help-41

[narrow night]

I added the vectors together and got 19.6 km/h

For x : 15cos(35) + 5cos(15) For y : 15sin(15) + 5sin(15)

then i found that the hypotenuse and got 19.6 km/h

now as for the angle I did tan-1(y/x)

but that gives me 31 degrees

and the answer is 41 degrees 19.6 km/h

idk what i did wrong

could somebody help me, thank you in advance

[mint nacelle]

so (r cos 35, r sin 35) is the resultant vector of the sailboat

and if you add the trajectory plus tidal current, you get (15 cos theta + 5 cos 15, 15 sin theta + 5 sin 15)

so if you equate the slopes of both vectors, you get tan 35 = (15 sin theta + 15 sin 15) / (15 cos theta + 5 cos 15)

,w tan(35 deg) = (15 sin theta - 5 sin(15 deg)) / (15 cos theta - 5 cos(15 deg)), solve for theta

take the acute solution cause it's the only one that make sense if you draw it

,calc 0.58353 * 180/pi

[grizzled pagoda]

Result:

33.433806219269

[mint nacelle]

oh weird

wait the current moves against the boat, so change the + to -

,calc 0.63820 * 180/pi

[grizzled pagoda]

Result:

36.566166485249

[mint nacelle]

that is weird

[narrow night]

I dont even get whats going on

lmao

[grizzled pagoda]

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[mint nacelle]

,calc 0.49661 * 180/pi

[grizzled pagoda]

Result:

28.453657063992

[mint nacelle]

oh whatever

[mint nacelle]

we don't know the resultant velocity of the sailboat, but we do know its trajectory

so (r cos theta, r sin theta) with theta = 35

and then the course of the sailboat is (15 cos theta, 15 sin theta), cause the velocity is 15

and then we add the wind vector, which is against the boat, so (-5 cos 15, -5 sin 15)

oh 15 km/h relative to the water

it's such a messy problem

[narrow night]

i give up

icl

i dont understand

[amber wasp]

@narrow night Has your question been resolved?

[narrow night]

<@&286206848099549185>

In good conditions, a sailboat can move at 15 km/h relative to the water. (a) What course must this sailboat follow (in other words, what orientation must the bow of the sailboat have) so that its trajectory is oriented at 35° north of east while it is experiencing a tidal current of 5 km/h directed 15° north of east? (b) What will then be the modulus of its speed relative to the earth?

[amber wasp]

@narrow night Has your question been resolved?

[amber wasp]

@narrow night Has your question been resolved?

[red river]

Hello @narrow night . In fact the vectors tidal current + sailboat must be equal to the trajectory vector (35° orientation)

,w (15sin(x)+5sin(15°))/(15cos(x)+5cos(15°)) = tan(35°)

[grizzled pagoda]

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[red river]

,calc 0.72512*180/pi

[grizzled pagoda]

Result:

41.546315640526

[red river]

Notice that this equation would be really hard to solve without this kind of calculator

Instead, you can use cosine law first in order to find the speed

Anyway, I suggest you first draw a correct diagram for this problem

[amber wasp]

@narrow night Has your question been resolved?

[narrow night]

icl you got it wrong i think i did it right

[red river]

I got 41°, which should be the correct answer according to what you mentioned. Did you get that too?

[narrow night]

oh nvm yeah you got it right

but idk how

[red river]

I wrote some suggestions above in case you're still interested

[narrow night]

alright so i did wrong

i thought it was 15 km/h of 35

that’s why

all righty thank you

hey man @red river

got a question

when i’m on a bus for example and i put a ball in the aisle

why does the ball go backward

[amber wasp]

@narrow night Has your question been resolved?

[narrow night]

how do i find the angle now

is this even good

can someone help me pls

In good conditions, a sailboat can move at 15 km/h relative to the water. (a) What course must this sailboat follow (in other words, what orientation must the bow of the sailboat have) so that its trajectory is oriented at 35° north of east while it is experiencing a tidal current of 5 km/h directed 15° north of east? (b) What will then be the modulus of its speed relative to the earth?

[deep karma]

i can do this shit in my head but i rlly wanna take a shit

ill help u once im back

[narrow night]

okay

thanks

icl this question is not a 12 th grade question

never seen some like that in my life

fuck this question

[narrow night]

this guy did not come back 😭

.close

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[deep karma]

hi

[deep karma]

😭

i just opened

discord

LMAO

[narrow night]

no way lmao

all good though

i found the answer

[deep karma]

yay

[narrow night]

using my peanut brain

[deep karma]

pepenut brain

[narrow night]

😂

have a good one man

[deep karma]

i think you just solve vectors here

[narrow night]

yeah

[deep karma]

yeah have a great dau

[narrow night]

i just didn’t understand the problem

it’s 1:40 am for me

am i cooked

.close

idk how to close this

[narrow night]

.reopen

[amber wasp]

✅

[narrow night]

.close

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[split sail]

solve this eq

[split sail]

pi^x = 4x

[primal holly]

,w (pi)^x = 4x

[grizzled pagoda]

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[split sail]

ok

.cllose

.close

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[keen pawn]

Trying to write this in its logical form

[keen pawn]

so $\forall \varepsilon >0 , \exists M>0 , G(M) \implies L( \varepsilon)$

Where G(M) = x>M

[grizzled pagoda]

Veni, vidi, perii is not f(wai)

[keen pawn]

and $L(\varepsilon) \equiv $|f(x)-b| < \varespilon$

[grizzled pagoda]

Veni, vidi, perii is not f(wai)
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[undone merlin]

$$\forall; \epsilon > 0 .; \exists; M .; x > M \implies \left| f(x) -b \right| < \epsilon$$

[grizzled pagoda]

Shuba

[keen pawn]

Yeah, I know that way

But I'm pretty sure my prof wants each of these to be represented by a logical statement

so like x>M should be a statement like G(M) where that means x>m

[gloomy tide]

wouldn't it be G(x)...

[fierce linden]

ah that explains why you insist on abbreviating statements

though to be completely safe doing that, you need to write stuff like L(f,x,b,eps)

[keen pawn]

Yeah should have probably done that

[keen pawn]

Why

[fierce linden]

meh that just looks annoying

[gloomy tide]

tbh it should be both: G(x, M) means x > M

but also that's very annoying to do

[undone merlin]

This?
$$G(x,M) \equiv x > M$$
$$L(f,x,b,\epsilon) \equiv \left| f(x) - b \right| < \epsilon$$
$$\forall.; \epsilon > 0 .; \exists; M > 0.; G(x,M) \implies L(f,x,b,\epsilon)$$

[keen pawn]

Something like that, yeah

[grizzled pagoda]

Shuba

[fierce linden]

did your prof explicitly say to abbreviate stuff

[keen pawn]

That's what we've done in class

[undone merlin]

you probably don't need to explicitly define the predicates like this

[keen pawn]

Why not just $G(M)$

[grizzled pagoda]

Veni, vidi, perii is not f(wai)

[fierce linden]

it should depend on everything you see

[gloomy tide]

because you need forall x \in R at some point

[keen pawn]

oh yeah, true

[undone merlin]

can be

for an implied forall x

[fierce linden]

imo unless youre explicitly told to do this on hw, i wouldnt bother

this kind of problem is focused on translating english to math, so whatever is already clear math should be left alone

[undone merlin]

imo, not much point in explicitly defining the predicates unless you need them for later use.

[gloomy tide]

implied forall when we're painstakingly writing out logical implications
absolutely not

[keen pawn]

Cool

Thanks!

[paper dawn]

1 + 1

[keen pawn]

Hope I do well on my test tomorrow, as is obvious, my logic is a bit shaky, p good at proof writing and set theory though

thanks everyone

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[atomic frost]

I’m confused on the answer to 6), I got pi not pi/2

[patent raptor]

I am also confused

[atomic frost]

yeah symbolab's calculator gave me pi as well, perhaps my professor made a mistake

[quick spoke]

,w int_{-infinity}^{infinity} 1/(1+x^2) dx

[patent raptor]

it's arctan(inf)

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[atomic frost]

interesting

[quick spoke]

wolframalpha agrees so ur probs right

[atomic frost]

hmm, i guess i'll ask my professor about it

thanks

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[unkempt grove]

The answer to 15 is (negative infinity, infinity), but i'm not sure how the teacher got there.
Including the previous question and solution for reference

[tidal meteor]

is the question -4 or 4?

@unkempt grove

[unkempt grove]

question reads as |10-4x| greater than or equal to -4

[tidal meteor]

i think it should be 4

as the value will always be greater than or equal to 0 anyways

it has a modulus

absolute value

[unkempt grove]

Definetly negative 4

[tidal meteor]

thats y im asking u to check it

bruh

hmmm

well

its gonna be greater than 0 for any value

so it is -inf,inf

basically

it doesent matter what value u take for x

its always gonna be greater than 0

thats y the range is -inf to inf

@unkempt grove

[unkempt grove]

interesting

give me a minute, gotta do something IRL

[tidal meteor]

aight

[amber wasp]

@unkempt grove Has your question been resolved?

[unkempt grove]

@tidal meteor
It's evident i'm doing something wrong, but what, i don't know

[tidal meteor]

listen

do you know what the function does?

the |x| function

[unkempt grove]

yes
Absolute value is always a positive number, and it's just the distance a number has from zero

[tidal meteor]

right

so doesent matter what x is

it will come out to be positive right?

[unkempt grove]

I'd think so, yeah.
But my math teach has kind of confused me on that concept.
Anyways, if it helps, these are some of my notes from the lesson

[tidal meteor]

ok

listen

|-2|=2

and |2|=2 right?

|any number| = positive number

or 0

[unkempt grove]

absoluteley

[tidal meteor]

So doesent matter what is in the middle

it will always be greater than -4?

as it is positive??

basically

any positive number is greater than -4 right?

and 0 is greater than -4

lmk if u did not understand

[unkempt grove]

Any positive number is greater, absolutely
but what about numbers -3 to -1?

[tidal meteor]

that is not a possible outcome right?

ok listen

do you know why -1 is multiplied?

ok

listen'

imagine you have |x-5|

now there are two possibilities

either x-5 is negative

or x-5 is positive right?

@unkempt grove

[unkempt grove]

understood

[tidal meteor]

ok

so now

is (x-5) is positive

then |(x-5)| is basically just (x-5)right?

[unkempt grove]

I'm with you

[tidal meteor]

now if (x-5) is negative

we multiply it with -1

to make it positive

so |(x-5)| = -1*(x-5)

@unkempt grove

ok imagine x-5 is -2

then mod x-5 is -2 x -1

so 2

[unkempt grove]

understood

[tidal meteor]

ok now

back to your problem

[tidal meteor]

.

|10-4x|>-4

now the |10-4x| is greater -4 always cuz it is always positive right?

[unkempt grove]

yes

I think i'm starting to see where you're going with this, but carry on, please

@tidal meteor

[tidal meteor]

So basically

no matter what u put in place of x

its always positive

therefore always greater than -4

so u can put anything in place of x

ranging for -inf to inf

thats it

[unkempt grove]

got it, thank you!

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[red needle]

Not sure how to go about this problem. Just factoring the problem completely and the answer given is 4(y+2)^2 *(y-2) anyone know how to go on this?

[red needle]

Didn’t post it right the first time sorry

[south harbor]

what are you trying to do?

[red needle]

Factor completely to that answer

[rancid raven]

(y²+y-2)²-(y²-y-6)² is this the question?

[south harbor]

yo my bad

there's an easier way

I'll let you figure it out

maybe try the naive way and brute force expanding, see what you get

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[atomic frost]

Can someone explain the blue part? I don't understand how that works at all

[elfin vale]

It's polynomial division

[atomic frost]

ohh

one sec

[elfin vale]

1 is the quotient and -1 the remainder

[lost star]

+1-1 in the numerator

[grizzled pagoda]

ren

[atomic frost]

alright so i got 1 with a remainder of -1

looks like the -1 goes up top and the 1 gets added on?

[pseudo crescent]

That means that x^2 = 1*(x^2+1) - 1, and so x^2 / (1 + x^2) = 1 - 1/(1+x^2)

[lost star]

heya meth

long time no see

[atomic frost]

oh okay i think i just didnt know how remainders work

[atomic frost]

ohh okay so you just divide every term by 1+x^2

that makes sense

thank you

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[pseudo crescent]

in general, you will get
Result of divison + remainder / divisor

[pseudo crescent]

Hello

[atomic frost]

gotcha thanks again

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[leaden mica]

pls help I want to learn how to do this quickly like in this picture

[leaden mica]

is there a concept which I should run through in my mind

[mint nacelle]

yes, reverse chain rule

note that the derivative of sin 2x is 2 cos 2x, so you are multiplying by 2

so to undo the differentiation (fundamental theorem of calculus)

you need to divide by 2

so -cos 2x * 1/2

[leaden mica]

wdym 'so you are multiplying by 2'

[mint nacelle]

chain rule when you differentiate sin 2x

cause the derivative of the inside function is d/dx (2x) = 2

[leaden mica]

yes, first we see sin=cos, then 2x =2

[mint nacelle]

you shouldn't use = in that way

but yes

[leaden mica]

I mean differentiation equal to

so what about it

[leaden mica]

..

[mint nacelle]

do you get that integration is the reverse of differentiation

[leaden mica]

yes

[mint nacelle]

how about observe that d/dx (-cos 2x) = 2 sin 2x

so d/dx (-1/2 cos 2x) = sin 2x

constant multiple rule

[leaden mica]

how is this different from substitution method

[mint nacelle]

we're not substituting anything here

we're doing the integral by inspection

cause you know that -cos differentiates to sin

[leaden mica]

yes

[mint nacelle]

and then we multiplied both sides by 1/2

to get the integral of sin 2x

[leaden mica]

you mean this?

[mint nacelle]

yeah, that is correct

it's not very useful however since it doesn't tell us about sin(2x) or sin(kx) in general

[leaden mica]

do you know a YouTube video on this @mint nacelle

[mint nacelle]

https://www.youtube.com/watch?v=4BwF1A-5aWg

try this

just more examples to get your head around the concept

[leaden mica]

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[gray pivot]

What is the general method/methods for doing these kind of qs?

[gray pivot]

,rotate

[mint nacelle]

oh that's clever

I'm pretty sure that the discriminant (-(4 + i))^2 - 4(1)(9 - 3i) is equal to -21 + 20i

so you can just apply the quadratic formula as normal
yep, it works the exact same using the complex numbers

then you just need to find $\frac{-b \pm \sqrt{\Delta}}{2a}$ where you already know $\sqrt{\Delta}$ from the previous part

[grizzled pagoda]

higher's secret twin brother

[gray pivot]

ohhh

ok

[mint nacelle]

the two square roots are a multiple of -1 apart anyways, so you still get 2 solutions

[gray pivot]

wait

i dont rlly get the wording of

[clear plinth]

5x5

[gray pivot]

find the square root of -21+20i

[mint nacelle]

the square roots

[gray pivot]

yeah

[mint nacelle]

in the complex numbers, there are n roots if you're doing the nth root

[gray pivot]

yep

[mint nacelle]

so yeah they would be z = 2 + 5i and -2 - 5i

both are equally valid

[gray pivot]

so do i find the square roots of -21 +20i using the (x+iy) formula

wait

[mint nacelle]

yes so you set (x + iy)^2 = -21 + 20i

[gray pivot]

oh

ok

[mint nacelle]

compare coefficients, so x^2 - y^2 = -21 for the real part

for example

[gray pivot]

and then the answer of that i just throw in the discriminant for the next step

[mint nacelle]

yes

[gray pivot]

ohhhhhhh

ok

tysm

[mint nacelle]

npnp!

[gray pivot]

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[pliant copper]

which algorithm is this

[red knot]

Send the orginal question if possible

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[bright kelp]

Hi any help?

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[sinful sapphire]

MP and MP

or idk

i guess that's the problem, which one is it exactly

i think HS is what i mean

assume D, then A follows, from which B → C follows and also B follows

hm

i guess not then

i want [B & (B → C)] → C

that's MP

then if you combine D → B and D → (B → C) into D → (B & (B → C)) you got it

but how

what's Conj?

@neat cliffcan you make C → C?

it does seem like you can't even start making anything with C

what's Equiv?

got it

@neat cliffdo the 10 on the left side too

~D V (~B V C)

idk maybe they combine into something afterwards

i give up

there's no way for me to try, right?

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[north dirge]

helpp

i have no idea how to solve it, and i need to explain why i solve it the way that i do

[twin terrace]

do you know how to solve simple equations (stuff like 5x-4=2)?

[north dirge]

yes

[twin terrace]

great

this is basically that

[north dirge]

but i should mention im bad at them

[twin terrace]

thats ok as long as you know how it generally works

[north dirge]

okay:,(

[twin terrace]

so the first task is already "done" it says 7.5 Tr is equal to 0 Tc so we can ignore that one

in the next one youre given Tc=20

you can put that 20 into the given equation, and solve it for Tr to find the temperature in Romer

[north dirge]

what

[twin terrace]

ok you said you would know how to solve 5x-4=2. Now imagine if instead of that, it said 5x-y=2 but i told you that y=4

its the exact same thing, because you can just change the y with a 4

do you get that

[amber wasp]

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[shadow parcel]

can someone explain how they do b) here? i can do a) and understand that far

[shadow parcel]

are they applying some formula for infinite geometric sums

[split sail]

do the ratio test

[shadow parcel]

Ok looking up what that is

[fickle ferry]

this

[shadow parcel]

wait is that the same as this ratio test thing

[fickle ferry]

when does sum x^n coverge?

[fickle ferry]

No

[shadow parcel]

when -1 < x < 1 i think?

[fickle ferry]

correct now when will sum (x/2)^n converge

[shadow parcel]

when -1 < x/2 < 1 or -2 < x < 2 i think

[shadow parcel]

right?

[fickle ferry]

correct so now can u use that idea in this question

[shadow parcel]

That just tells me it will converge or not

thats not what im after

its question b

[shadow parcel]

"Find x such that S = 1/2"

or am i wrong and its applicable there too?

[fickle ferry]

ok do u know the formula for S

if s= sum of x^n

|x|<1

[shadow parcel]

i just remember some geometric formula that looks like

a1 / 1-R

first term/ 1 - rate or something

if thats not it then i do not remember it by heart

[shadow parcel]

Is that it or am i thinkin of some other formula?

[fickle ferry]

mhm

use that

[tawny rover]

can someone help in help-26

[shadow parcel]

like this one i think

[fickle ferry]

yep

[tawny rover]

yep

[shadow parcel]

i always struggle with what is a and R here tho

is the 1st term just the expression plug in n = 1

so that is a

and r is something? idk what R is tbh

i think one can find r by dividing the 2nd term by the first but that just gives the expression inside the parenthesis too

that means a and r are equiv?

[marsh cloud]

It is actually an infinite GP series here 'a' the the first term of the series and 'r' is the common ratio

[shadow parcel]

Then for my expression above a and r will both be 2/(3sqrt(x+3)) no?

if im right(prob not) then i can write it like this and solve for x somehow?

[fickle ferry]

ye

[shadow parcel]

wtf ok, lemme try to solve that, not sure where to begin

[shadow parcel]

ok it worked, is this a good way to solve it or did i miss a step that could make it shorter/easier?

thx!

.close

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[shadow parcel]

What does it mean tha for these conditions does the system have a unique solution, infinite or no solutions

[shadow parcel]

i can get as far that if i calculate the det and set it to 0 and solve for a i get 3 and 5

[pallid canopy]

https://www.onlinemathlearning.com/image-files/system-equations-solutions.png

[shadow parcel]

i can get as far as to see and read this but i dont understand how to apply it to my problem

[pallid canopy]

in the 2d case, x + y = 1 and 2x + 2y = 2 are the same equations

[shadow parcel]

how are the relative slopes of these lines related to the determinant of this equation and system and number of solutions

makes sense

if that was system they have infinite solutions or?

[shadow parcel]

or is that line of thinkin wrong?

[fickle ferry]

rank

[pallid canopy]

yes when two equations are the same line, then they have infinite solutions

[shadow parcel]

okay how do i apply this to my problem? I'm having trouble going from the simple to complex example. Atleast its complex comparatively to me

[fickle ferry]

do u understand what it means to have no solutions or infintie solutions in 3D

[shadow parcel]

i guess if i have 3 equations describing three three dimensional shapes where the three never intersect in one place it has no solutions and if they only intersect for every point in their domain they have infinite solutions?

is that right? idk

[fickle ferry]

well these are only planes

so like only consider tht

[shadow parcel]

arent they 3d? its x y and z

oh i just put it into desmos

they are planes

[fickle ferry]

ax+by+cz+d=0 is a plane

now what do u get if 2 planes intersect each other

[shadow parcel]

then what functions in 3d are like bucket shapes and stuff

[fickle ferry]

you dont have to worry about it rn

[shadow parcel]

they have a common solution there

?

[fickle ferry]

geometrically

u get a line correct?

[shadow parcel]

what

really?

[fickle ferry]

ummm

[shadow parcel]

oh yeah

[fickle ferry]

think about it

[shadow parcel]

if they slice eachother

[fickle ferry]

yes

[shadow parcel]

its when 2d lines slice that i get a point and in 3d i get a line/curve

ok

[fickle ferry]

now what does it mean for a line and a plane to intersect each other

what do u get

[shadow parcel]

and im lost again

you mean geometrically or ?

[fickle ferry]

geometrically

again

[shadow parcel]

a point i guess

[fickle ferry]

yep

ok

now

[shadow parcel]

couldnt the line go through at an angle so its several points?

[fickle ferry]

it can eitehr a)

be on the same plane

(so u have infinite solutions)

or go through

(so u have 1 solution)

a point

is this clear

[shadow parcel]

ok so either its a line/curve shooting through the plane at one point

or it can go through it across the entire plane giving

infinite solutions

[fickle ferry]

ye

p

now suppose u have 3 equations

ax+by+cz+d=0

and whatever whatever

u have 3 variables

[shadow parcel]

ok

[fickle ferry]

what are the number of solutions u expect for (x,y,z)

[shadow parcel]

no idea

they will be planes

[fickle ferry]

alright so like its easy to see geoemtrically

[shadow parcel]

that either cut or dont cut eachother

[fickle ferry]

ill try to find a diagram or else explain it to you

[shadow parcel]

how many lines if its 3d or points in 2d they intersect at i don't see how that info would tell me how many solutions but im prob thinking of it wrong

[fickle ferry]

so like firstly ill just tell you it can have 1 unique solution

and geometrically u can see it like this

consider plane 1 and plane 2

they intersect and have a "common" line of intersection

and now this common line of intersection intersects plane 3

so u have a point which is common to plane 3 plane 2 and plane 1

which is why its unique

[shadow parcel]

ok so in this scenario

either the 3 planes cut in one point = unique, never intersect = no solution or the 3rd plane joins along/across the line so infinite??

maybe i grapsed that wrong

[fickle ferry]

yep

yep

exactly that

u also understand geometrically the never intersecting part right?

[shadow parcel]

that just means the 3 planes never touch any one point right?

[fickle ferry]

like suppose the common line is "parallel" to the plane

not necessaruly

[tulip jasper]

Need help please

[shadow parcel]

really?

how is that possible

[fickle ferry]

consider a prism

[shadow parcel]

please move to another help channel sir

[fickle ferry]

u see the 3 planes?

the common line of intersection is "parallel" to the base plane

[shadow parcel]

a prism is lots of planes more than 3 but you mean the bottom, left and right one?

[fickle ferry]

yes

[shadow parcel]

here they never all 3 touch in one spot right?

[pallid canopy]

https://i.sstatic.net/XDenC.jpg

[fickle ferry]

yes

[shadow parcel]

only 2 at a time

[fickle ferry]

this works fine

so i hope u understand what it means "geometrically"

[shadow parcel]

wait so here 1, 2 and 7 have infinite solutions?

[fickle ferry]

yeo

yeo

yep

**

[shadow parcel]

6 is unique

and the rest are none

or nah?

[shadow parcel]

if what i just said is true we've gotten as far as for me to understand what it means geometrically for shapes to intersect and have a certain amount of solutions i think, which is kinda a revelation. This is just for planes btw? There are 3d functions that have crazy shapes. Do the same kinda rules of intersections and solutions apply there? i would think so

[fickle ferry]

its just for planes

now do you know what an augmented matrix is?

[shadow parcel]

prob seen it gotta look it up as the term doesn't ring any bells directly

[fickle ferry]

like all u need to do is find the rank of the augmented matrix and the coeffecient matrix

u can check how to find the rank of a matrix

[shadow parcel]

is this one of those where i do gauss elimination

[fickle ferry]

its just gauss elimination

yeah

and this dirrectly corelleates to what we learnt geometrically

since suppose u get something like

1 0 0

0 1 0

000

[shadow parcel]

almost identity

[fickle ferry]

this means the 3rd plane is a linear combination of the above two planes

since like in gauss elimination

all u do is multiply and subtract or add

almost like how u just try to find solutions for

ax+by=c

dx+ey=f

[shadow parcel]

hmm ok

[fickle ferry]

and like

when u have one plane to be a liner combination of two other planes

you get the 7th situation

[shadow parcel]

not sure what happened here you said " youget something like" and wrote an almost identity? was this to related to the determinant or something else?

[fickle ferry]

like

like

after apllying

gauss elimination

to a matrix

u get row echelon yeah

or an almost diagonal matrix if we are being lazy

abc

0de

00f

[shadow parcel]

im still not following sry

all ive done with gauss elim is solve for ax = b so or inverses

or solve determinants faster

[fickle ferry]

yeah well

[shadow parcel]

thats all i know

[fickle ferry]

in the gauss elimination

have u used matrixes?

like

without writing

equal signs

[shadow parcel]

i've def used matrices but

not sure how deep into it you define as having used tehm

i know basic operations like, add, sub, matrix multiply, find the inverse

solve for det

what identity is kinda

im thinking

[shadow parcel]

writing up and aug matrix

like that

[fickle ferry]

this is the aug matrix yea

[shadow parcel]

when i have a system of linear eq

[fickle ferry]

now can u convert it into like row echelon form

[shadow parcel]

ok what is that

[fickle ferry]

or a diagonal matrix

like

u multiply the second row

by d/a

and subtract it from the first row

so u get

a b c

0 ed/a-b fd/a - c

g h i

and then u do it agian

for the third row

and in the end

[shadow parcel]

holup

[fickle ferry]

u get something like

a b c

0 stuff stuff

0 0 stuff

[shadow parcel]

wdym by multiply by d/a

and here you have 9 diff variables

[fickle ferry]

uhhhh

https://www.youtube.com/watch?v=Sc7OY62lQ9U

[shadow parcel]

I've heard of rank and maybe calculated rote once or twice

[fickle ferry]

https://www.youtube.com/watch?v=cSj82GG6MX4

[shadow parcel]

is that what will help me solve the problem?

[fickle ferry]

yeah that essentially

yes

[shadow parcel]

like the rank of the matrix given a value of a

[fickle ferry]

i was just trying to explain how rank geometrically relates

[shadow parcel]

will tell no/inf/unique solutions

[fickle ferry]

yes

[shadow parcel]

i mean i wanna understand that too but i guess maybe im behind on that too much to explain in one sitting

[fickle ferry]

the method is like

u find the rank of the augmented matrix

[shadow parcel]

what u were doing here is that rank calculation

orm just

[fickle ferry]

yes

rank calculation

[shadow parcel]

okok

[fickle ferry]

to reduce it to row echelon form

so that i can find the rank

u find the rank of matrix A

[shadow parcel]

ok how many ranks are there, wathcin vid rn

is it just a numerical value like det(A)

[fickle ferry]

yeah

for like a 3x3 matrix

u can have rank 1 2 or 3

just look up

"rank of a matrix and consistency of linear equations"

and u should get ur answer

after u learn the algorithm try understanding what it means geometrically

i need to go now

[shadow parcel]

ok ill try i have the exam in 12hrs

wish me luckl

[fickle ferry]

its 5 am and i have class in 3 hours

[shadow parcel]

thx

[fickle ferry]

good luck

[shadow parcel]

yea sry

bb take care ty for help

<34

❤️

[fickle ferry]

yw

[shadow parcel]

.close

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• After 15 minutes, feel free to ping <@&286206848099549185>.
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[split sail]

How to find inverse of 7x+1/5-x

[mighty lagoon]

you didn't go from the 2nd to 3rd line right

otherwise you're on the right track

[split sail]

Wait like how?

I did cross multiply

[split sail]

Use f(f-1(x)) = x

If h(x) is inverse of f(x)

Use f(h(x)) = x

Easily

[mighty lagoon]

you cross multiply when you're adding fractions

in this case you're solving for a = b/c and you're getting rid of the denominator

you do the same thing to both sides of the equation

[split sail]

Wait so what do I do now

[mighty lagoon]

you just did algebra wrong but you set up the first equation correct

[split sail]

I have x(5-y)=7y+1

[mighty lagoon]

good

[split sail]

Take y together and x on other side

And manipulate to get coeff of y=1

[mighty lagoon]

so to find the inverse what you set up was you switched x and y

and then you solve for y

[split sail]

Can you like show me?

Do I distribute x(5-y) or divide by x for both sides

[mighty lagoon]

so from here, how do you solve for y = ... ?

[split sail]

No that’s where I am confused on because I can’t put y on other side after divide by x on both sides

[mighty lagoon]

you need to isolate y

distributing is indeed a start

[tulip jasper]

Need help please

[split sail]

How do I isolate like how do I simplify -xy subtract with 7y

[mighty lagoon]

if you have different terms containing y like ay + by, then you could factor out y

to get (a + b)y, then you'll have just one y left

[split sail]

This?

[mighty lagoon]

that looks correct so far

[split sail]

Do I divide (x+7)?

[mighty lagoon]

you could try and see how it goes
at this point you just need to do algebra, the question about inverses you already do how to do

[split sail]

I have an another question what is range for the original function

Which is 7x+1/5-x

[mighty lagoon]

you can simplify first

and then try to recognize what type of function this is

[split sail]

I know that domain can be anything other than 5

[mighty lagoon]

ah

if the range of the original function doesn't reach certain values, what does that say about the inverse?

[amber wasp]

@split sail Has your question been resolved?

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[stone quarry]

$$\frac{2sec(x)tan(x)}{x^3} - 6sec(x)$$

[grizzled pagoda]

Devil Wears Prada

[stone quarry]

this is what i got

is it right?

only using product rule btw

[shell osprey]

d/dx (1/x^3) is not 3 ( rather not -3)

[stone quarry]

?

i know that

what does that have to do with the problem though

[shell osprey]

your answer suggests otherwise though?

[stone quarry]

what

shouldve i kept it as $$2x^{-3}sec(x)tan(x)$$

[shell osprey]

no the second part

[grizzled pagoda]

Devil Wears Prada

[shell osprey]

$-6sec(x)$ should be $\frac{-6\sec(x)}{x^4}$

[grizzled pagoda]

someone1010

[stone quarry]

why

[shell osprey]

why not?

[stone quarry]

i used product rule

$$(2x^{-3})(sec(x)tan(x)) + (-6)(sec(x))$$

[grizzled pagoda]

Devil Wears Prada

[shell osprey]

and in no way is $\frac{d}{dx}(2x^{-3}) = -6$

[stone quarry]

when $$f(x)g'(x) + f'(x)g(x)$$

[grizzled pagoda]

someone1010

Devil Wears Prada

[shell osprey]

$\frac{d}{dx}(2x^{-3}) = -6x^{-4}$

[grizzled pagoda]

someone1010

[stone quarry]

$$\frac{2sec(x)tan(x)}{x^3} - \frac{6sec(x)}{x^4}$$

[grizzled pagoda]

Devil Wears Prada

[stone quarry]

is this right

[shell osprey]

probably

[stone quarry]

also another question

how would i multiply $$3x^4(e^x)$$

[grizzled pagoda]

Devil Wears Prada

[shell osprey]

wdym?

[stone quarry]

how

[shell osprey]

$$f(x) = 3x^4$$
$$g(x) = e^x$$

[stone quarry]

do i just leave it as $$3x^4e^x$$

[grizzled pagoda]

someone1010

Devil Wears Prada

[stone quarry]

yes

[shell osprey]

oh as an answer

yea, you can't simplify it any more

[stone quarry]

ok

[shell osprey]

nope you meant as a question

just product rule

[stone quarry]

bro

no way ur real

[shell osprey]

go to #help-14 and read #❓how-to-get-help

[stone quarry]

is this $$3x^4e^x + 12x^3e^x$$

[grizzled pagoda]

Devil Wears Prada

[shell osprey]

pretty sure it is

,w d/dx 3x^4e^x

[grizzled pagoda]

Results provided by WolframAlpha

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

Requested by someone1010

[stone quarry]

nah dont use wolfram

[shell osprey]

eh wolfram being a goof

[stone quarry]

it uses other rules

ye

now this

[shell osprey]

basically the same question lol

[stone quarry]

oh yeah

this is harder

[shell osprey]

easier than your first one

and basically the same as it

[stone quarry]

ok let me do it hold on

$$-5x^2sin(x) + 10xcos(x)$$

[grizzled pagoda]

Devil Wears Prada

[shell osprey]

seems right

[stone quarry]

and thois one

what is the derivative of 2e^x

just 2e^x?

[shell osprey]

yea

[stone quarry]

so then its

wait

$(2e^x)(\frac12x^{-\frac12})$

[grizzled pagoda]

Devil Wears Prada

[stone quarry]

how

is it $$\frac{e^x}{\sqrt{x}}$$

[grizzled pagoda]

Devil Wears Prada

[shell osprey]

I mean, not sure what you're saying but it's prod rule?

[stone quarry]

like when those terms are multiplied

does it equal that

[shell osprey]

uh, yea

[stone quarry]

ok

so the full answer is

$$\frac{e^x}{\sqrt{x}} + 2e^x\sqrt{x}$$

[grizzled pagoda]

Devil Wears Prada

[shell osprey]

i would think so

[stone quarry]

okay well i dont think i need much help

ill open another one when i do

i think theres more challenging ones coming up

thanks

.close

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[split sail]

Where did I go wrong ?

[split sail]

I’m doing rref

[amber wasp]

@split sail Has your question been resolved?

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@split sail Has your question been resolved?

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[verbal lark]

One message removed from a suspended account.

[verbal lark]

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[shadow stump]

divide numerator & denominator by x

that's the same as multiplying by (1/x) / (1/x) = 1

[verbal lark]

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[shadow stump]

|x| = sqrt(x^2) (but we only care about positive x so |x| = x)

[verbal lark]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[wispy kestrel]

Gr 12 Data Management

[pallid canopy]

Can you enumerate all the ways Elena can win

<@&268886789983436800>

[wispy kestrel]

uh

?

hi

[wispy kestrel]

ya

like

she can win all 5 : W W W W W

win ad lose 1: W W W W L

and W W W L L

[pallid canopy]

No

Best of 5 means the rounds are over after Elena wins 3

[wispy kestrel]

oh

umm ok

so how do i do it then

ermm we didnt learn that yet

is there another way

we just learnt odds

well what i wrote was:

W W W W W

and

W W W W L

And

W W W L L

but then someone said that it ends after 3..

oh

nvm

itys ok

ty

.close

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[thick seal]

pls

[thick seal]

idk how its marked wrong

omg

i put 6.022 x 10^22 instead of 23

idk why i catch my mistakes when im asking for help always

.close

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[thick seal]

huh

nope

still wrong

im showing my work 1 sec

[cedar plaza]

.reopen

[amber wasp]

✅

[thick seal]

sry it took a while i was looking for my phone

they;'re saying the answer is 1.08x10^22

that cant just be 1.08

1.07 make mroe sense

1.07 and a 3 to its right

so surely cant just round up

😕

[shadow stump]

your answer looks right, their answer is probably a little off due to rounding in the middle of calculation

[thick seal]

o ok

thnxs

.close

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[leaden mica]

here when I solved tan inverse, then it was zero point answer so it's in radian

[leaden mica]

but here it's bigger number then it's in degree?

[amber wasp]

@leaden mica Has your question been resolved?

[leaden mica]

<@&286206848099549185>

and we have to find max and min T, so we are finding max and min value of theta. but it came negative angle, so we will add 90° to it? I don't understand this

@help

<@&286206848099549185>

I'm waiting for 45 mins on this question

please

.close

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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>.
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[split sail]

hi'

[split sail]

two congruent circle are set such O1 and P1 are the centers of the circles

find the perimeter of QORP

oh

yeah

and the radius is 4

[fresh ocean]

So what is the perimeter? Can you write the sides involved in it?

[split sail]

what

sides?

[fresh ocean]

does anything special about the sides stand out to you?

[fresh ocean]

yes. Perimeter is sum of lengths of sides

[split sail]

Oh shi

i am talking about the curve

[fresh ocean]

Oh ok

[split sail]

the curve QORP

yeah

[fresh ocean]

still

segment OQ and all such are still somewhat relevant

what can you say about the angle QOP?

[split sail]

idk

[fresh ocean]

What about the triangle QOP? And its sides?

[split sail]

OP = 4
QO = 4
QP = 4

damn

[split sail]

60 deg

[fresh ocean]

yes