotal=WF+Wg+Wfriction+Wnormal
so Wtotal=23.86J

The initial speed of the suitcase is given as v0=0 m/s . The work-energy principle states that the total work done on an object is equal to the change in its kinetic energy(itk that u know this law):

Since v0=0, the equation simplifies to:
Wtotal=12mv^2

solve for v

and v=1.545m/s

[fierce violet]

and that's it itk ^^

[naive steppe]

ok ty but what abt my drawing

whats wrong whats correct

[fierce violet]

well the direction of vector of gravity is correct

Direction of F (applied force parallel to the ramp)

is correct

but itk that u should draw the components of gravity's vector

and for the friction force it should opposes the motion of the suitcase

(ps: the drawing is not that clear for me so i gave u some "general" informations about how to draw'em)

[naive steppe]

what i dont get is that for the grav force we use the friction force

like yeah that also pulls u down but its not the same as the literal Fg

in my eyes

[fierce violet]

the friction force doesn't pulls u down it resist the motion

[calm coral]

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[potent sorrel]

Might be a silly q .. Can someone elaborate this part a bit?

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[sand jackal]

Is my solution correct?

[potent lotus]

roloplays_1

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[calm coral]

@sand jackal Has your question been resolved?

[dry nymph]

no

[sand jackal]

<@&286206848099549185>

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@sand jackal Has your question been resolved?

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[formal tartan]

Could I plz get some help w/ verifying this?

I mean, like, proving it

[dull wagon]

What have you tried

[formal tartan]

@dull wagon

But idk where to go from here

[dull wagon]

simplify by multiplying the numerator and denominator (of the main fraction) by something

[formal tartan]

I just don’t know by what

[dull wagon]

with the goal of ridding yourself of nested fractions

[formal tartan]

Is there any way u could write it out for me?

If possible

I just don’t know how to do that

[dull wagon]

in the presence of nested fractions, it'd be useful to use the LCM of the denominators of the smaller fractions

[formal tartan]

So cos^2 x?

[dull wagon]

here you only have one small fraction, so you'd use the denominator of that

yes

[formal tartan]

Wait, so

Like this?

[dull wagon]

no

[formal tartan]

Oh

[dusky pasture]

no just multiply the denominator by cos^2x

[dull wagon]

just multiply by cos^2(x) would change the value

[formal tartan]

Like this, then?

[dull wagon]

no

[dusky pasture]

no

[formal tartan]

Or just the denominator of the small fraction?

[dusky pasture]

move the denom to the other side

[dull wagon]

both numerator and denominator

of the main fraction

[formal tartan]

I’m confused

[dull wagon]

i.e multiply by
cos^2(x)/cos^2(x)

[formal tartan]

Isn’t that what I was doing here?

Oh

[dull wagon]

that's indicating multiplication by cos^2(x), which would change the value

[formal tartan]

Like this?

[dull wagon]

yes

[formal tartan]

Oh, okay

Where do I go from here?

[dull wagon]

simplify the denom

[formal tartan]

Do I multiply it all out?

Oh

[dull wagon]

don't bother expanding the numerator component

when simplifying fractions it's usually better to keep stuff in factored form, expanding the numerator here works against you

[formal tartan]

But I thought you said this was right

I feel like this is rlly wrong

[formal tartan]

Right

Where am I going wrong?

Wait

[dull wagon]

I don't know your thoughts process to get that

your current goal is to expand the
$$\br{1- \frac{\sin^2(x)}{\cos^2(x)}}\cos^2(x)$$

[potent lotus]

ℝαμOmeganato5

[formal tartan]

Okay

Lemme try again

[formal tartan]

But I thought you said this was right

@dull wagon

[dull wagon]

the top line yes

I'm telling you what to focus on

[formal tartan]

Is this correct?

[dull wagon]

not what I recommended, but that is a valid way to approach this

[formal tartan]

I think this is also the method my teacher went with in class

[dull wagon]

just make sure that top fraction line is longer than the bottom one if going this route

or use () to make the order of operations clearer

[formal tartan]

Yeah, I think I got it

But was my process correct?

Bc this is what I did

It’s at the very top

Was I supposed to multiply cos^2 by both the numerator and denominator?

@dull wagon

Wait, no, I think it’s right

[dull wagon]

the way you wrote it no

they way i recommended was that you multiply the entire thing by $\underbrace{\frac{cos^2(x)}{cos^2(x)}}_{1}$
$$\frac{\cos^2(x)-\sin^2(x)}{1 - \frac{\sin^2(x)}{\cos^2(x)}} \cdot \frac{\cos^2(x)}{\cos^2(x)}$$
your teachers approach seems to be combining the terms in the denominator first by doing this
$$\frac{\cos^2(x)-\sin^2(x)}{\blue{1} - \frac{\sin^2(x)}{\cos^2(x)}} = \frac{\cos^2(x)-\sin^2(x)}{\blue{\frac{cos^2(x)}{cos^2(x)}} - \frac{\sin^2(x)}{\cos^2(x)}}$$

[potent lotus]

ℝαμOmeganato5

[formal tartan]

Oh, okay

So you just multiply the denominator by cos^2 x/cos^2 x?

@dull wagon

Bc that seems to be what the teacher did

[dull wagon]

no

you express the 1 in the denominator as cos^2(x)/cos^2(x)
or multiply that 1 ONLY by cos^2(x)/cos^2(x)

[formal tartan]

You can do that?

[dull wagon]

yes

[formal tartan]

Oh, okay

[dull wagon]

same principle being applied to simplify something like
1/2 + 1/6

[formal tartan]

Got it

Could u plz help me prove this?

@dull wagon

Idk where to begin

It’s my last one

[dull wagon]

there's a famous identity that allows you to simplify part of it immediately

[formal tartan]

Oh, I got it

Thank you

[calm coral]

@formal tartan Has your question been resolved?

[calm coral]

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[blazing coyote]

could I have a hint

[blazing coyote]

if S and T were subspaces this would 'ave been easier

[unique lynx]

Use the definition of span

[unique jackal]

the span of S is the set of all linear combinations of vectors in S

[blazing coyote]

yeah, I know

[unique jackal]

and likewise for T

[blazing coyote]

I thought of that

[unique jackal]

but S is a subset of T…

[unique lynx]

if v is in S, then it can also be written as a li ear combination of vectors in T

[blazing coyote]

but does it make sense to talk about a non-finite spanning list

[unique lynx]

The rest should follow

[unique jackal]

yes, yes it does

in fact, the whole set V (which is infinite over R or C) is a spanning set for V, for obvious reasons

[blazing coyote]

hmm, okay

let me try

okay

got it

thanks

.close

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[dreamy pendant]

Hi I am specifically confused about part C but in general i am not feeling confident

I believe this is certainly a permutation problem

In part A I believe I am correct that its just 12! cause 12 books, 12 spots, no replacement - simple

Part B I just made it so theres 12 books but we are only arranging 9 out of those books so adjust the permutation formula, its a group fo 9 out of the 12

part C i am the least confident - it seems way too easy

i just pretndeded it was 4 sepearte problems and I dont think that works

[shadow zenith]

how is it 12? i feel like its way more than that

even on the first row, it could be ABC, DEI, AEI, JBC, etc.

[dreamy pendant]

no its 12 factorial my friend

[shadow zenith]

oh shoot 😆

[dreamy pendant]

Basically question I have is can you just add up pemrutation equations

Like I did for part C is that legal

[shadow zenith]

yeah, i think its just permutation inside permutation

[dreamy pendant]

inside?

or multiply 6 * 4

[shadow zenith]

inside

[dreamy pendant]

i dont know hwo to do that

.close

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[shrewd herald]

physics problem: 18.3 is the wrong answer

[shrewd herald]

it was one of my previous attempts

[proper elk]

What have you done already, where do you feel stumped, what do you think you want to try next?

[shrewd herald]

one second...im sending a pic

is it ok to assume that the accleration when its frictionless is 1.0 m/s^2

[proper elk]

I wouldn’t

[shrewd herald]

how would i go about to solve this q

[proper elk]

You may want to treat it as a system

Your approach didn’t seem bad though and you did good drawing it out

[shrewd herald]

wdym

[proper elk]

You know with one set of forces (one with and one without friction) you get the result that when MuK=.22 the proportion between the two accelerations is 3/5

[shrewd herald]

so i would set up a proportion where a(friction)/a(frictionless) = 3/5

[shrewd herald]

a(frictionless) = gsintheta

a(friction) = gsintheta - MuKgcostheta

[calm coral]

@shrewd herald Has your question been resolved?

[calm coral]

@shrewd herald Has your question been resolved?

[calm coral]

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[woven mulch]

im in my quadratics unit and I am not understanding how they got these 2 answers

[calm coral]

@woven mulch Has your question been resolved?

[azure rose]

I'm assuming with it being "quadratics unit" that you do not know calculus?
For this question I would assume it is on earth and follows earth gravity. You can calculate the time taken with the equation f(x) = (velocity) - (acceleration)x. You would set it equal to 0 in order to figure out when it's velocity is 0 and than solve for x (when velocity is 0 it can either mean the object has stopped moving or has turned around and will than start moving again (aka the highest point it will get before "stopping" and turning around to start heading down))

[woven mulch]

no

no calc

lemme show you the equation they gave

h(t) = -16t^2 + vi t + ho

[calm coral]

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[woven mulch]

.reopen

[calm coral]

✅

[azure rose]

This is the "position" function

vi = initial velocity and ho = acceleration

If you wanted to calculate time with this equation you will plug in the given info and solve the equation for t while it is equal to 0

$0 = -16t^2 + v_it + ho$

[potent lotus]

dragonbreath

[woven mulch]

Lemme try

[azure rose]

I am not sure as to -16 being a good value though

[azure rose]

Oh yeah, I guess I misinterpreted the 16, so yeah I would use h0 as initial height which is 215

actually, hmm, given this equation setting equal to 0 won't solve for the same thing. It's a quadratic equation and the vertex of the parabola is where the highest point will be, so using the vertex equation will solve for the highest point in time and than simply plug in in that time (x) will solve for the highest distance (y)

@woven mulch

[woven mulch]

This si what I got so far

The bottom one

This is so much harder than my ap physics projectile motion 😅

[azure rose]

Algebra based physics is gross tbh

[woven mulch]

The bottom one is what I got

So wouldn’t it be 199

Honestly dk atp

[azure rose]

wdym?

[woven mulch]

Lemme retry this

[azure rose]

use your -b/2a equation to determine the vertex point (which will give a value (x or t) of time)

[woven mulch]

.close

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[karmic ridge]

i know what im supposed to do, but im not sure how i would find such matrix

[swift laurel]

if a set of vectors is linearly dependent, then at least one of them can be written as a linear combination of the others

[karmic ridge]

ohh

i thought it said linearly independent

thank you!

.close

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[halcyon ore]

would be 201,145 g?

and is sig figs needed for something like this

[swift laurel]

yes, and yes

[halcyon ore]

hm

so 201g?

[swift laurel]

no

[halcyon ore]

i should use the originals from the word problem?

[swift laurel]

but this is not the same measurement

[halcyon ore]

yeah

True

[swift laurel]

it is 1000x smaller

[halcyon ore]

hm

so 201145 g

isnt it like

exact number

[swift laurel]

but we still need to apply sig figs

[halcyon ore]

oh

[swift laurel]

neither of the provided numbers are exact

[halcyon ore]

oh true

well are we using the most uncertain one then

[swift laurel]

yes

[halcyon ore]

so 245000

have 3 sf

OH

OMG

201000

[swift laurel]

it has 3 sig figs but it's rounded wrong

[halcyon ore]

oh

[swift laurel]

yes

[halcyon ore]

thnx ok perfect

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[cedar arrow]

im doing a project (making a game) and i have to write documentation, most of it is explaining the math behind it would anyone help me out and read it and tell me if it makes sense

[calm coral]

@cedar arrow Has your question been resolved?

[calm coral]

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[halcyon ore]

can i multiply them by 2 as well

like

32 g oxygen

4 g hydrogen

and so on

or not needeed

[ionic valley]

works yeah

as long as the ratio is still 81

8:!*

8:1*

[halcyon ore]

ok perfct

thnx

.close

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[green cosmos]

Problem 27

[green cosmos]

Would my absolute max be at 0 because there is a value at 2?

Or is it -1

I’m unsure

[swift laurel]

do you have the sketch?

[swift laurel]

there should be an open circle at (0, 2)

[green cosmos]

Oh damn

So would my abs max be -1

Local min 0

That’s it?

[swift laurel]

no, it can still be bigger than that

[green cosmos]

Apologies I don’t understand

[swift laurel]

f(0.1) = 2 - 3(0.1) = 1.7, so 1 can't be the maximum

[green cosmos]

,w graph tanh

[potent lotus]

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Requested by light945850

[green cosmos]

,graph tanh

,,graph tanh

[potent lotus]

Light

[green cosmos]

,

Aisha’s

.close

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[drifting seal]

,w graph tanh(x)

[potent lotus]

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[drifting seal]

@green cosmos

[calm coral]

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[subtle mortar]

guys how do i solve this

[remote mural]

Calculus allowed ?

[upper sparrow]

What's definition 1.1?

[remote mural]

I wonder what that reaction means

[upper sparrow]

The cat one? It's me saying hi

[glacial steeple]

You don’t need to know definition 1.1 to get the answer

[glacial steeple]

What work have you done? If none, how would you approach this

[upper sparrow]

(you may not, but as they explicitly tell you to use it, I'm very sure you would get marked down if you didn't)

[glacial steeple]

Agreed

[subtle mortar]

i solved it but my friend says my steps r wrong

[remote mural]

Absolutely right

[glacial steeple]

Last step is wrong

[subtle mortar]

how so

[glacial steeple]

If you want it all on the right side it would have to be 1/2(x+4)

[remote mural]

Wait up

Last step is wrong

Yeah

[subtle mortar]

so it should be x - 2y + 4 = 0 ?

[glacial steeple]

Why are you manipulating the form

You were done at point slope

[subtle mortar]

wait so is my first answer correct?

[glacial steeple]

Just keep equations in point slope to reduce error

[subtle mortar]

oh aightt ty 🫶

[calm coral]

@subtle mortar Has your question been resolved?

[calm coral]

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[amber valve]

I'm not sure what I did wrong here

I started by isolating / multiplying both sides

ended up with it 17 = 64 - 8 x 6^x 8 x 6 = 64

6^x = 47/8

6^x = 47/8 x = logv6 (47/8)

x = logv6 (47/8) = 0.572 ???

!

[upper sparrow]

,calc log(47/8)/log(6)

[potent lotus]

Result:

0.98824986860156

[amber valve]

oh

I tried that at first

but they said it wasn't right'

The answer isn't 0.988

[upper sparrow]

Hmmm, strange

[shell rune]

Should be

Photomath approves

[amber valve]

its not i entered it alrady

[upper sparrow]

Thing's tweakin' then

[amber valve]

its not multiple answers either?

thats the only thing i can thnk of

[shell rune]

Nope

[amber valve]

sigh must be tweaking then

thx

anyways

.close

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[rain yew]

Can someone help me solve this idk how to do it

A graphing calculator is recommended.
For the limit
lim
x → 3
(x3 − 4x + 1) = 16,
illustrate this definition by finding the largest possible values of 𝛿 that correspond to
𝜀 = 0.2
and
𝜀 = 0.1.
(Round your answers to four decimal places.)
𝜀 = 0.2 𝛿 =
𝜀 = 0.1 𝛿 =

[calm coral]

@rain yew Has your question been resolved?

[rain yew]

<@&286206848099549185>

[calm coral]

@rain yew Has your question been resolved?

[rain yew]

.close

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[wintry trench]

Currently doing test corrections and I wanted to double check, is the answer for d DNE?

[dark fox]

It is 4

[wintry trench]

i got a point marked off on that

[leaden thunder]

,rotate

[dark fox]

Go argue with your teacher

With confidence

[wintry trench]

i most definitely will, thank you for the help

.close

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[wintry trench]

Can someone double check my work and let me know if I have made any mistakes

[dark fox]

No

[wintry trench]

thanks

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[violet grail]

hello, i solved for vf and got this. I use chat gpt to check my answers and it is saying i got this wrong can someone help me to see why im wrong

[leaden thunder]

Show your work

Also don't use chatgpt for math

[violet grail]

alright

my phone is dead my camera is bad

i will explain what i did

i split it into x and y

i know x has values of V D T

v being 230

and Y has vf vi a d and t

d being 800

a being - 9.8

and vi being 0

i solved straight for vf since thats what i thought the question was asking

and i used forumla of d eqials vit + 1/2 a t ^2

and thats where i got my answers of 125

[leaden thunder]

This is not legible

[violet grail]

it would be if my phone camera was used

[leaden thunder]

Your acceleration and velocity should be the same direction

But looks like one is positive but the other is negative

[violet grail]

i have it split up into x and y

and i thought a is always -9.8

and the other v is horizontal

[upper sparrow]

(also the question is not asking for just the vertical velocity)

[violet grail]

yes my teacher taught us a way to solve for stuff int he vertical way while having vf vi a d t

but the horizontal velcoity is what i solved for

and i dont know if its orrect or not

[leaden thunder]

v is a vector with x and y components

What you choose to be positive vertical is up to you, you just have to be consistent

You can't have acceleration negative and towards the earth while velocity point up towards space

[violet grail]

alright

so was my answer correcct

[leaden thunder]

No

For the reason I stated above

#help-42 message

[violet grail]

could you help me through the questions please

question*

.close

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[royal raft]

Im wondering if there is an easier way to go about this question

[royal raft]

sorry for the mess

but I did not know how to solve that integral off the top of my head

you may have to click on the picture to view the entire question

the answer was just arctan(y/x) + pi/2 if I did my math right

there’s probably some simpler logic to get to that I’d assume

[calm coral]

@royal raft Has your question been resolved?

[strange tinsel]

theres a mistake, Vx and Vy are wrong

and that changes everything

[royal raft]

oooof

I always make these mistakes

[strange tinsel]

:p

you shouldnt get another arctan

[calm coral]

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[versed osprey]

How do I solve quadratic equations using the squaring method

[calm coral]

@versed osprey Has your question been resolved?

[worthy igloo]

is it completing the square

?

[versed osprey]

Yes

[worthy igloo]

hm ok

[versed osprey]

I have no idea where to start

[worthy igloo]

can you give an example tho let me solve and explain

[worthy igloo]

ok

ok so the first step is to divide both sides by the coefficient of x^2

but the coefficient is 1 so it wont show

[hearty hamlet]

thats so ez

[worthy igloo]

the next step is to square the coeficient of x which is 3 and add to both sides

[worthy igloo]

will you explain?

[versed osprey]

Please continue

[hearty hamlet]

explain what

?

[worthy igloo]

ok

so we have to factorize them cos now the equation is

x^2 + 3x + [3/2]^2 = 10

idk if you understand tho

i moved the constant to the rhs

[versed osprey]

I sort of get it

[worthy igloo]

what is confusing?

[versed osprey]

The brackets

Should they be (b/a)

[worthy igloo]

i had to cos 3 over 2 is all squared

[worthy igloo]

yes but all squared

dyg?

[versed osprey]

Okay please continue

[versed osprey]

Yeah

[worthy igloo]

ok

[worthy igloo]

you also add [3/2]^2 to the other side

cos whatever you do tho the lhs you must do to the rhs

maths rule

dyg?

so it will be

[versed osprey]

Ye so far

[worthy igloo]

x^2 + 3x +[3/2]^2 = 10 + [3/2]^2

so now we factorize

which is we forget 3x

so we get

[x+ 3/2]^2 = 10 + 9/4

[versed osprey]

X^2+[3/2]?

Or do we also remove the 2

[worthy igloo]

no the purpose of factorizing is to bring the ones raised to power together

so its

[x+3/2]^2

so they are both squared

shld i countinue?

[versed osprey]

Yes please

[worthy igloo]

we need to add the two numbers at the rhs

so we will multiply the rhs by 4 to get common denominator'

so thats 40/4 + 9/4

so we add

we can even use 4 to cut 40

but the denominator wont be same

so we add

so its 49/4

• if we divided 4 by 40

we would have gotten 10 over one

and the lcm will be 10 + 9/4

19/4

dyg?

[versed osprey]

Ye

So I shouldn't necessarily divide the 40

I should add it then divide it

[worthy igloo]

yh

its more straight forward

[worthy igloo]

it will give a decimal

lets countinue

[versed osprey]

Will affect my results though if I happened to divide it first?

[worthy igloo]

yes it will vary

buttt

[versed osprey]

As in would the teacher see it as incorrect

[worthy igloo]

if you have a situation like 18/6 + 4/3

if you divide 18 and 6 by 3

you get 6/3

and the denominator are the same dyg?

[worthy igloo]

i dont know but from my view its better to not divide

are you there?

[versed osprey]

Yes

[worthy igloo]

ok

[versed osprey]

And yes

[worthy igloo]

so we have [x+3/2]^2 = 49/4

so what do we do when we have a square we want to cancel out?

we square root both sides

[versed osprey]

Square root?

[worthy igloo]

do yk the square root sign

[versed osprey]

Ye

[worthy igloo]

yes a symbol

[worthy igloo]

send a pic so i can confirm its not on my keyboard

√

this is it

[worthy igloo]

yesss

good so we do

√[x+3/2]^2= √49/4

so for the lhs square cancels square

[versed osprey]

Do we divide the 49 before we find the square root?

[worthy igloo]

and the square of 49 is 7

[worthy igloo]

no divide will give a decimal

and decimals have no square

[worthy igloo]

while that of 4 is 2

so now its

[versed osprey]

7/2

[worthy igloo]

correct

so now we have x + 3/2 = 7/2

oh wait

[versed osprey]

Which is 3+1/2

[worthy igloo]

how?

[versed osprey]

7/2

Or am I wrong?

Wait

3.50

[worthy igloo]

no i made an error

before we square we take 3/2 to the other side

can u wait for a min let me solve again so i dont mis lead you?

[versed osprey]

Thank you for ur effort lol

Yes please do

[worthy igloo]

ok

i am back

i am not wrong a rule in maths

is that one you square root a no +or - is add as its coefficient

so its x + 3/2 = + or - 7/2

dyg?

[versed osprey]

Ye

[worthy igloo]

so we start with negative

x + 3/2 = - 7/2

multiply both sides by -3/2

in layman word im taking 3/2 to the other side

so x = -10/2

= -5

and for positive

so we have x = 7/2 - 3/2

7-3=4

so thats 4/2

which is 2

sooo

x= -5 or x=4

dyg?

[versed osprey]

They're both equally correct?

[worthy igloo]

they are both the answers

you cant chose one

[versed osprey]

Confusing but I get it

[worthy igloo]

in completing the square

you have two answers

positive and negative
positive and positive or negative and negative

[versed osprey]

I'm gonna try out some problems from my own textbook to make sure that I really understand the method

If I have any more issues is it fine if i contacted u?

[worthy igloo]

because once you √ + or - is automatic

[worthy igloo]

yh sure let me add you?

[versed osprey]

Ty btw

[worthy igloo]

good also learn quadratic formula method

its straight forward

[worthy igloo]

my pleasure

[versed osprey]

It's very easy to understand

Thanks for all ur help

[worthy igloo]

ok so if a question dosent specify what formula to use go with the formula one

its worth it'

[worthy igloo]

you welcomeeeeeee

[versed osprey]

How do I close the channel?

[worthy igloo]

.close

i think

try

[versed osprey]

Ty again

.close

[calm coral]

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[reef raft]

can somone explain this to me? when text is put into big blocks like this it confuses me alot

[reef raft]

here is the question itself

[torpid summit]

Hello!

I'll explain it in steps

[minor gust]

zero rule

=0

=0

=0

[torpid summit]

Step 1
They use the "zero product property", which basically means that since they multiply to 0, each equation equals to 0
This gives you the equations
3x+p = 0
5x^2 - 45 = 0
2x^2 - 16x + 6p = 0

Step 2
They first modify the equation 3x+p=0
This gives you x=-p/3

Step 3
They then modify the equation 5x^2 - 45 = 0
This gives you x = +3, -3

Step 4
They then get the x value from 2x^2 - 16x + 6p = 0
Using the quadratic formula, they get x =

4+sqrt(16-3p) and 4-sqrt(16-3p)

Step 5
Lastly, they add all these x values together, set it equal to 20/3, and solve for p

For any of the detailed algebra look at the paragraph im lazy

[reef raft]

very sorry but one last question what do you mean x=+3,-3 or is that a typo?

[torpid summit]

Since x^2 = 9, x can equal both positive or negative 3

[reef raft]

oh

[reef raft]

once again thanks alot and ur certainly not lazy

[torpid summit]

lol np

[reef raft]

.close

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[strange trail]

.close

Sorry idk why my discord is being weird and not sending the question

[strange trail]

Let me resend it

I’m just confused as to what to input as the elementary matrices

[upper sparrow]

Remember yesterday I mentioned each row operation corresponded to an elementary matrix?

[strange trail]

Yessss

And in this case I did 2 row operations

So there will be 2 elementary matrices

[upper sparrow]

Yep

[strange trail]

Are the elementary matrices just going to be
[1 0 0]
[0 1 0]
[0 4 1]

And

[1 0 0]
[0 1 0]
[0 0 1]

[upper sparrow]

Not the second one, that’s just the identity matrix, so corresponds to no change

[strange trail]

Ahh I see

[upper sparrow]

Remember the idea that we could basically build up a matrix from its RREF form from yesterday?

[strange trail]

Yes

[upper sparrow]

Cool, and of course our RREF here is the identity matrix (which is just as well, as they claimed that the matrix was invertible)

[strange trail]

Yes

[upper sparrow]

And of course, the first matrix you have is fine, that’s an elementary matrix as is

But now, taking that first matrix you have, we wanna turn that into the original matrix, and we know we can do that by multiplying the last row by -2 (because in reverse, we multiplied it by -1/2 to get to that step)

You know the elementary matrix that corresponds to multiplying the last row by -2?

[strange trail]

Wait so

I take my
[1 0 0]
[0 1 0]
[0 -8 -2]
Matrix and multiply it by -2??

[upper sparrow]

Noo, as in

You want the
[1 0 0]
[0 1 0]
[0 4 1]
matrix, and then you want to multiply the last row of it by -2

And there’s an elementary matrix that does that for you, what does it look like?

[strange trail]

Ohhh okay

I would get my original matrix

[upper sparrow]

You would, yep

[strange trail]

My original matrix?

sorry I’m a bit confused like

Wait actually I think

I got it

[1 0 0]
[0 1 0]
[0 -2 -2]
Because if you multiply this with the other matrix, you would just get your original one

Is that correct?

[upper sparrow]

Not quite, you’re quite close though!

In particular;

,w [[1, 0, 0], [0, 1, 0], [0, -2, -2]] * [[1, 0, 0], [0, 1, 0], [0, 4, 1]]

[potent lotus]

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[strange trail]

Oh sorry, I forgot that matrix multiplication was different

[upper sparrow]

That multiplies the last row by 2, but also subtracts 2* the second row from the third

[strange trail]

That it’s like a different process you’re not just multiplying each element with each other

[upper sparrow]

Yep, it’s slightly different-

Remember yesterday I said what they all would look like #help-19 message

[strange trail]

Yes I see

[upper sparrow]

so to multiply the last row by -2, we just need the diagonal matrix where the last diagonal entry is -2, and all the rest of the diagonal entries are 1

[strange trail]

Ahhh so
[1 0 0]
[0 1 0]
[0 0 -2]

[upper sparrow]

Yep exactly that! Now

,w [[1, 0, 0], [0, 1, 0], [0, 0, -2]] * [[1, 0, 0], [0, 1, 0], [0, 4, 1]]

[potent lotus]

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[strange trail]

So those are my two elementary matrices 😁

[upper sparrow]

our original matrix back, and it’s a product of elementary matrices!

[strange trail]

Thank you so much!!

That makes sense

Oh I got it wrong 😅

[upper sparrow]

Oh

[strange trail]

What 😅

[upper sparrow]

Other way around, order matters!

[strange trail]

Oh 😅😅

[upper sparrow]

You can’t swap the order of matrix multiplication

[strange trail]

Yeah, I completely forgot that

It’s OK I have another attempt on the question

This is the other one

Am I going right about it so far

[upper sparrow]

You removed it

[upper sparrow]

Yep that’s all good

Well, wait-

[upper sparrow]

(Technically that should remain as -1/2 for a while longer, but anyways at this point you could even clear out all the entries above that bottommost 1 as well )

[strange trail]

Oh yeah sorry let me fix that and then continue to clear

[upper sparrow]

Yep cool

[strange trail]

A little bit lost, where do I go from here on out 🫠

[upper sparrow]

All you need for the identity is to get rid of that -1/2

[strange trail]

R1 + 1/2R3 ?

[upper sparrow]

Saw that

[strange trail]

And there it’s good now

[strange trail]

I’m blind 😞😞

[upper sparrow]

Happens to us all

[strange trail]

But there it’s the identity now !!

Now what are my steps to figuring out the elementary matrices

With the other one, it was easy because it was just two

[upper sparrow]

Basically writing out the corresponding "reverse" elementary matrices from right to left

The last operation we did was to add 1/2 of the last row to the first, so the first one we'll do is subtract 1/2 of the last row to the first

And that one corresponds to the matrix
[
\pmqty{1 & 0 & -\frac12 \ 0 & 1 & 0 \ 0 & 0 & 1}
]
which, of course, is incidentally the last matrix we got before the identity, and was already in elementary form (so really we could have actually stopped when we got to the step before, and just had the last matrix be our rightmost one!)

[potent lotus]

@upper sparrow

[strange trail]

OK, so that’s one of our matrices

[upper sparrow]

Yep, the rightmost one!

Now, the second operation we did was add row 2 to row 1, so now we wanna subtract row 2 from row 1

So the (1, 2) entry of the elementary matrix should be -1 (to add -1 times the second row), and the diagonals all 1, everything else 0, right?

[strange trail]

Yes I guess so

[upper sparrow]

Guess so?

,w [[1, -1, 0], [0, 1, 0], [0, 0, 1]] * [[1, 0, -1/2], [0, 1, 0], [0, 0, 1]]

[potent lotus]

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[strange trail]

I’m just writing it out to see how it plays out !

But yes

[upper sparrow]

fair fairs

[strange trail]

That is our original matrix

Well, by original, I just mean

slowly getting there

[upper sparrow]

I got what you mean

Now, I'm gonna make you tell me what the elementary matrices I should pick next should be

So, the next operation was multiplying the second row by -1/2, so now we wanna multiply the second row by -2, what does the elementary matrix that does that look like?

[strange trail]

Sorry I’m a little lost I’m just trying a few things out

I just forgot which matrix I should be multiplying by two for the second row

Maybe
[1 0 -1/2]
[0 2 0]
[0 0 1]

[upper sparrow]

Almost, not quite! (also I mistyped, I meant -2 rather than 2)

That's what you get when you multiply by the appropriate matrix, but rememeber the elementary matrices have certain forms

In particular, these ones

So, what would the elementary matrix that multiplies the second row by -2 look like, from here?

[strange trail]

Diagonal entries are all 1 except the second row where it’s -2 and the rest is 0?

[upper sparrow]

Yep, that's it! The matrix
[
\pmqty{1 & 0 & 0 \ 0 & -2 & 0 \ 0 & 0 & 1}
]

[potent lotus]

@upper sparrow

[upper sparrow]

And, to check what we're doing

,w [[1, 0, 0], [0, -2, 0], [0, 0, 1]] * [[1, -1, -1/2], [0, 1, 0], [0, 0, 1]]

[potent lotus]

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[strange trail]

Yess okay

[upper sparrow]

Just as we wanted

[strange trail]

Now we do r2-2r1?

[upper sparrow]

Yep, and the matrix that corresponds to doing that would be?

[strange trail]

1 0 0
-2 1 1
0 0 1?

[upper sparrow]

Careful, all the diagonal entries are 1

[strange trail]

I edited it. Does it look right now?

I hope it works out

[upper sparrow]

Remove the non-diagonal 1 entry as well

Only the (2, 1) entry and the diagonal entries are nonzero

So, basically

,w [[1, 0, 0], [-2, 1, 0], [0, 0, 1]] * [[1, -1, -1/2], [0, -2, 0], [0, 0, 1]]

[potent lotus]

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[upper sparrow]

Which is this

[strange trail]

Ohhh I see

OK, that makes sense

Also, when were calculating the number of elementary matrices

We’re just counting the amount of row operations we did right

Or is that not how it works

Because that would mean we have seven matrices

[upper sparrow]

It is [how it works], each row operation corresponds to multiplying by one elementary matrix!

[strange trail]

So I’m gonna have to fill out all of these individually

Damn 😞

[upper sparrow]

Yep, and so far we have three of them already, from right to left...

[upper sparrow]

...this one, then...

[upper sparrow]

...this one, then...

[upper sparrow]

...the one on the left here

[strange trail]

Kk yes

Does the swapping row elementary matrix really have to be put in

Okay so the next one would just be

1 0 0
0 0 1
-2 1 0

[upper sparrow]

No -2 there should be 0

Other than that, all good

,w [[1, 0, 0], [0, 0, 1], [0, 1, 0]] * [[1, -1, -1/2], [-2, 0, 1], [0, 0, 1]]

[potent lotus]

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[teal crag]

hello on the other line

[strange trail]

Oh okay okay makes sense

[teal crag]

do y'all need help

martix i see

matrix

[upper sparrow]

(remember the elementary matrices are independent of the matrix you're operating on, and only depend on the operation you're doing)

[strange trail]

Okayy we have 3 left

The next operation was

We have to do R2-4r1

[upper sparrow]

Yep

Remember that all the diagonal entries for this one are 1, and there's only one entry that's not on the diagonal that's nonzero

[strange trail]

Is it 4 😅

[upper sparrow]

it's -4, but where should that entry be?

(because you're adding (-4) * row 1 to row 2)

[strange trail]

1 0 0
-4 1 0
0 0 1
?

[upper sparrow]

Well, let's find out

,w [[1, 0, 0], [-4, 1, 0], [0, 0, 1]] * [[1, -1, -1/2], [0, 0, 1], [-2, 0, 1]]

[potent lotus]

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[upper sparrow]

[strange trail]

Okayyy wohoooo

Last one 🙏

[upper sparrow]

Yep and that one would be?

[strange trail]

-2 1 0
0 1 0
0 0 1 ?

[upper sparrow]

One of the entries in that is wrong

Do you know which one?

[strange trail]

Hmm

The 1

In the first row

[upper sparrow]

Correct, what should it be?

[strange trail]

0

[upper sparrow]

Yep, now, our final drum roll

,w [[-2, 0, 0], [0, 1, 0], [0, 0, 1]] * [[1, -1, -1/2], [-4, 4, 3], [-2, 0, 1]]

[potent lotus]

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[upper sparrow]

Our original matrix

[strange trail]

Let’s goooo 🙏

This looks good right 🙏

Thank you very much I know this took a while thank you for your time

[upper sparrow]

Also I think I mixed myself up we only need 6 of them I think, rather than 7

The identity matrix isn't really needed (though technically, you could argue that the identity matrix is elementary as well, multiplying each row by 1 )