help-10

[scenic lynx]

thank youu

.close

[obtuse pebble]

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[grave halo]

hey hope y’all are doing good. was wondering if someone could help me find where i went wrong in this pic.

[tardy epoch]

can you show more steps. like i get you're doing \int u dv = uv - \int v du.

what are u, dv and how did you evaluate \int v du

[grave halo]

I’m doing integral of ln(x)*1/x^2 and am using the method

integral of f’g = fg - integral of f*g’

with f(x) being 1/x^2

and g’(x) being ln(x)

it works when using opposites for f and g’

but im curious as to why it doesnt work the first way

the last line here is where it works

[high lily]

because you integrated both the ln(x) and the 1/x^2 in the first approach which isn't what you're supposed to do

[obtuse pebble]

@grave halo Has your question been resolved?

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[tacit lava]

can someone help me solve this

[tacit lava]

so far the working i have done is that Q3 = 213180, Q4 = 426360.

<@&286206848099549185>

[timid silo]

q1+4/3q1 = total-q3-q4

[tacit lava]

so what do i do next

[grave halo]

if q3 is 17%

q4 is 34%

[tacit lava]

yeah i have worked out q3 and 4 its just i cant seem to work out q1 and 2

[grave halo]

then q1 + q2 = 100 - 17 - 34

[tacit lava]

qarter 3 is 213,180

and quarter 4 is 426360

[grave halo]

and 4/3q1 = q2

[grave halo]

use these two equations in a system

and you will find the answer

[tacit lava]

so what do i do next

im sorry im just clueless at the moment

[grave halo]

do you know how to solve systems of equations?

[tacit lava]

nope

[grave halo]

q1 and q1 put together is 49%

and q2 is 1/3 bigger than q1

so in total q2 is worth 4/3 (q1's) and q1 is ofcourse worth 1 q1 so 3/3

4/3+3/3 = 7/3

[tacit lava]

so quarter 2 is 27%

and quarter 1 22%?

[grave halo]

I don't believe so

but with what i just said we have 7/3 q1 = 49

[tacit lava]

okay

[grave halo]

and therefore q1 = 21%

and since we have q1, q3, and q4

we can do 100% (the total) - q1 - q3 - q4 = q2

which means q2 = 100 - 21 - 17 - 34

[tacit lava]

okay thanks

i understand now

[grave halo]

no worries

if you're done you can type ".close"

[tacit lava]

.close

[obtuse pebble]

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[fallow stump]

https://i.gyazo.com/56a5b55bbbeda6c137e0f990348f1844.png

[fallow stump]

How can I solve this?

[grave halo]

I believe it is the exact same problem as yours

I don't know why the picture cut

[fallow stump]

AH thanks

cosine rule yeah

.close

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[vale warren]

Having some trouble determining the derivative of f(π/4) by the definition

[brave bramble]

Have any work so far?

Well, honestly you can just google "derivative of cos(x) by definition"

I wouldn't expect you to figure it out without knowing it beforehand

[vale warren]

I got this for now

Im stuck because i cant get rid of the π/4+y

to get to the notable limit

[brave bramble]

Use the h → 0 definition, almost always easier.

[vale warren]

K ill try it

[brave bramble]

Don't do the trig identity

Instead, use the summation identity

[vale warren]

trig identity?

[brave bramble]

The cos ²(x) - sin²(x) will make your life harder

[vale warren]

so turn it into 1 - 2sin^2(x)

[brave bramble]

Leave cos(2x) as is haha

[vale warren]

im gonna have to get rid of it tough

could you solve it and send me the solution please

<@&286206848099549185>

[timid silo]

are you allowed to use Lhopital ?

[vale warren]

No

[timid silo]

Ok

[timid silo]

First use difference of square formula then √2 comes out now change cosx to sin(π/2-x) then use sinC + sinD formula

@vale warren

[vale warren]

ye hold on

[timid silo]

k

[vale warren]

so

cos^2(π/4+y) - sin^2 (π/4+y) = [cos(π/4+y) + sin(π/4+y)][cos(π/4+y) - sin(π/4+y)] right?

@timid silo

[timid silo]

Yes

Go ahead

[vale warren]

But where does the √2

come

from

[timid silo]

you can solve the first bracket as it is finite and not equal to 0 so you can take it out of the limit

[vale warren]

OH

[timid silo]

put y=0 in it

[vale warren]

youre right

i can separate the limits

[timid silo]

Yes

[vale warren]

but im gonna have to undo the variable change right

so i can change the cos

[timid silo]

No why

You don't have to

[vale warren]

i cant change co(π/4+y) to sin(π/2-y)

[timid silo]

you change it sin(π/4-y)

Yes you can why can't you

[vale warren]

why

[timid silo]

Why not what's wrong

Wait

Its π/4 -y

[vale warren]

isnt it cos(π/2+y)=-sin(y)

[timid silo]

Bruh what's this same argument

How can they have same arguments and be equal

[vale warren]

oh fuck nvm

i edited

it

i just copy and pasted the argument so i wouldnt g«have to write it again

and forgot to take the parts

[timid silo]

No there would be no negative sign outside
It was on the inside but cos(-x) = cosx

[vale warren]

are you talking about the second one?

im talking abt this

[timid silo]

No I'm talking about cosx = sin(π/2-x)

Bro I gotta go sleep so ask someone else please

But I've already told how it would be done

[vale warren]

Ah ok

Alright thx so much m8

Sleep tight

[obtuse pebble]

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[low patrol]

When is a system of differential equations linear?

[grave halo]

if all of the terms are to the power of one

no x^2 or y^2 etc.

[low patrol]

wut

like

y'(x) = -y(x) is linear then

but y'(x) = y^2(x) is not?

[grave halo]

my bad I read your question too quickly

I didn't see the "differential"

[low patrol]

oh ok, thanks tho

[grave halo]

"The linear differential equation is of the form dy/dx + Py = Q, where P and Q are numeric constants or functions in x. It consists of a y and a derivative of y. The differential is a first-order differentiation and is called the first-order linear differential equation."

you can just google it and that comes up first

tell me if you have more specific questions that a simple google search doesn't answer

[low patrol]

Still looking for an answer. Not from you, thanks

Looking for someone to help me understanding the above definition of a linear system of differential equations

Is it that we can "condense" all the system to an equation of the form y^(k)(t) = A(t)x+ b, where A is a matrix of functions of t and x(t) is a vector of "derivatives" of x?

like x = (t, x, x^(1), ..., x^(k-1))

[tardy epoch]

"linear" refers to linear in the terms with derivatives

[tardy epoch]

You're mixing variables here. why do you have both y and x as a function of t?

[tardy epoch]

"ordinary" refers to only one dependent variable and its derivatives

[tardy epoch]

this would be a PDE

[low patrol]

I mean y, instead of x

sorry

let me rewrite: y^(k)(t) = A(t)y + b, with y a vector of the form (t,y,y^(1), ... , y^(k-1))

[tardy epoch]

this is awful notation

pick something more obvious

https://people.qatar.tamu.edu/tingwen.huang/math308_spring13/4.1.pdf

You can write the left hand side as a dot product

[low patrol]

How would you do it?

[tardy epoch]

just write it like this. it's not that hard

matrix multiplication and dot product and all this summation saves like 5 seconds of work

[tardy epoch]

if you have a system, then you have m of these for each y_1, y_2, ... y_m

if you insist, then it's in section 6.2 here
https://bookdown.org/vshahrez/lecture-notes/introduction-to-systems-of-odes.html

and page 1 here
http://www.math.utah.edu/~gustafso/2250systems-de.pdf

[low patrol]

Thank you

.close

[obtuse pebble]

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[low patrol]

what's n(P)?

negation?

[lilac grove]

Ik this means that the set P intersection Q is equal to the set P

[lilac grove]

Number of elements in the set P

[low patrol]

oh ok

P is a subset of Q tho

[lilac grove]

Green=P, Yellow=Q

[lilac grove]

How

[versed turret]

Read n(P n Q) = n(P) as "everything in P is also in P n Q, which is also all in Q."

"So all of P is in Q" → P ⊂ Q

[lilac grove]

How did u interpret those wordings

Interpret*

[versed turret]

Imagine the sets, P and Q.

P n Q is defined as all the elements in both P and Q

If something from P is in P n Q, then it must also be in Q, because otherwise it wouldn't be in P n Q

Does that make sense?

[lilac grove]

Yes

[versed turret]

So then if EVERYTHING from P is in P n Q, then everything from P n Q is also in Q

On the venn diagram, this is the same as the intersection being completely inside Q

Which is only possible if all of P is inside Q

⟹ P is a subset of Q

[lilac grove]

I see "everything in P is also in P and Q"

Why can't Q be a subset of P then

[versed turret]

I guess technically we don't know that P ≠ Q

[lilac grove]

Ye

Q can have other elements

Not similar to the one in P

[versed turret]

P ⊆ Q

[lilac grove]

Ye that makes sense

[lilac grove]

But that says P = Q

Can we remove that line under the symbol

[versed turret]

The question doesn't say anything else, so from what I gather we don't really know. P could be equal to Q

If it is, the original statement still holds true

Because the only constraint that the question applies is that |Q| >= |P|

[lilac grove]

Idk that

But

Shoukd I check the answer

[versed turret]

Yes

[versed turret]

Yeah I think any diagram where P is inside Q would've worked

[lilac grove]

n(...) : number of elements
n : and
= : is/are

I read that as "number of elements (that are) in P and (in) Q are (the) number of elements in P"

[versed turret]

Yes.

[lilac grove]

I still don't understand tho

[versed turret]

What do you not understand?

[lilac grove]

Just says about the number of elements

Not the elements themselves

[versed turret]

The number of elements in P refers to all the elements of P

We don't care about what the elements themselves are.

[lilac grove]

Is there a proof for this in raw set (roster) form?

[versed turret]

I mean just think about it, if you add an element to a set, the number of elements of that set goes up by 1, no?

[lilac grove]

Ye it does

[versed turret]

That's just how n(P) is defined

The number of elements of P

If n(P) = n(P n Q) and everything in n(P n Q) is in P, then because of the equality everything from P has to be in P n Q

[lilac grove]

I am getting it better now

We can say that Q is a proper subset of P

Can't say P is a proper subset of Q as we don't have enough information abt Q and it can have more number of elements than $n(P\cap Q)$

[warm shale]

JUGisMUG

[lilac grove]

Am I right abt this?

[obtuse pebble]

@lilac grove Has your question been resolved?

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[copper latch]

L’hospital?

[haughty coyote]

No need

Just compute it. It's fairly simple

[copper latch]

Yea that works fine too

[haughty coyote]

l'Hospital's is not even possible here

[obtuse pebble]

@lime flicker Has your question been resolved?

[uncut ridge]

Limit does not exists, you can approach x to 4 from the right and the left to get infinity with opposite sign

[haughty coyote]

No there's a square

So it's -1/0+

[uncut ridge]

right

it goes to -infty

[lime flicker]

i cannot plug in 4 right

[haughty coyote]

No but the denominator has a nice limit

[uncut ridge]

Try to factor the denominator

[lime flicker]

I did factor it, but it didnt help me

[uncut ridge]

what did you get when factoring?

[haughty coyote]

The denominator goes to 0, and since it's a square, it has to be positive

There is no need to factor anything

[lime flicker]

(z-6)(z-4)

[uncut ridge]

it's just to see why we can't just plug 4 in

[pallid flame]

we can tho

-1/0+

= -∞

[lime flicker]

why is it 0+ and not just 0, which it would normally be when we plug in to the denominator?

[haughty coyote]

Just plugging gives (-1)/0 though. You need more than that to say 0+

[rotund pecan]

Plug in 4

Literally just plug it in

[haughty coyote]

That allows you to say 0+

[lime flicker]

i see

[pallid flame]

then plug in x = 4 + ε

[lime flicker]

it wouldnt work if it was 0, the answer is infinity because its 0+ right?

[haughty coyote]

That's not plugging 4, that's doing extra work

[pallid flame]

dual numbers

[haughty coyote]

Yes otherwise you don't know the sign of the infinity

[pallid flame]

x = 4 + ε with ε^2 = 0

[haughty coyote]

Beyond the score of the problem. This is beginner limits, not more advanced methods

[uncut ridge]

You can also prove this using the formal definition of limit but i think that just approaching 4 from both sides you can see that it goes to -infinity

(delta-epsilon definition)

[lime flicker]

i see.

.close

[obtuse pebble]

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[obtuse pebble]

@gleaming pivot Has your question been resolved?

[obtuse pebble]

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[gleaming pivot]

.reopen

[obtuse pebble]

✅

[gleaming pivot]

<@&286206848099549185>

[burnt scroll]

Which parts did you do so far @gleaming pivot

[obtuse pebble]

@gleaming pivot Has your question been resolved?

[obtuse pebble]

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[slow owl]

what would happen if i created a composite function g(g(x)) where g(x) = |x|

[slow owl]

how would i write the absolute of x twie

twice*

[timid silo]

just writ ethe absolute value twice

[slow owl]

so | |x| |

[timid silo]

then get rid of one of them since the absolute value of absolute value is equal to the absolute value

[slow owl]

thanks bro

.close

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[autumn oak]

how this work??

[nocturne minnow]

You can create a system of equations with the given info

If t = 2 years, the price is 42750, correct?

You can plug in that info into the formula

You have one equation you need

The other, what is the value of t when it was 57000?

[autumn oak]

how do I find A and k tho???

[nocturne minnow]

Solving a system of equations

Like I mentioned

[autumn oak]

But how do i plug a and k in there if i dont know what it is??

[nocturne minnow]

Have you ever done something like this?
2x + y = 4
-5x - 3y = 9?

To find x and y?

Because if you have, that is the same process you are applying to your problem

[autumn oak]

oh yeah

i wonder if that the way we meant to work out this question

but ill give it a go

[nocturne minnow]

That is the way you need

By creating the two equations, you can then find the values of A and k

[obtuse pebble]

@autumn oak Has your question been resolved?

[obtuse pebble]

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[timid silo]

Write the equation in standard form for the ellipse with center at the origin, vertex (
–
10,0), and co-vertex (0,4).

[timid silo]

ion even know where to start

[royal basin]

do you know in general what the equation of an ellipse looks like?

[timid silo]

nah

[royal basin]

so this: $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ is foreign to you?

[warm shale]

Ann

[timid silo]

ok

[royal basin]

maybe it's just me, but hearing an answer of "nah" to "do you know how to do X in general?" feels like you're being purposefully dismissive of people's attempts to help you.

in any case

[timid silo]

my fault lemme turn my black self white and talk all proper

[royal basin]

maybe it's just me

no need

oh, they left.

[vagrant gale]

I had this whole ass explanation planned out but that's now cut

[royal basin]

.close

[obtuse pebble]

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[rough berry]

rip

[scarlet locust]

f

[obtuse pebble]

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[shadow lava]

Alright so I know this answer is wrong but I am just trying to understand why?

[shadow lava]

That last part how come I’m not allowed to remove the 7 on the numerator and denominator?

[tardy epoch]

$\frac{a+b}{c + b}$ does not equal $\frac{a}{c}$

[warm shale]

riemann

[scarlet locust]

take a simple example of $\frac{1+1}{2 + 1} \ne \frac{1}{2}$

@warm shale

[tardy epoch]

no spaces on the inside of $

[shadow lava]

Hmmm 🤔

[warm shale]

Swas.py

[scarlet locust]

thanks

[shadow lava]

How do you remember this rule? I get very confused. Here it is allowed even with addition included

[scarlet locust]

it's not a rule

[scarlet locust]

oh this?

[shadow lava]

Here we are just simplifying right?

[scarlet locust]

its just $\frac{9}{-3} + \frac{3\sqrt{7}}{-3}$

[warm shale]

Swas.py

[shadow lava]

Right

[high lily]

$\frac{a\pm b}{c} = \frac ac \pm \frac bc$

[warm shale]

ℝamonov

[shadow lava]

OK I think I understand if broken up into two rational numbers

[tardy epoch]

fractions* not rational numbers

[shadow lava]

But with this example it must be broken up into (a / c+b) + (b / c + b)

[shadow lava]

Oh Ok I always think rational for fractions for some reason

[tardy epoch]

then do it? just stop making the mistake i pointed out

[shadow lava]

I think rational it is the global term for almost all numbers

Aside from pi and infinity

[tardy epoch]

ai yai

[high lily]

rational numbers can be expressed in the form p/q
where p and q are integers

[scarlet locust]

🤷‍♂️

[high lily]

irrational numbers can not

[scarlet locust]

sqrt(2) be like

[shadow lava]

Ty all

I will keep that in mind

I think I am trying to simplify or something so I remove from num and den but I can’t do that

Will keep practicing and hopefully get it right

.close

[obtuse pebble]

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[obtuse pebble]

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[tight kettle]

(x+2)(3-x)+6=0

how do you cross multi

[tight kettle]

help

[heady cipher]

X times 3 and 2 times 3 and x rimes x and 2 times x

[tight kettle]

ok wait

what next

i wrote it in paper

[heady cipher]

Oh btw its x times -x

I didnt see

[tight kettle]

ok

[heady cipher]

Damn wait

X times 3

X times - x

2 times 3

2 times - x

There we go

Fixed rhe negatives

What did you get when you did that?

[tight kettle]

lol

i dont know how to cross multi it

i just wrote down facotrs

[heady cipher]

Lemme show u with a pictures

[tight kettle]

factors

ok thank you

[heady cipher]

The blue lines

You multiply x by both 3 and -x

[tight kettle]

oh

[heady cipher]

And the purple

Yoi multiply 2 with 3 and -x

[tight kettle]

ok

how do you cross multiply it using factors

it said i have to factorize it it

then cross multiplication

ayo

[acoustic bronze]

isn't it already partially factored

[tight kettle]

what

i mean its kinda my fault because

i didnt listen to clas

<@&286206848099549185> please help

[acoustic bronze]

what u want to do is do foil so that you expand the entire thing

and then combine like terms, and then factor

[tight kettle]

ok

.close

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[coral parcel]

Ive attempted subbing this into the base equation but I just can't see that working

<@&286206848099549185>

<@&286206848099549185>

[obtuse pebble]

@coral parcel Has your question been resolved?

[coral parcel]

<@&286206848099549185>

[tight kettle]

rip lol

[coral parcel]

ikr

<@&286206848099549185>

[swift marsh]

Sorry what are you actually trying to do?

I can't tell if part (b) is related in some way to the top and what you're actually currently stuck on

[coral parcel]

Umm

[obtuse pebble]

@coral parcel Has your question been resolved?

[timid silo]

Well after brushing up my quadratics a bit

You have the maximum/minimum right? Thats turning point? (English second)

In that case u can just use the formula

In green

And then find a by plugging in another value, which you have(?) (4,6)

I dunno bout dilation tho. That means the parabola gets wider/narrower?

I guess youd have to multiply the whole formula you got by -2?

I could be wrong! If there are examples in your book/material then maybe take a look if u havent already

Actually no

The graph passes thru (4,6) after the dilation so after u plugged the top into the formula in green, then you multiply by -2 and then u plug in (4,6). To be sure. I could be totally wrong tho

[obtuse pebble]

@coral parcel Has your question been resolved?

[obtuse pebble]

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[timid silo]

How do I start solving this?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[cedar kiln]

Hi

[cedar kiln]

can you help me find two problems with two ways solutions

?

anything

just example

this helps me to mutch plz

[warm canopy]

You need to say what level of maths you're at

[cedar kiln]

no prob just examples

if you can help

[high lily]

$\int_{-2}^2 \sqrt{4-x^2} \dd{x}$

[warm shale]

ℝamonov

[high lily]

solve
x^2 = 1

[cedar kiln]

bro something simple i don’t understand that

[thick oracle]

u mean a problem with 2 solutions?

[cedar kiln]

2 or more ways to get solution

sorry 😞

[thick oracle]

a quadratic

1. quadratic formula

2. factoring

[cedar kiln]

ok

[nocturne minnow]

3. Completing the square

[cedar kiln]

ok send me

[lethal sand]

4. Guessing

[thick oracle]

haha

[cedar kiln]

examples

[nocturne minnow]

You can look them up

[cedar kiln]

2

[nocturne minnow]

Google has quadratics everywhere

[vast wren]

5. using graphs

[cedar kiln]

4

[glossy ibex]

4.5 checking

[vast wren]

4.5.1 simply using wolfram

[nocturne minnow]

6. Getting a friend to do the problem for you

[thick oracle]

$$x^2+x-12=0$$

[warm shale]

Mike Oxbig

[vast wren]

Show that cos(pi/7) - cos(2pi/7) + cos(3pi/7) = 1/2

There’s both an algebraic and geometric solution

[thick oracle]

done

[vast wren]

3. use wolfram lol

[thick oracle]

yeah

[vast wren]

I actually like its geometric solution a lot, it involves a regular heptagon

[cedar kiln]

Hello People if you can think of two examples of problems that have two or more ways to solve

[vast wren]

[cedar kiln]

😩

[thick oracle]

a geometry problem

[vast wren]

if you know logarithm:

[thick oracle]

1. pythagoras

[cedar kiln]

anything

[thick oracle]

2. trigonometry

[vast wren]

if 3^x = 15^y, then find the value of 5^((2x-y)/(x-y))

1. logarithm

2. exponential conversions

[cedar kiln]

what is that

!close

[thick oracle]

lamo

[vast wren]

buh

[thick oracle]

use this

[vast wren]

^

[thick oracle]

u have 2 ways

to solve for ac

pythagoras or trygonometry

[cedar kiln]

ok cool

thanks

[thick oracle]

np

[cedar kiln]

do you have and one more

[thick oracle]

um

idk

lol

[vast wren]

lemme think

[thick oracle]

proving pythagoras theorem

[vast wren]

you can look up proofs of the menelaus theorem

[thick oracle]

there are over 300 proofs

[cedar kiln]

Tell myself "Leave!", while I'm still strong
"Don't look back!", 'til I'm ten miles gone🤟

[vast wren]

ok bro

[thick oracle]

huh

[vast wren]

Let a, b, and c be positive integers. For a+b+c=10, how many trios of (a, b, c) exist?

1. trial & error

[thick oracle]

haha

[vast wren]

2. permutation

idk dude

[thick oracle]

any question can be solved in multiple ways

just so u know

👍

[alpine bison]

the worst way is induction remeber

[vast wren]

P(x) is a third degree polynomial. Find P(7) where P(1)=1, P(2)=3, P(3)=5, and P(4)=6.

[thick oracle]

bro he has no idea what that is

[vast wren]

yea lmao

[thick oracle]

)))

[vast wren]

❤️

thanks for the heart tho

[thick oracle]

https://tenor.com/view/the-wolf-of-wall-street-clap-clapping-clapping-hands-leonardo-dicaprio-gif-22829304

[cedar kiln]

Thanks a lot ❤️

.close

[obtuse pebble]

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[vast wren]

You’re welcome ❤️

lmao

[thick oracle]

bro u in love or something

[vast wren]

what a romantic dude

[thick oracle]

)

np bro

[obtuse pebble]

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[wintry topaz]

In a 4-digit number 4x7y replace x and y to get a smallest number divisible by 12

[wintry topaz]

I got a result of x=1 and y=6 and my book says x=2 and y=2 is correct.

What am i missing?

[obtuse pebble]

@wintry topaz Has your question been resolved?

[drifting wraith]

lmao

maybe it's a typo, like it's actually 4x8y idk

[wintry topaz]

ye probably a typo ive been staring at it for 10 minutes, no way im reading it wrong lol

[obtuse pebble]

@wintry topaz Has your question been resolved?

[obtuse pebble]

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[hardy fulcrum]

in how many ways can we send six urgent letters if we can use three messengers and each letter can be given to any of them ? so that each has at least one.

[hardy fulcrum]

plz help

<@&286206848099549185>

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[obtuse pebble]

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[dreamy ember]

I need help

[robust sleet]

what have u tried

[obtuse pebble]

@dreamy ember Has your question been resolved?

[fierce lagoon]

They gone

They asked and then dipped

[dreamy ember]

The answer

[fierce lagoon]

Haha no we're not gonna give you the answer right off the bat

What have you tried

Show us your work first

[dreamy ember]

When I said the answer I meant that my answers shows in the picture that what I tried

[fierce lagoon]

I mean probably use a comma instead of a +

[obtuse pebble]

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[vale bobcat]

I used zero product property here to get x={3/2, 4, and -3} Would that be correct? I'm not sure if I have to do anything with the two here.

[scarlet locust]

that is correct

2*n = 0
means either 2 is 0 or n is 0

clearly 2 isn't 0

so you can ignore the 2

and just imply that n is 0

[vale bobcat]

oh ok!

I see

[fierce lagoon]

It's an amplifier, so you can ignore it

[vale bobcat]

makes perfect sense

[fierce lagoon]

If you divide both sides by 2 then you'll just get the same equation except without the 2

[vale bobcat]

right on! Thanks for the help guys

.close

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[teal timber]

<@&286206848099549185>

[teal timber]

i need help with these 7

i didnt understand anything

[lucid totem]

Is this canvas

[teal timber]

yes

[grave halo]

the growth factor of a function is the number that is getting raised to the power of something so if f(x) = 2^x, 2 is the growth factor

[teal timber]

hmm

[grave halo]

the initial value of every function would be f(0)

so you need to calculate these

[teal timber]

so it would be b

right?

[grave halo]

are you asking for q1?

[teal timber]

yes

[fierce lagoon]

Hey buddy might wanna read the rules

[teal timber]

mb

[fierce lagoon]

Also ewwww canvas

[teal timber]

ikr

this is algebra 1

[grave halo]

you are looking for a function that has 3 to the power of a variable

[fierce lagoon]

Well let's look at 1

Growth factor is basically "what has $3^x$ in it"?

[warm shale]

Umbraleviathan

[teal timber]

oh so function three

[fierce lagoon]

Yeah

But initial value of 2?

Uhhhh

Do they mean y intercept of 2?

[teal timber]

idk

wouldnt it be like the same for the 3^x

[fierce lagoon]

No

[teal timber]

hm

we can skip that for now

q3

is question three

function two

4,5,6,7

[unique kettle]

The first four functions are in the form $y=f(x)=I(P)^x$ where I is the initial value, P is the growth or decay factor

[nocturne minnow]

Is this a test?

[warm shale]

clip

[teal timber]

its a quiz

[nocturne minnow]

Are you allowed outside help?

[teal timber]

yes

[lucid totem]

Sorry man this is Pre Calc 12 curriculum in Canada, I’m only in Pre Calc 11

[teal timber]

ohhhh

so it would be function three

[wispy shoal]

yes

here are some rules you should practice

[teal timber]

@wispy shoal so would Q5 be function two?

[wispy shoal]

yee

[teal timber]

ayyy ok

how do i find q 6 and 7

?

[wispy shoal]

i dont know xD

[teal timber]

lmao ok

[obtuse pebble]

@teal timber Has your question been resolved?

[obtuse pebble]

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[sleek sinew]

We are allowed to combine exponentials like this?

I dont think ive ever seen this rule before

They have different bases, same exponents

Does this work for any exponentials with diff bases, same exponents?

[fickle shell]

yes

2^x=e^(xln(2)) and 5^x=e^(xln(5))

[pallid flame]

what

a^x * b^x = (ab)^x

[pallid flame]

this is a basic exponential property @sleek sinew

[drowsy girder]

$$a * a * a *...... a * b * b * b...... *b$$
Imagine you have a being multiplied x times and b being multipled x times. When you rearrange it you get
$$ab * ab * ab *....*ab$$
this will also be mutiplied x times.
Therefore
$$a^x * b^x = (ab)^x$$

[warm shale]

Pluton

[sleek sinew]

Thanks that makes a lot of sense

.close

[obtuse pebble]

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[little hill]

hi everyone , here's the problem i have . i have 3 points, A,B,C, in 2d space (x,y). im looking for a new point D wich is the furthest point that is on the line that goes from A to C , but the segment AD cant exceed 600 units. at the same time, this point D cant be at more than 1000 units from point B. any idea how to solve this?

[obtuse pebble]

@little hill Has your question been resolved?

[obtuse pebble]

@little hill Has your question been resolved?

[obtuse pebble]

@little hill Has your question been resolved?

[royal solar]

have you heard of vectors chum

@little hill

[little hill]

yes sir, i do

but vector whise i dont know how to write an equation that solves the conditions

@royal solar

[royal solar]

$\vec {AC} = \lambda (\vec{AD})$

[warm shale]

Rylo

[royal solar]

|AD| < 600

|BD| <1000

@little hill

does that make sense

[little hill]

yes , thats the conditions

@royal solar

[royal solar]

yeah, that’s it

I don’t have any coords

[little hill]

yeah well, how to find the D coords knowing the A cords, B coords, and C coords

A B C are known

[royal solar]

use the conditions man

or at least tell the condition

[little hill]

those are the conditions, knowing A(xa,ya) , B,(xb,yb) and C,(xc,yc) , that AC=lambda AD and |AD| < 600 and |BD| <1000, how to calculate D(xd,yd)

[royal solar]

1. D isn’t fixed by the looks of it

just work with them as you’d normally man

[little hill]

no its not fixed, its conditionned by the conditions above. thats why i want to express D in function of A,B,C

[royal solar]

AC for example is (xc - xa)i + (yc-ya)j

[little hill]

what i,j correspond to here

[royal solar]

perpendicular vectors,

i is for x direction

j is for y direction

[obtuse pebble]

@little hill Has your question been resolved?

[obtuse pebble]

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[timid silo]

how do i do this

26 = -1+(27x)^3/4

[obtuse pebble]

Please don't occupy multiple help channels.

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[mental peak]

Question:
A resort owner wants to close a beachfront area for swimming activities. Based on her plan only 3 sides will be fenced with 270 meters rope and floats, while the shoreline part will be open. Determine the dimensions of the rectangle that will give a maximum.

[mental peak]

I just wanna ask if my solution is correct
GIVEN:

P = 270m

FORMULA:

P = x + 2y

A = xy

SOLUTION:

P = x + 2y

270 = x + 2y

270 - 2y = x

A = xy

A = (270 - 2y) y

A = 270y - 2y^2

Get Derivative

A’ = 270 – 4y

270 – 4y = 0

-4y/4 = - 270/4

y = 67.5

P = x + 2y

x = 270 - 2(67.5)

x = 270 – 135

x = 135

ANSWER:

135m by 67.5m will give us the maximum

[high lily]

can you remove the unnecessary line gaps

[mental peak]

sorry

P = x + 2y
270 = x + 2y
270 - 2y = x

A = xy
A = (270 - 2y) y
A = 270y - 2y^2
Get Derivative
A’ = 270 – 4y
270 – 4y = 0
-4y/4 = - 270/4
y = 67.5

P = x + 2y
x = 270 - 2(67.5)
x = 270 – 135
x = 135

135m by 67.5m will give us the maximum

here is my solution

i just wanna ask if this si correct?

[high lily]

looks ok

[mental peak]

ohh im glad to hear that

thank you!

.close

[obtuse pebble]

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[obtuse pebble]

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[stiff cosmos]

Stoaches are fictional creatures with distinctive octarine stripes. Unfortunately they are vulnerable to the chirax virus (which isn’t actually harmful, but causes them great embarrassment).
In the general population of stoaches worldwide, 23% of stoaches carry the chirax virus. The local habitat has recently been devastated with both fires and floods. A researcher would like to know whether this has affected the local population’s health, and in particular, its rate of chirax infection.
A sample of 65 local stoaches is tested for chirax, and 6 of them have this humiliating condition. The researcher does a zz-test to see whether the local chirax rate has changed.
Determine the pp-value from this test. (Give your answer to 4 decimal places.)

[stiff cosmos]

help

[forest sinew]

zz test

👀

thats not a thing, right

[forest sinew]

assuming its just a z test do you know how to do it

[obtuse pebble]

@stiff cosmos Has your question been resolved?

[forest sinew]

too good to respond

[stiff cosmos]

isnt it two sample z test

?

zz

And no idk how to do z test

[forest sinew]

so like difference of means?

[stiff cosmos]

yeh

[forest sinew]

and then maybe binom

cheatsheet come in handy

😌

[stiff cosmos]

thanks lol

[forest sinew]

does that actually help

i think this is the set up

[stiff cosmos]

kinda

idk the answer it's hwk

[forest sinew]

like

your MLE for p_i is just x_i bar

so youd substitute in there each of your averages

for the hats

[stiff cosmos]

right yep

[forest sinew]

this is dist standard normal

so out pops a z

[obtuse pebble]

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[timid silo]

Help

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[swift marsh]

Consider taking $\log_x$ of both sides, getting $y > x\log_x(y)$

[warm shale]

dk.dkn

[swift marsh]

Now you could use the fact that y > x, and that y and x are positive integers greater than or equal to 2 to deduce a fair amount

[timid silo]

Uh huh

I dont understand

[swift marsh]

Perhaps even better would be to just take the natural log of both sides instead

ln(x^y) > ln(y^x) so ylnx > xlny and y/x > lny/lnx

Try to get the answer out of that ratio

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

Thank you very much

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[drowsy girder]

Pretty sure area of that is ef/2

You got e

f = f_1 + f_2

,w sqrt(15^2 - 9^2) + sqrt(41^2 - 9^2)

[warm shale]

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[drowsy girder]

,w 52*18/2

[warm shale]

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[drowsy girder]

Seems about right

[obtuse pebble]

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[obtuse pebble]

Please don't occupy multiple help channels.

[obtuse pebble]

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[stray lichen]

Sally and Marko are two surveyors that have become separated out in the wilderness. Sally is due east of Marko. Marko radios Sally "The distance from me to the top of Kitt's Peak is 7.8km. It is at an angle of elevation of 32 degrees". Sally radios back "I am 6.5km from the top of the peak". Calculate all possible distances Marko must hike due east in order to reach Sally. Round to the nearest tenth of a kilometer.

[stray lichen]

I've calculated 1.6km, but I'm unsure of my answer

<@&286206848099549185>

[nocturne minnow]

You have to wait at least 15 minutes before pinging helpers

[stray lichen]

sorry boss

[urban patrol]

There's two answers since sin can be two things when doing law of sines

Use your answer for your law of sines as a reference angle

To then find the second possible angle

[stray lichen]

would the answer of 1.6km be wrong?

i thought that the other possibility would be to create an obtuse scalene triangle and solve that way to find the second answer

because im looking for the side length and not the angle

<@&286206848099549185>

[obtuse pebble]

@stray lichen Has your question been resolved?

[stray lichen]

close

!close

.close

[obtuse pebble]

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[obtuse pebble]

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[lost void]

For the distance on the first one I’ve got 10 I just need help trying to figure out the method on solving this. I'm stuck on question 6.

[obtuse pebble]

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<@&286206848099549185>

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[lost void]

<@&286206848099549185>

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@lost void Has your question been resolved?

[urban mirage]

diagonals of a rectangle are congruent

[obtuse pebble]

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[ashen needle]

trying to figure out how to write a rule for this

[ashen needle]

The interest in each year

all I can think of is 10,000(0.055) which only works for the first line

[nocturne minnow]

Oh

[dusk turret]

Hint: each year, you get 5% more than the last.

That's your "algorithm".

How would you write that in numeric terms?

[ashen needle]

Thats what im trying to figure out lmao

[obsidian isle]

Compounding interest

The biggest evil in our society

[ashen needle]

didn't Einstein say it was the 8th greatest wonder or something?

[tall tangle]

e

[kindred moss]

So knowing that you decided to post ur problem

[ashen needle]

what

[kindred moss]

Nvm

So ur trying to make a rule for interest?

[ashen needle]

I'm trying to use the compounding interest rule

A=10,000(1+0.055)^t

[kindred moss]

Yh I get that

What you trying to output

Like just the value after a year?

Cuz it seems to work just fine

[ashen needle]

Just thinking atm

I also want the interest

[kindred moss]

x . (a + b)^y
Example
x = 10000
Let’s say we want interest
a = 1
b = 0.055
y = 2

I guess that would be a rule to follow

[ashen needle]

Thats just this up here

[kindred moss]

Yh but as a rule

Using algebra it turns into a rule doesn’t it?

[ashen needle]

yeah, I've already got that part

im trying to find a rule for the interest now

again just thinking

[kindred moss]

final amount( 1 + annual rate . time)

Maybe

[ashen needle]

that wont work

[dusk turret]

Let the amount of money at the end of the nth year be represented as $a_n$.

[warm shale]

Wishes

[dusk turret]

$a_{n+1} = (1.05)a_{n}$

[warm shale]

Wishes

[dusk turret]

Can you figure out a closed-form expression for this?

If you want a rule for the interest given a starting amount, a final amount, and time, then use (1+i)^t = final amount/initial amount

[ashen needle]

I still dont understand what your saying @dusk turret

[dusk turret]

I think this issue is mutual! Before I say anything else, could you explain once more for me what you're trying to find?

[ashen needle]

The interest in each year

[dusk turret]

The interest in the nth year is the difference between the amount at the end of the nth year and the amount at the end of the n-1th year.

Suppose you want to find how much interest was earned in the fifth year.

That's the amount you have at the end of the 5th year MINUS the amount you had at the end of the 4th year.

Make sense?

[ashen needle]

Thats makes sense but I feel like there could be some rule or form that might be a lot more simple

Otherwise its just going nth year - nth year-1 = interest

[dusk turret]

I don't know if it gets any simpler than that!

[ashen needle]

hmm okay then

Thanks for the help

.close

[obtuse pebble]

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[low fossil]

I have proved that:

(A):

$\frac{1}{1-|\zeta|}\ge 1+|\zeta|\ge |1+\zeta|\ge 1-|\zeta|$

(B):

$|(1+\zeta^m)\cdot (1+\zeta^{m+1})\cdot …\cdot (1+\zeta^{m+n})$

$=|1+\zeta^m|\cdot |1+\zeta^{m+1}|\cdot …\cdot |1+\zeta^{m+n}|$

$\ge (1-|\zeta|^m)\cdot (1-|\zeta|^{m+1})\cdot …\cdot (1-|\zeta|^{m+n})$

(C):

$1-|\zeta|^m-|\zeta|^{m+1}-…-|\zeta|^{m+n}$

$\le (1-|\zeta|^m)\cdot (1-|\zeta|^{m+1})\cdot …\cdot (1-|\zeta|^{m+n})$

$\le \frac{1}{1+|\zeta|^m+|\zeta|^{m+1}+…+|\zeta|^{m+n}}$

Then I stuck here. Can anyone give me some more hints?

[warm shale]

Trenton

[strong vale]

$\frac{1}{1-|z|}\ge 1+|z|\ge |1+z|\ge 1-|z|$

(B):

$|(1+z^m)\cdot (1+z^{m+1})\cdot …\cdot (1+z^{m+n})$

$=|1+z^m|\cdot |1+z^{m+1}|\cdot …\cdot |1+z^{m+n}|$

$\ge (1-|z|^m)\cdot (1-|z|^{m+1})\cdot …\cdot (1-|z|^{m+n})$

(C):

$1-|z|^m-|z|^{m+1}-…-|z|^{m+n}$

$\le (1-|z|^m)\cdot (1-|z|^{m+1})\cdot …\cdot (1-|z|^{m+n})$

$\le \frac{1}{1+|z|^m+|z|^{m+1}+…+|z|^{m+n}}$

[warm shale]

azeem321

[obsidian isle]

Wtf are yall doin

[strong vale]

idk might be easier to look at for any future takers

who wants to see zeta

[low fossil]

Lol

I guess it is close to an end

But I stuck

[obsidian isle]

Did you already use geometric series formula

[low fossil]

Nope

(C) is Weierstrass Product Inequality

[low fossil]

$\sum_{i=m+1}^{m+n} |z|^i=\frac{|z|^{m+1}(1-|z|^n)}{1-|z|}$

Or do you mean this

[warm shale]

Trenton

[obsidian isle]

Yes

[low fossil]

(C) (rectified):

$1-|z|^m-\frac{|z|^{m+1}(1-|z|^n)}{1-|z|}$

$1-|z|^m-|z|^{m+1}-…-|z|^{m+n}$

$\le (1-|z|^m)\cdot (1-|z|^{m+1})\cdot …\cdot (1-|z|^{m+n})$

$\le \frac{1}{1+|z|^m+|z|^{m+1}+…+|z|^{m+n}}=\frac{1}{1+|z|^m+ \frac{|z|^{m+1}(1-|z|^n)}{1-|z|}}$

[warm shale]

Trenton

[low fossil]

Umm but how does it help

[strong vale]

was a mistake made

[obtuse pebble]

@low fossil Has your question been resolved?

[low fossil]

Mm

[obtuse pebble]

@low fossil Has your question been resolved?

[obtuse pebble]

@low fossil Has your question been resolved?

[outer steeple]

Where did you get this question @low fossil ?

[obtuse pebble]

@low fossil Has your question been resolved?

[low fossil]

Exercise in my year one course

[obtuse pebble]

@low fossil Has your question been resolved?

[low fossil]

.close

[obtuse pebble]

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[grizzled belfry]

hello

[orchid otter]

like that

[grizzled belfry]

finally

[orchid otter]

post your question here

[grizzled belfry]

now who will be here to ans my question?

[orchid otter]

and when you get your question answered you type .close

[grizzled belfry]

this one

[orchid otter]

you post it here and you wait until someone knows how to answer

[grizzled belfry]

thank you so much

[dense imp]

first off, do you understand what an additive inverse is?

[grizzled belfry]

yeah i learned in online class but now i completely forgot it

can u give some info

[robust sleet]

google

[dense imp]

so generally, its the object that you can add to another object to get the "Zero object" of whatever context youre in

so for example 2 is the additive inverse of -2 if you're talking about whole numbers

its quite simple

5 is the additive inverse of -5

since 5 + -5 = 0