help-49

[midnight plank]

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[bold pine]

I just started my course and I'm very confused on this question. Does this mean that I should give values 0 or 1 (true or false) depending if they're the same as p -> q?

[last slate]

If p->q is false then what does that mean?

in fact, when is any implication false?

[bold pine]

if p is not true?

[last slate]

No

p->q is false when p is true and q is false

[bold pine]

i want to make sure im understanding this right..

so then for a) it would be 0 because that means p and q?

[last slate]

what do you mean by 0?

[bold pine]

it said to give the truth value so i assumed it would mean true and false?

[last slate]

Yes

so if p is true and q is false then what would be the truth value of p and q?

Oh

i see what you wrote above

i didn't know you meant 0 as false

but yes it is false

[bold pine]

no worries

[last slate]

what about b) then?

[bold pine]

is it because they're in like an "and statement"

[last slate]

Yup

[bold pine]

ohhhhh

[grand pond]

jswatj

[bold pine]

and there is no exact true or false to p and q

[last slate]

well we know p->q is false which means the only possible choice for p, q are that p is true and q is false

so there is an exact truth value to p and q

[bold pine]

ohhh

[bold pine]

would it be 0 because its. not p and q

[midnight plank]

@bold pine Has your question been resolved?

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[bold pine]

.close

[midnight plank]

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[last slate]

someone just give me the answer for this. im straight done.

[midnight plank]

@thin zealot Has your question been resolved?

[terse frigate]

Hint: triangle LKM is similar to triangle ONM.

[midnight plank]

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[midnight plank]

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[oak rampart]

i need to find the expression equivalent to the expression below. i tried but i honestly have no clue what to do can anyone help?

this is what i got so far

[dull yoke]

why didnt u just leave it at 10x-5

whyd u do it again?

the first time u did it, it was right

[oak rampart]

that’s not an answer choice so i figured i was supposed to do smthn else

[dull yoke]

alright, well try simplifying the answer choices and see what u get

cuz the 6(3x-5)+3 thing is wrong

i see what u tried to do, its not mathematically correct though

[oak rampart]

gotchu i’ll try it out thx man

[dull yoke]

sure 👍

[oak rampart]

figured it out thanks

.close

[midnight plank]

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[midnight plank]

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[vivid cipher]

how does this simplify to this?

[vivid cipher]

isn't cosh(x) =( e^x + e^-x )/ 2

[warm magnet]

Yeah

But lemme see

[foggy cave]

oh this is sneaky

[tired shadow]

Write 2 as 2e^(2x)e^(-2x) and do some wacky manipulations

[warm magnet]

Ayo RBG latex

How

[vivid cipher]

omg

[foggy cave]

@hard umbra

😁

[warm magnet]

Show

Wait will it work on iOS

[hard umbra]

ayy

[foggy cave]

i invented my own primitive version from observing snow

[warm magnet]

Well

Something that I realized

[warm magnet]

$$\sqrt{e^{2x} + e^{-2x} + 2} = ex + e^{-x}$$

[grand pond]

Umbraleviathan

[warm magnet]

Fuck

[foggy cave]

🤩

[reef pelican]

so pretty

[warm magnet]

I forgor a ^

[foggy cave]

generalized rainbow tool

[hard umbra]

colours are kinda ass

[warm magnet]

$\shading {left color=y, middle color=m, right color=z} {$\paren {u + \f1u}^2 = u^2 + \f 1 {u^2} + 2$}

This works right

I'm crying

[foggy cave]

[warm magnet]

But anyways that's a thing that will happen

[foggy cave]

innovation is a wonderful thing

[warm magnet]

The thing is that the 2 comes from a pair of u/u

So that's just soemthing you gotta know

[vivid cipher]

gotcha

thanks!

[warm magnet]

And something I wished I was fucking taught

[foggy cave]

i only know it because i saw it somewhere before

[warm magnet]

This would've saved me brain space in precalc

[vivid cipher]

.close

[midnight plank]

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[umbral turret]

help

[umbral turret]

i dont know where to go from here

[midnight plank]

@umbral turret Has your question been resolved?

[umbral turret]

<@&286206848099549185>

[midnight plank]

@umbral turret Has your question been resolved?

[umbral turret]

no im figuring this shit out on my own

thanks for the help guys

need more eu time zone people

.close

[midnight plank]

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[midnight plank]

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[open fractal]

can someone help me with these problems? i got the first one right but the others i get wrong.

[velvet compass]

do you know what does f(3+h) mean

[open fractal]

f of 3 which is 9 plus the value of h?

[slender walrus]

no

[velvet compass]

no

[slender walrus]

how did you get the first one

[open fractal]

i plugged in 3 for x

[warm magnet]

Holy shit it's Obama

[open fractal]

yes

[slender walrus]

apply the same princple for f(3+h)

[warm magnet]

For b, your answer should have h in it

[slender walrus]

plug (3+h) for x

[open fractal]

so 3 +h squared

[slender walrus]

()

[open fractal]

which is 9 plus h squared?

[slender walrus]

no

$\fdream$

[grand pond]

ℝamonov

[open fractal]

(3 +h)squared?

[slender walrus]

yes,

[open fractal]

so

(3+h) squared plus 9 for the second problem?

minus*

[slender walrus]

yeh,
and it seems they want you to expand and simplify

[open fractal]

i simplified and got just h but i think its wrong

[slender walrus]

defintiely wrong

how exactly are you "simplifying"

show what you're doing after
$$(3+h)^2 - 9 $$

[grand pond]

ℝamonov

[open fractal]

add 9 to the other side and square both sides

[slender walrus]

what other side

there is no other side

[open fractal]

i add 9 so its like 9 = (3+h)squared

[slender walrus]

whut

[open fractal]

its wrong

[slender walrus]

there is no other side

[open fractal]

oh

[slender walrus]

expand and simplify, don't overthink

[open fractal]

so

[slender walrus]

can you expand (3+h)^2 properly

[open fractal]

(3+h)(3+h) -9

?

[slender walrus]

keep expanding

[open fractal]

9 + 6h +hsquared -9

?

[slender walrus]

pls use ^ to denote exponentiation
^2 to denote squared

[open fractal]

so 6h + h^2

or is that wrong

[slender walrus]

yes, that's good

[open fractal]

ok so do i simplify more to that?

[slender walrus]

do you think that you can?

do you have any like terms?

[open fractal]

no

[slender walrus]

are you gonna try and forcibly do some stuff to it because you don't like it?

[open fractal]

i dont see a way i can simplify problem b anymore

unless there was an h at the bottom

which there isnt

[slender walrus]

that part comes later,

[open fractal]

ok

[slender walrus]

they're really breaking this problem down for you

dividing by h is the next question, part c)

[open fractal]

can you do h(6+h)

to simplify it more

[slender walrus]

simplicity is around the same, which is more useful depends on what you're doing

for part b) its fine to leave it as 6h + h^2

and then do the factorisation for simplifying c) if you want

but it isn't really needed

[open fractal]

ok so with c we divide 6h + h^2 over h

and we get 6 +h

[slender walrus]

yes

[open fractal]

for d

do we plug in 0 for h?

[slender walrus]

yes

[open fractal]

wouldnt that just be 0?

[slender walrus]

wdym

[open fractal]

oh wait

[slender walrus]

wdym by "that"

[open fractal]

it would be 6

as the limit apporaches 0 the limit would be 6

because 6 + 0 =6

or am i overthinking again

hello?

[prime geyser]

damn it's full, I need help

[open fractal]

whats up thanos

[prime geyser]

what's up Obama

[open fractal]

you can have this channel i figured it out i think

.colse

.close

[midnight plank]

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[prime geyser]

@open fractal you + me

[prime geyser]

reopen it, I already got one

[open fractal]

omg 😩

[midnight plank]

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[halcyon trellis]

Try express those numbers as power of 2 and 3

[prime geyser]

can u check if it's 12?

[halcyon trellis]

Give me a sec

[lyric charm]

bruh if you just want the answer go ask WA

,calc (81/16)^(-3/4) * (25/9)^(-3/2) * (2/5)^-3 * 12

[grand pond]

Result:

12

[slender walrus]

show work, are there any parts of what you did that you are unsure about

[lyric charm]

or do you want the sub-100-millisecond solution

[prime geyser]

no, if I made mistake I can clear it

[lyric charm]

(because i don't have it)

[prime geyser]

because it was confusing

ok i will look it later

.close

[midnight plank]

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✅

[halcyon trellis]

Is there a context to this or is this the whole question.

[prime geyser]

this is whole

[prime geyser]

what does this curly bracket mean??

oh wait

.close

[midnight plank]

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[prime geyser]

ohh it's not clear yet

.reopen

[midnight plank]

✅

[velvet compass]

it's a set in which we write particular elements

generally finite amount of elements

[prime geyser]

oh

yea

thanks

like it's the values which satisfy

not the range

ye

ok

[velvet compass]

yes

not an interval

just values

[prime geyser]

yea

I knew that, just wanted to confirm

thx

.close

[midnight plank]

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[prime geyser]

wait one more

[midnight plank]

✅

[prime geyser]

idk how to begin with

[sick bramble]

help with b only 🙏

[prime geyser]

hello can u open new channel, its my doubt channel

[sick bramble]

sorry

[prime geyser]

.close

[midnight plank]

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[midnight plank]

@prime geyser Has your question been resolved?

[midnight plank]

@prime geyser Has your question been resolved?

[midnight plank]

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[compact gulch]

Hello

I need help in Boolean Algebra.

The problem at hand is that there are 3 levers (x,y,z) that control a lamp. if an even number of levers are activated the lamp turns on (i.e., if a lever gets a value of 1 it counts as activated).
The issue arises with the table of my values, since in the solution when x=0, y=0 and z=0 the lamp is still turned on? Does someone perhaps know why?

[carmine void]

Some consider 0 an even number

[compact gulch]

but none of my levers would be activated in any way in this case, so I don't understand/see how my lamp could be turned on?

[terse spear]

It's just a truth table, it doesn't logically have to make sense. The goal is to convey the concept of truth tables/boolean algebra

[compact gulch]

Ahh, so basically since 0 is sometimes counted as an even number, we have an even number of levers, so the lamp is on

or in other terms an even number of 1

.close

[midnight plank]

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[midnight plank]

@shadow osprey Has your question been resolved?

[shy citrus]

well of course you get two, you solved a quadratic

you have to decide which one is the "right" one

what are the two answers for x that you got?

if one is negative, you can discard that one because distance is always positive

if you have two positive values, you can use triangle inequalities to find the correct x

the sum of two sides must be greater than the third side

lol wait aren't the two solutions just the lengths from s to the intersection points

[midnight plank]

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[midnight plank]

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[carmine void]

What do you need help with?

[unborn tree]

The bottom problem solving

[tribal carbon]

Do you get what slope is?

[unborn tree]

Ehh not so much

[tribal carbon]

Hmm

https://youtu.be/MeU-KzdCBps

[midnight plank]

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[sonic compass]

<@&286206848099549185>

[midnight plank]

@sonic compass Has your question been resolved?

[sonic compass]

<@&286206848099549185>

.close

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[silver eagle]

can someone explain the boxed part? it seems like a syntax error to me, lining up two row vectors

[midnight plank]

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[silver eagle]

<@&286206848099549185> ^

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@silver eagle Has your question been resolved?

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[midnight plank]

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[midnight plank]

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[forest shoal]

So I have a question

How does dy/dx = ny

[midnight plank]

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[forest shoal]

Using differentiation btw

[midnight plank]

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[forest shoal]

Any hint to start

Btw using differentiation

[last slate]

chain rule ig

f'(g(x)) * g'(x)

[forest shoal]

Is this right?

[fallow scarab]

that's not equal to 1

otherwise your whole thing in the beginning would be (sec(x) + tan(x))^n = 1^n = 1

[forest shoal]

Hmmm

Any hints?

[fallow scarab]

don't make that mistake?

[forest shoal]

Wym

Im saying any hints to solve this

[fallow scarab]

also you dropped the (x) here

[fallow scarab]

you were on the right track. just fix your mistake

[forest shoal]

Oh okay ty

.close

[midnight plank]

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[midnight plank]

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[tidal osprey]

hi, so i feel like this solution is not too complicated but i seem to be stuck

i got to a point where i wrote that there are infinitely many k-numbers for a given k if the equation $x(x+k) = y^2-k$ has infinitely many solutions

[grand pond]

jn3008

[tidal osprey]

in x and y

infinitely many integer solutions*

[midnight plank]

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[glass hearth]

Idk if this is math based but its worth a shot

[glass hearth]

The mass of an empty flask is 149.04 g. What is the mass of the flask filled with ethyl acetate (d = 0.902 g/mL) if the same flask weighs 275.12 g when filled with water?

[next rover]

empty + 1 × V = 275.12
empty + 0.902 × V = Eks

[glass hearth]

What does the eks mean

[next rover]

the latin X

th required number

[glass hearth]

Oh ok

Alright hold on

[next rover]

you just find V from the first equation

[glass hearth]

What does the V stand for

[next rover]

flask volume

[glass hearth]

Oh

So the mass of the empty flask is 149.04 and then multiply 1 which is the volume of the water to that to get the flask volume?

Actually nvm ill figure it out myself I think I got it

.close

[midnight plank]

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[dark cliff]

hi guys quick question

on problems like these, whats the point of saying x>= -11? does that "or equal to" mean anything in that second part?

like is there any use for it in limit problems?

[midnight plank]

@dark cliff Has your question been resolved?

[elder swan]

hmm

[elder swan]

I'm not understanding ur question too well

[dark cliff]

Why not have the question set up where the first part is x<-11 and the second part is x>-11, rather than x>=-11 for the second part

will there be a case where it matters, whether that interval also includes -11, or its only values greater than -11?

[elder swan]

well i guess if x >=-11

[dark cliff]

so whats the point of it

[elder swan]

i honestly dont know, prob to make the function exist there

lol im not too experienced sry

[dark cliff]

no worries

[midnight plank]

@dark cliff Has your question been resolved?

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[pale nexus]

i got the equation x^2 + y^2 = 16 but i dont understand how they got a z of 3

[midnight plank]

@pale nexus Has your question been resolved?

[pale nexus]

<@&286206848099549185>

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[broken wren]

Im a bit confused on where to start. Can someone help me solve A then I can figure out the rest.

[broken wren]

ping me

[graceful roost]

Use this identity $\sin (x) = \sin (2\pi + x)$

[grand pond]

no.

[graceful roost]

@broken wren

[broken wren]

ahh ok

[broken wren]

so our equation would be sin(pi/3) = sin(2pi + pi/3)

?

[graceful roost]

Yeah

[broken wren]

ok cool

sqrt3/2

[broken wren]

@graceful roost for question 2, can I leave it as degrees or do I have to convert to to radians

[graceful roost]

U can leave it as degrees, but notice that in this case u should have 360 in the identity instead of 2pi

[broken wren]

so something like that

[graceful roost]

Yes

[broken wren]

so its just 80 degrees

[graceful roost]

What?

[broken wren]

one side is equal to the other

so its sin(80)

wait

[graceful roost]

They are asking u to find an angle that is not 80 which has the same sine value

[broken wren]

Oooh

280?

[graceful roost]

Why?

[broken wren]

it would have the same sin value wouldnt it, just negative.

or I guess 120 is the same

[graceful roost]

360+80

[solemn fable]

They want the angle to be between 0 and 2pi

[broken wren]

oh 440 ._.

[solemn fable]

Use sinx=sin(pi-x)

[graceful roost]

Oh lol forgot about that

Use this identity instead
$\sin (\pi - x) = \sin (x)$

[grand pond]

no.

[graceful roost]

Thx

[broken wren]

yes, its true

I dont know how im supposed to derive the angle from this

[graceful roost]

Let's say u have 80 degrees, u compute 180-80

[broken wren]

yeah

100

wait if its that easy why do we need the formula

[graceful roost]

This one?

[broken wren]

yeah

[graceful roost]

Because how would u otherwise know that the sine values of 80 and 180-80 are equal?

[broken wren]

I mean, look at the unit circle but I guess on paper that makes sense.

.close

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[lucid forum]

make it two fractions first

$(A+B) / C = A /C + B/C$

[grand pond]

gloom

[grim sigil]

.close

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[weary pendant]

do you know how to find the area of a rectangle?

[grim sigil]

Yh

[weary pendant]

have you found the area of both rectangles?

[grim sigil]

Yh

[weary pendant]

do you know how to find out how much bigger one number is compared to another?

[grim sigil]

How do u do that

[weary pendant]

for example 10 is 5 times bigger than 2 because $$\frac{10}{2}=5$$

[grand pond]

Duh Hello

[weary pendant]

how would you apply that to this problem?

[grim sigil]

I get it now

[weary pendant]

[grim sigil]

I was doing - not divide

.close

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[digital quail]

Can someone explain how I go about question 4c

[digital quail]

Question a) is 2(x-1)^2 -10

[weary pendant]

find points where y=8 and check the points around it if they are above or below 8. alternatively you can use the derivatives of y to check the surroundings instead

since it is a quadratic you know that it can only have 2 intersections hence there is at most 3 parts that need to be checked, the double derivative to check if it is concave or convex also works

[digital quail]

I got x=4 and x=-2

[weary pendant]

then i would check easy to check points both between and outside of those. alternatively you can simply see that the x^2 term is positive and will overtake the rest of the equation, and hence know that it will be outside these points where y>=8

[digital quail]

Cant say I understand clearly what that means

Or what I have to do

[weary pendant]

i think the easiest way to do it is the first option. so check easy points, for example what is y when x=0?

which is in between the two points where y=8

[digital quail]

When x=0, y=-8

[weary pendant]

so when x=0 then y is not greater than or equal to 8 correct?

now lets check a point which is outside of the two points you found, for example x=-3

what is y then?

[digital quail]

I got y=22

[weary pendant]

which is clearly bigger than 8 yes?

[digital quail]

Yea

[weary pendant]

now since it doesnt become 8 for any value lower than x=-2 then we know that for all values where x<=-2 then y>=8

does that make sense?

[digital quail]

Yeah

[weary pendant]

ok, now we have to check the other side as well

there is one last side

what x value do you think it would be good to check?

[digital quail]

Don’t we only have 2 x values

[weary pendant]

yes, but we need to check what happens between these values

and outside these values

[digital quail]

We’ve found outside the values

[weary pendant]

we are still missing 1 outside value

[digital quail]

5?

[weary pendant]

perfect

[digital quail]

=22

[weary pendant]

so what would u say are the values for $x$ where $y\geq 8$?

[digital quail]

Smaller than 4

[grand pond]

Duh Hello

[digital quail]

Larger than -2

Anything between them right

[weary pendant]

not quite, remember that we found that for x=-3 we had y=22, and for x=5 we had 22, and for x=0 we had y=-8

and we know that for x=-2 we have y=8 and for x=4 we have y=8

[digital quail]

Yeah

[weary pendant]

so where is y larger than 8?

it might help to make a graph of it

and make a line for y=8

[weary pendant]

now, for what values of x is y greater than 8?

[digital quail]

-3 and 5

[weary pendant]

those are examples, but since we know that it is smaller than 8 in between -2 and 4, what can we say about everywhere else?

[digital quail]

It’s greater than 8 ergyerwher else?

[weary pendant]

yep

[digital quail]

Oh yo

[weary pendant]

the reason we wanted to check the points both outside and inside is because it could have been the opposite

for example

so because of this, you need to check two points outside of where y=8 and a point between the points where y=8

[digital quail]

Yeah makes sense

[weary pendant]

the big thing to note is that when u find 2 possible values for x where y=8 then it means that y cannot cross the line y=8 anywhere else

and because of this we know that it must follow that for anything outside or inside they will follow the same rule as the example choices for x we made

[digital quail]

What does it mean y cannot cross the line y=8 anywhere else

[weary pendant]

the points where $y=8$ and $y=2(x-1)-10$ meet are as you found, $x=-2$ and $x=4$. Since there are no more solutions then these are the only points where they cross. This can be represented as points $(-2,8)$ and $(4,8)$

[grand pond]

Duh Hello

[digital quail]

Yea

[weary pendant]

if u make a point on the graph you made where they meet u will see that those two points are the two points i mentioned

[digital quail]

Yeah

[weary pendant]

so those are the only solutions for where they cross, not really sure how to explain it anymore than that

[digital quail]

Yup

[weary pendant]

thats just what i meant when i say they cannot cross, or intersect anywhere else

[digital quail]

Yeah makes sense

[weary pendant]

so from that you can find the answer to the problem

[digital quail]

So back to the question, are the answers not just 4 and -2

Those are the only points

[weary pendant]

well, you need to say which sides of it are correct since it asks for GREATER or equal

not just equal

[digital quail]

Yeah

[weary pendant]

which was what we checked once we checked each side. so it is everything less than or equal to -2 and everything greater than or equal to 4

[digital quail]

Is that the answer

[weary pendant]

yes

[digital quail]

Ahh thank u bro

Finally

I think I got it

[weary pendant]

no worries

[digital quail]

I do get it

.close

[midnight plank]

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[digital quail]

.reopen

[midnight plank]

✅

[digital quail]

.close

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[last slate]

How would I prove this given the following two axioms a and b?

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[grim carbon]

How would you prove by contradiction: There is no smallest irrational number.

What I have tried so far is: Suppose on the contrary that there is a smallest irrational number, call it x. Then |x/2| < |x| is a smaller number, and so must be rational. This means that x/2 = m/n for some integers m, n, where n != 0. But then we get x = 2m/n, where 2m and n are integers, so x is rational and we get a contradiction.

[grim carbon]

This is accounting for "smallest irrational number" and not "smallest positive irrational number" so it accounts for negatives too

[junior flower]

that doesn't really make sense

|x/2| is not always less than x and I don't think being less than |x| is helpful

if you want to stay with that idea you could first assume x is positive and get rid of the absolute value bars and that would be fine

actually that's probably not helpful either, so I'd probably abandon that idea lol

have you maybe learned than sqrt(n) is irrational whenever n is an integer that isn't a perfect square?

[midnight plank]

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[sour reef]

https://en.wikipedia.org/wiki/Leibniz's_notation

[hallow olive]

It's just the notation to say you're taking the derivative of r with respect to t

There are many different notations, this is Leibniz's notation. Another one often used is Lagrange's notation, which in this case would be r'(t)

[sour reef]

The question of "why use d like that" and "are dr and dt their own entities" is a big question that dips into the history of calculus and the philosophy/ontology of calculus. Probably best to just think of it as a black box notation, as a beginner at least.

(sorry, I know that's an unsatisfying answer)

[hallow olive]

I agree, just think about it as notation for now

[sour reef]

Yes, you'll see it written both ways

$\frac{dr}{dt}$ and $\frac{d}{dt}r$ mean exactly the same thing.

[grand pond]

RSB

[midnight plank]

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[smoky trench]

Could anyone please help?

[flat halo]

hihi

so

the new price is

what pc of the original

[smoky trench]

What

@flat halo

[flat halo]

whats the new price

in terms of

%

of the original

[smoky trench]

Idk

That’s why I’m asking u for help

[flat halo]

ok

the original price is

100%

so

thrs a

35% discount

whats the new price in terms of

%

of the orig price

[smoky trench]

I don’t know

[flat halo]

do u know what a 35% discount means

[smoky trench]

Yes

[flat halo]

what does it mean

[smoky trench]

I did a same question on it like this

And I divided it by 100

So it becomes a decimal then times by the amount

[flat halo]

ok so

whats that amt

[smoky trench]

And I did that for this question and it said it was wrong that’s why I’m asking your

[flat halo]

ok what did u do

[smoky trench]

The amount is £49.40

Could you just work out the answer

[flat halo]

no

[smoky trench]

I need help

[flat halo]

what did u do

[smoky trench]

I got it

Bye

[flat halo]

well

good job

[hallow olive]

.close

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[quasi meadow]

When a quantity q of units is demanded,
the price for each unit is p = q2 + 20q dollars. When a quantity q of units is
supplied, the price for each unit is p = −2q2 + 10q + 3000 dollars. Find the
equilibrium quantity and equilibrium price.

[quasi meadow]

i know you set the two equations =

so q^2+20q=-2q^2+10q=3000

which equals -3q^2-10q+300

what do i do with exponets how do i get rid of them

[hallow olive]

It's a quadratic equation, right?
-3q²-10q+3000=0

You can solve it with the quadratic formula

[quasi meadow]

(x-30)(3x+100) you can factor too right?

i used x

because easier but i know its q

[hallow olive]

Yes, you can factor as well

[quasi meadow]

than what would i do with this factored form to figure out the equilibrium quianity?

[hallow olive]

,w factor -3x^2-10x+3000

[quasi meadow]

so equilibrium quaninty = 30\

[grand pond]

Results provided by WolframAlpha

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

Requested by Andrea276#3928

[hallow olive]

Just checking if your factorization is correct, and it is

[hallow olive]

Yes

30

[quasi meadow]

ok i have one mor equesiton if you dont mind?

[hallow olive]

Since it's definitely not negative, like the other solution to the equation

[quasi meadow]

yeah haha it cant be negative

[hallow olive]

Send it, if I can help I'll do it

[quasi meadow]

you able to?

[hallow olive]

Have you already tried something?

[quasi meadow]

i belive y = 2154.92 and y0 is 620

so 2154.92=620b^t

[hallow olive]

Ok, so y0 is correct

y, instead, is just a variable that changes each month depending on t. You need to find b instead

Wait, y0 is not correct since 620 is not the initial balance, but the balance after a month

Ok, so let's start from the beginning. You need to find an exponential function, $y=y_0b^t$. Your goal is to find $y_0$ and b.

[grand pond]

Andrea276

[quasi meadow]

alright

[hallow olive]

You have plenty of data from your table to do that. You know y_0 always remains the same, and b is raised to different powers based on the month. So you're going to get y=y0b¹ after the first month, y=y0b² after the second month etc.

You need to find the value of two variables, y0 and b, so a system of two equations should be more than enough for that.

Remember that y is the debt, and t is the number of months passed

You want to find y0 and b instead

[quasi meadow]

im so confused on how i would set something like that up

i know for first month for example 620=y0b^1

[hallow olive]

Yes

Now try with the second month

[quasi meadow]

761.88=y0b^2

would i use these two equations?

[hallow olive]

Yes

[quasi meadow]

how would I use them to help find the solution tho?

set them equal>

?

[hallow olive]

You can set up a system of equations with them

[quasi meadow]

so like elimantion subsution

[hallow olive]

Two equations and two variables, you should be able to solve it

Yes

620=y0b
761.88=y0b²

[quasi meadow]

so
620=y0b^1
761.88=y0b^2

[hallow olive]

I'd suggest to start by solving for y0 in the first equation

Then substitute it into the second

[hallow olive]

I meant y0, not b, sorry

[quasi meadow]

how would you slove for it though i am sorry im so consfused i was sick for this lecture when my prof went over so im trying to learn all this new material by myself

she doesnt post any notes to help us online

[hallow olive]

Have you ever solved a system of equations?
Like:
2x+y=6
3x-2y=5

[quasi meadow]

yes thats a lot easier

that makes sense in that form

times top by 2

4x+2y=12

subtract y

[hallow olive]

That's the same, the variables are y0 and b instead of x and y

[hallow olive]

Don't solve it, I'm sure you can do it ahah

620=y0*b
761.88=y0*b²

[quasi meadow]

left with 620=b
and 761.88=b^2

?

wait nvm

mutiply top by a negative 1

so you can subtract?

to do elimation i dont understand substution lmao never done it

[hallow olive]

No, it's a multiplication, you can't solve it that way

You should use substitution (or I think that's how it's called).
Solve for y0 in terms of b using the first equation, then substitute it into the second equation

Start by solving for y0 in terms of b

[quasi meadow]

but howww im so confsued

[hallow olive]

620=y0*b

Isolate y0

Example:
x*y=5
Isolate x

[quasi meadow]

-y from each side

?

[hallow olive]

No, y it's multiplied, not added

[quasi meadow]

x=5-y?

[hallow olive]

If it were x+y=5 you would hsve subtracted y from both sides

[quasi meadow]

so would you divded it out?

[hallow olive]

But its multiplied, so you should divide by y both sides instead

[hallow olive]

Yes

[hallow olive]

So x=5/y, right?

[quasi meadow]

yes

so 620/b=y0

?

[hallow olive]

Correct, now substitute it into y0 in the second equation

761.88=y0*b²

y0=620/b

761.88=(620/b)*b²
right?

[quasi meadow]

one sec

seeing how you did it

lmaoo

ohhh ok

i see it

yes you're right

just replaced y0 with the 620/bb

[hallow olive]

Then you need to simplify and solve for b

[hallow olive]

Yep

[quasi meadow]

so

761.88 = 620/x * x2 than you would cancel out common factor?

so cancel out x on each

[hallow olive]

Yes

[quasi meadow]

so

761.88=620x

[hallow olive]

👍

[quasi meadow]

divded both sides by 620?

[hallow olive]

Correct

[quasi meadow]

1.228

would you round up

1.23?

[hallow olive]

Yes, keep it as 1.23

[quasi meadow]

so b=1.23

?

[hallow olive]

That's an approximation anyways

[hallow olive]

Yes

[quasi meadow]

ok dope

[hallow olive]

Then you had this

[quasi meadow]

620/1.23

504.07

[hallow olive]

👍

[quasi meadow]

y0=504.07

[hallow olive]

put everything together into the exponential equation now

y=y0*b^t

[quasi meadow]

620=504.07*1.23^t

[hallow olive]

You shouldnt substitute 620

[quasi meadow]

just leave it as y

[hallow olive]

Since that value depends on the month, so on t

[hallow olive]

Yes

[quasi meadow]

ok so the exponetial equation is y=504.07*1.23^t

and thats part a done

[hallow olive]

Correct

For b, how many months in a year?

[quasi meadow]

12

so y=504.07*1.23^12

[hallow olive]

Yep

[quasi meadow]

6044.38

that seems really high

in 4 months to go up by 4k

[hallow olive]

Exponential functions grow really quickly

[quasi meadow]

ok that is it

thank you for the help

i apprecaite it a lot

[hallow olive]

Yw

[quasi meadow]

take care man

.close

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[silver terrace]

Im having trouble understanding what i should do with the x of the equation as we are figuring for delta(im pretty sure thats the variable shown) and i also dont know how to do this without a table of values or without a graph

[strong lava]

It's not delta it's a small sigma (capital sigma is the summation sign)

Also here they don't specify the value of x and sigma for the limits in which they don't intervene, because you can show that the result doesn't depend on their value

[silver terrace]

How would i go about finding the result if i dont need to know the value of either of the variables?

Wait is it more like since that sigma is 0 it would be undefined and that if x is 0 it would be 1?

[strong lava]

For x = 0 you can indeed do it using continuity

For sigma -> 0 you can do it by composing limits

Since exp is continuous

lim exp(stuff) = exp(lim stuff)

Then you can compute it

[midnight plank]

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[normal citrus]

i don't understand, for me it's 1/8cos(4x) - 1/4cos(2x) + 1/4

[sand bluff]

redo your work

you probably did something wrong when converting e^4ix + e^-4ix

[normal citrus]

ah ok i got it

.close

[midnight plank]

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[normal citrus]

hum why?

[elder zephyr]

calculate the right side and you will see why

[normal citrus]

i mean how do we pass the left to the right

because it's pretty useful but i don't understand

[pearl idol]

wait what's your question?

if you calculate out the right side, do you see how it equals the left side

[normal citrus]

yeah the question wasn't clear sry

[elder zephyr]

by factoring by exp(ipi/12)

[normal citrus]

and why exp(ipi/12) is there a rule?

[pearl idol]

that depends on what you're trying to achieve

Is there more context to this question

[elder zephyr]

writing this thing in exponential form

rexp(iα)

[normal citrus]

.close

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[midnight plank]

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[eager plover]

Hello

[eager plover]

anyone know how to find the inverse of this functions

[obtuse basin]

Just isolate y in x = 100/(8 + 6^(-y))

(you can start by isolating 6^(-y) is you're having trouble on how to start)

[eager plover]

how can i isolate that 6^-y

like how can i bring it down from being exponentiated

[flat halo]

use ln

[obtuse basin]

log_6 would be more convenient for this

[flat halo]

6^(-y) = e^ln(6^(-y)) =

hm ye

LOL

[hard umbra]

all logs are the same

[obtuse basin]

Any logs would work ye, but what's the point of writing e^ln(6) or a^log_a(6)

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[obtuse basin]

So you have x + 10sqrtx + 25 = 2x + 1, right?

(also don't forget to write that x needs to be positive)

[obtuse basin]

So from here you need to isolate the sqrt

10sqrtx = x - 24

Before squaring both sides again notice that x - 24 needs to be positive

So x > 24

After squaring both sides you get 100x = x^2 - 48x + 24^2, which is just a quadratic equation

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[normal citrus]

How can I solve that

[normal citrus]

Since the 2 last line are the same I'm blocked

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[normal citrus]

<@&286206848099549185>

[small turtle]

Write the coefficients as a matrix and do row simplifications

[normal citrus]

i wanna do it without matrice since it's not the theme of the chapter

[small turtle]

Ok since last 2 are both = 1 you can also set them equal

[normal citrus]

plus i can't use the calculate

[small turtle]

x+y-z = 7x+y+5z

6x=-6z

x=-z

[normal citrus]

well see

[small turtle]

x-y+3z=-1

x-y-3x=-1

-2x-y=-1

Can you do the rest

[normal citrus]

x = 1/2 -y/2

[small turtle]

From second equation:
y=1+z-x = 1-2x

Then you automatically get z too as z=-x

[normal citrus]

then x = 1/2 - 1/2 -x

[small turtle]

Then plug these in to get y

[normal citrus]

useless x)

[small turtle]

Oh yeah I made a mistake sorry😄

[normal citrus]

well it's no helping too

[small turtle]

Yeah it’s mistake 1 sec

z=-x is correct though

And y is just y = 1-2x

Putting x=-z in first equation:
x-y-3x = -1
gives -2x-y=-1 so y = 1-2x

[small turtle]

Wait it was the solution already here see

[normal citrus]

and that's it?

we can't go futher?

[small turtle]

Yeah can’t do any further except for specific values

[normal citrus]

ok

thx for your help

.close

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[hot stirrup]

How to find Velocity of y?

[hot stirrup]

idk if i did it right

I just wanna know how u guys would do it

[lilac kayak]

what is y?

i guess x is displacement

and so is y?

[hot stirrup]

Y is the graph on the xy plane

it said something like that

[dull yoke]

What do the 2 equations represent

[hot stirrup]

sorry

the x represents the x in respect to t

and the y is y in respect to x

idk

it was physics problems

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[last slate]

how does it simplify to that

[tropic nexus]

what is the common denominator?

[last slate]

ik that u multiply the left side by (t^4-2t)^3

[tropic nexus]

are you sure that's what you want to do?

[last slate]

wait

not cubed

[tropic nexus]

yeah

just t^4-2t

[last slate]

yes

and then multiply everything out?

[tropic nexus]

then you have to expand it and simplify

yeah

really messy

[last slate]

yeah lol let me do that rn

@tropic nexus

for the simplified version

in the pic i sent

how did the t^4 become t^3

[tropic nexus]

it looks like they took out a t^4 from the entire denominator

and the t^4 became a t^3

[last slate]

yeah i did everything and still didnt get the same answer

could u try to do it on paper for me and send me it please

i cant get it to work for some reason

@tropic nexus

[tropic nexus]

I dont want to 😭

try wolfram alpha

,ask (t^4-2t)*9t^8

[grand pond]

Results provided by WolframAlpha

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

Requested by SlowBrosRUs#9180

[tropic nexus]

ok 9t^12-18t^9

,ask 9t^12-18t^9-3t^9(4t^3-2)

[grand pond]

Results provided by WolframAlpha

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

Requested by SlowBrosRUs#9180

[last slate]

lol

[tropic nexus]

-3t^12-12t^9

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[tranquil turtle]

help here please

[midnight plank]

@tranquil turtle Has your question been resolved?

[tranquil turtle]

<@&286206848099549185>

I have tried to make a perpendicular line, and measure the distance from the pie shop and the point of intersection

[midnight plank]

@tranquil turtle Has your question been resolved?

[tranquil turtle]

I tried solving, and my final answer is that the pie store is 9.8 km away from the shop

<@&286206848099549185>

!close

?close

+close

;close

.close

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[normal idol]

uh

[normal idol]

LMAO

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@normal idol Has your question been resolved?

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[normal idol]

.close

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[vast swift]

I'm quite sure I forgot how to do this someone give me some general info?

[midnight plank]

@vast swift Has your question been resolved?

[vast swift]

<@&286206848099549185>

[tender linden]

Use the product rule, quotient rule, and chain rule

For example, h'(x) for the first question would be f'(x)g(x)+f(x)g'(x)