help-23

[lean otter]

so for example, the bottom triangle has base 8 and height 6

(or base 6 height 8 if you look at it the other way)

[cinder breach]

It says "draw a net" is it like this? This was solved in note book but I couldn't understand

[lean otter]

all four faces would be triangles, so a net of a pyramid would look something like this

but for your example, you need to make sure the right angles are in the right place

[cinder breach]

Ah

If let's say I needed to find height and only have volume and base how would I get it?

[lean otter]

Look at the 2nd formula on your sheet

[cinder breach]

I want height tho

Not volume

I have some more Qs on how to do btw

[lean otter]

but you have volume, and base area

so you can rearrange to get height

[cinder breach]

Ummm

What's volume

😐

[lean otter]

^

[cinder breach]

I forgot everything

[lean otter]

you said you have volume and base from height

volume is how big something is in 3D

[cinder breach]

Ah

How do I rearange them to be height

[lean otter]

$V=\frac{1}{3}Ah$, divide both sides by $A$ to get $\frac{V}{A}=\frac{1}{3}h$ then multiply by 3 to get $h=\frac{3V}{A}$

[flat frigate]

Toby

[cinder breach]

Ok

Yk how I could solve this?

[lean otter]

,rotate

[lean otter]

you can use the forth equation on your sheet to get the surface area of a cone

note that diameter=radius*2

[cinder breach]

Oo

It's pretty confusing a b c ngl lol

[lean otter]

yeah

[cinder breach]

Ok this is confusing I can't lmao brain fried

It's gonna be such a pain in the ass to memorise all those shit

[safe radish]

@cinder breach Has your question been resolved?

[safe radish]

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[tender mountain]

when factoring large expressions like $x^2+3x^2-1x+1x^2$ would you simplify it to be smaller to be able to use strategies that require 4 numbers?

[flat frigate]

kangaroo rat

[limber smelt]

strategies that require 4 numbers??

[tender mountain]

I edited my message

[limber smelt]

no I mean what does 4 numbers mean ?

so it has 4 solutions?

[tender mountain]

no like $4x-3x+8x^2$ see how there is four numbers like 8x or something?

[flat frigate]

kangaroo rat

[limber smelt]

oh I think I get what you mean.

[tender mountain]

yes

[limber smelt]

0x^4+0x^3+5x^2-x

is that what you mean?

4 numbers?

[tender mountain]

$0x^4+0x^3+5x^2-x$

[flat frigate]

kangaroo rat

[tender mountain]

yes

[tender mountain]

also have you got a better name for it?

[limber smelt]

is that what you mean by "4 number"?

[tender mountain]

yes

[limber smelt]

okay, it's called "terms"

[tender mountain]

yes

[tender mountain]

so back to my question

[limber smelt]

erm

$0x^3+5x^2-x+0$

[flat frigate]

張嘉棋

[limber smelt]

4 terms

[tender mountain]

yes so

you know with factoring strategys?

[limber smelt]

yes

[tender mountain]

there are like not ones with 8 terms allowed?

[limber smelt]

please be clearer in what you are trying to ask help with

[tender mountain]

so in factoring strategies there are not strategies for every number?

[limber smelt]

sure there is

[tender mountain]

there is

[tender mountain]

so you had to memorise infinite strategies you are still learning?

[limber smelt]

above like a certain term, theres like one way I know of

there are some tricks to help solve it

[safe radish]

@tender mountain Has your question been resolved?

[tender mountain]

so do you think I could simplify a expression to work for a 4 term strategy?

[limber smelt]

can you tell me what equation you have there?

[tender mountain]

I am not trying to figure something out just trying to learn

so can you help

[limber smelt]

and I am teaching

[tender mountain]

yes

[limber smelt]

so can you tell me what type of equation you have there?

[tender mountain]

polynomial and quadric

[limber smelt]

a quadratic

yes

and we want 4 solutions

what does does that tell us?

[tender mountain]

wdym?

[limber smelt]

you said we want 4 numbers.

so are we solving a quadratic or a quartic?

[tender mountain]

yes

[limber smelt]

what?

[tender mountain]

quadratic

[limber smelt]

in that case only two solutions can arise from this.

do you know the method

[tender mountain]

yes

[limber smelt]

then what does x equals?

[tender mountain]

idk

[limber smelt]

but you told me you know the method.

[wise shell]

You should add the 1x ^2

[tender mountain]

I thought i knew the method

[wise shell]

Then have x^2 + 4x^2-1x

Then 5x^2 - x

[limber smelt]

oh was his problem simplifying?

he sure didn't seem to have asked that

[wise shell]

Factor out an x, and get it x(x-1)

[limber smelt]

where the 5 go?

[wise shell]

My bad

x(5x-1)

[limber smelt]

👍

[tender mountain]

My question is one of the methods I know of factoring can only be done when there are 4 terms. If there were 5 terms could I make the 5 terms 4 terms to do it in the method I know. If so can you tell me how to simplify it?

@limber smelt

[tender mountain]

hello?

[limber smelt]

you can split it into quadratic

[tender mountain]

what if it wasnt a quadratic in the first place?

[limber smelt]

give me a polynomial with 5 terms in it

and we will work through it

[tender mountain]

$x^2+9^2-y^2+x^2-9x^3$

[flat frigate]

kangaroo rat

[tender mountain]

Is that good

[limber smelt]

that is not a polynomial.

erm

[tender mountain]

oh yea

[tender mountain]

can you write one?

[limber smelt]

a polynomial is like this

[tender mountain]

ok

so can you write one?

@limber smelt

[limber smelt]

bruh the picture is literally an example \

stop pinging alot

thnx

[tender mountain]

sry

Im just in a bit of a rush

[tender mountain]

ok so can we go through it

[limber smelt]

in a rush for what?

[tender mountain]

i just gtg soon

[limber smelt]

hm

how long you got?

[tender mountain]

15 minutes

[limber smelt]

sorry, I don't think I can teach it in that time.

but pls watch this video when you find the time: https://www.youtube.com/watch?v=JJNi8kZ-SHU

[tender mountain]

ill ask if i can be a little longer

how is 25 minutes?

[limber smelt]

the video is literally the way I would do it so.

if there are questions after, then feel free to ask

[tender mountain]

I already know how to factor by grouping is that all this is?

[limber smelt]

pretty much for a quartic

[tender mountain]

oh

[tender mountain]

So back to the question then Is there a way to simplify the polynomial to become 4 terms? And will it still be effective for answering the larger then 4 term equation.

[tender mountain]

before I go?

[limber smelt]

the polynomial you provided simplifies to a quadratic

[tender mountain]

does every polynomial?

[limber smelt]

what? NO

[tender mountain]

oh

[tender mountain]

so i there a way for every polynomial to become 4 terms?

[limber smelt]

I don't get what you mean by 4 terms

4 terms as in the solution

[tender mountain]

as in the question/equation

[limber smelt]

if it's a QUADRATIC, then you can't simplify into 4 terms

[tender mountain]

anything not a quadratic?

[tender mountain]

yes or no?

[limber smelt]

that was not a question

[tender mountain]

what types of polynomials can be simplified to 4 then?

[limber smelt]

I seriously don't get what you mean by 4?

you have to explain what you have already done

[tender mountain]

4 terms like this 4x

[limber smelt]

go on

[tender mountain]

what do you want me to say

do you still not understand what I mean by 4?

[limber smelt]

no

[tender mountain]

then what do you want me to say

[limber smelt]

factorising? 4 numbers?

you just simplify and factorise

I don't get what's the problem with 4 numbers?

[tender mountain]

I know a way to factor that requires you to only have 4 terms is there a way to simplify 5 or 6 terms to 4 terms so i can apply my strategy?

[limber smelt]

explain your strategy

[tender mountain]

oh actually it requires 3 terms

[tender mountain]

$1x^2+7x+12$. 12 the last term in the equation and 1 the first coefficient. Multiply them to get 12 factors of 12 that add up to 7 the middle terms coefficient 3x4=12, 3+4=7 the answer is (x+3) (x+4).

[flat frigate]

kangaroo rat

[limber smelt]

yh

you factorised it

what else are you stuck on?

[tender mountain]

so that is 3 terms the equation has three terms how do you do that with something with 4 or 5 terms do you simplify it to 3 terms or what?

@limber smelt

[limber smelt]

if you want 4or 5 then the rest is just 0

[limber smelt]

x=3, x=4, x=0

[tender mountain]

wdym?

[limber smelt]

what you mean by this question then?

[tender mountain]

I mean is it possible to apply this strategy to an equation with 4 terms or more?

[limber smelt]

it is possible/

just extra steps

just use the youtube video

[tender mountain]

so I can use this exact strategy

[limber smelt]

yh with steps before it.

like the youtube video

[tender mountain]

the yt video will help me?

[limber smelt]

just try it/

he is a great teacher.

[tender mountain]

and if it doesn't work can you tell me what those steps are called?

[tender mountain]

like a name or something?

[limber smelt]

have you tried a textbook?

[tender mountain]

I have been googling

so do you know a name?

[limber smelt]

a name for what?

[tender mountain]

for the steps anything I can work on?

[limber smelt]

just google "how to factor an n term polynomial"

[tender mountain]

a n term?

[limber smelt]

oh my bad

[tender mountain]

so what?

[tender mountain]

what was it meant to be?

[limber smelt]

i fixed it now

[tender mountain]

ok

thanks

ty

bye!

.close

[safe radish]

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[lean otter]

Hey hey, I need help with this two questions. I have previously tried to do them like I tried to do tan (75) = tan (45+30)

And use the formula tan x + tan y / 1 - (tan x) (tan y)

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

<@&286206848099549185>

[plucky elk]

use pythagorean theorem to find $\tan(\alpha)$ and $\tan(\beta)$

[flat frigate]

riemann

[lean otter]

Based on the questions what quadrants are they specifically referring to?

[plucky elk]

draw a picture

[safe radish]

@lean otter Has your question been resolved?

[safe radish]

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[flat frigate]

Fawkes

[safe radish]

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[lean otter]

Wha-

.reopen

[still charm]

Why did you delete

[lean otter]

I clicked 3rd reaction one to remove LaTeX code ._.

[lean otter]

Btw the question is here

[safe radish]

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[lean otter]

If $\dfrac{\tan(\theta+\alpha)}{a} = \dfrac{\tan(\theta+\beta)}{b} = \dfrac{\tan(\theta+\gamma)}{c}$

Show that,

$\dfrac{a+b}{a-b}\cdot \sin^2(\alpha-\beta) + \dfrac{b+c}{b-c}\cdot \sin^2(\beta-\gamma) + \dfrac{c+a}{c-a}\cdot \sin^2(\gamma-\alpha)=0$

[flat frigate]

Fawkes

[safe radish]

@lean otter Has your question been resolved?

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

Not at all

<@&286206848099549185>

Sed

[safe radish]

@lean otter Has your question been resolved?

[trim trench]

what have you done or what are you stuck on

[safe radish]

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[lean otter]

the approach

[plucky elk]

Do this next time

[hexed night]

@lean otter is using complex representations of trigonometric functions allowed in the solution?

Have you been taught that yet?

It's easy to solve using that approach

[safe radish]

@lean otter Has your question been resolved?

[safe radish]

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[lean otter]

Do you mean compound angle and all? Yes

[hexed night]

No I mean complex number representations of trig functions

Like Euler's formula

[hexed night]

Exactly

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

Nay, idk anything like that

I only know some trig. Funcs and identities

[glass sonnet]

@hexed night what exactly are you stuck on

oh wait @lean otter is the one who needs help

my bad

[lean otter]

Yes, and the question is here @glass sonnet

[glass sonnet]

have you tried splitting using compound angles?

[lean otter]

Yep

[trim trench]

apply compound angle formula for tan

ig

[glass sonnet]

what did you get?

[lean otter]

Result was that my brain burst out in the middle

[glass sonnet]

can you show your working?

[lean otter]

According to me... I am breaking the whole proving part into 3 parts

[flat frigate]

Fawkes

[lean otter]

And in the similar question which i solved( not exactly same, question is different) i agot a relation between tan(A+B)/tan=(A-B) = sin(A+B)/sin(A-B) and in this i am now more confused

[glass sonnet]

why don't you first just start with tan(theta + alpha)/a

try expand that on its own first

[safe radish]

@lean otter Has your question been resolved?

[safe radish]

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[solemn drum]

how to do this?

the quickest way

[vast obsidian]

Hint: 77=7*11 🤯

[solemn drum]

umm

OH

[strong grove]

Then some things will cancel out from numerator and denominator. After that you can do trial and error, test the choices

[solemn drum]

390

right only 26 is divisible

ty

.close

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[pine tulip]

I got A as the answer, can someone please confirm?

[pine tulip]

😢

ill send what i did

[tribal grail]

it's B actually

I don't know how paradox got B

Ignore the thing at the top

I was doing a diff problem before

[pine tulip]

,rotate

[pine tulip]

this is what i did

[tribal grail]

Yeah

your solution is correct

How did u read A from that tho

You need to find the group of solutions

for which

the both equations are correct

not only 1

and both are only correct where their groups of solutions

intersect

[pine tulip]

i see, i thought that x<=-2

[tribal grail]

oh nvm

[pine tulip]

was equal to what is written on solution A

[tribal grail]

u graph isn't correct

in the 1st place

Do you now how to solve for this?

[pine tulip]

I thought so

[tribal grail]

$x^2-4$

[flat frigate]

Frojodo

[pine tulip]

yeah is x=2 or -2

[tribal grail]

wait

$x^2-4<0$

[flat frigate]

Frojodo

[tribal grail]

do u know how to solve for this*

[pine tulip]

x<2 or x<-2

[tribal grail]

incorrect

just think about it

plug in

-3

in x

[pine tulip]

then x<-2 becomes false

[tribal grail]

wym

wait

[pine tulip]

if x=-3

[tribal grail]

yeah

so the solution clearly isn't

x<2 and x<-2

what do u think u got wrong?

[pine tulip]

Honestly, i can try guessing, but how i did the problem is how i was thought at school here in italy lol

maybe i should have wrote x>-2? i have no idea

[tribal grail]

nono u weren't taught that way

many students mess it up

and it's very common

I even thought it was the same way you think it is rn

when I was first learning it

let me explain

[pine tulip]

i see

[tribal grail]

$x^2-4$

[flat frigate]

Frojodo

[tribal grail]

is the same as

$(x-2)*(x+2)$

[flat frigate]

Frojodo

[pine tulip]

yes

[tribal grail]

so you need to find for this

$(x-2)*(x+2)<0$

[flat frigate]

Frojodo

[tribal grail]

does it make more sense now?

[pine tulip]

Isnt it the same as before?

like from x-2=0

we get x=2

[tribal grail]

yes

[pine tulip]

from x-2<0, we get x<2

i just have to move the 2

and change its sign

[tribal grail]

but what is the requirement for this to be smaller than 0

here I'll help you out a bit

one of those 2 brackets

have to be negative

and the other has to be positive

if both are negative or positive

their end result will be >0

you need to find the group of numbers

for which one of the brackets

is negative

and the other is positive

the solutions is

X is the element of (-2,2)

[pine tulip]

so you mean that -2<x<+2

[tribal grail]

yes

let's try a few more problems!

$x^2-16>0$

[flat frigate]

Frojodo

[pine tulip]

so the correct way to draw it would be like this

correct?

[tribal grail]

yes

[pine tulip]

i see

[tribal grail]

try solving this one now

[pine tulip]

-4<x<+4

[tribal grail]

r u sure?

[pine tulip]

ups you are right

[tribal grail]

how can you look at this problem?

[pine tulip]

(x-4)(x+4)>0

[tribal grail]

so

what are the requirements for those

to be bigger than 0

[pine tulip]

so they have to be both x>4 and x>-4

[tribal grail]

if x>-4

then let's take a random number

that's higher than

-4

-2

per say

that would be

(-2-4)*(-2+4)

that's

-6*2

-12

-12 is not bigger than 0

[pine tulip]

Okay, so just like before the number has to be -4>x>+4

right?

[tribal grail]

no

I'll tell u the solution so u can maybe visualize it better

the solution is

X is the element of (-infinty,-4) U (4, +infinity)

or X is the element of R/(-4,4)

[pine tulip]

So its x<-4 and x>+4

[tribal grail]

yes!

[pine tulip]

Okay now i understand

[tribal grail]

that's great!

[pine tulip]

I have another question if you dont mind

[tribal grail]

shoot

I'll tell u if I know

[pine tulip]

Lets say that a disequation is something like this (-x^2+3x+2)/(x-4)>0

on the denominator we will have x>4

[tribal grail]

what do u think u need to find then!

[pine tulip]

but on the nominator we will have to change > to <

[tribal grail]

$((2x+2)/(x-4))>0$

[flat frigate]

Frojodo

[pine tulip]

my bad

i meant to write an example where you had to change the "direction" of >/< symbol

[tribal grail]

$(-x^2+3x+2)/(x-4)>0$

[flat frigate]

Frojodo

[tribal grail]

do you know how to solve a quadratic equation?

[pine tulip]

yes

[tribal grail]

okay

le'ts solve -x^2+3x+2

[pine tulip]

(x-2)(x-1)<0

wait

[tribal grail]

I don't think so?

[pine tulip]

im wrong

[tribal grail]

wait

[pine tulip]

lets put +3x or +2 with minus instead of +

so we can put the equation on that form

[tribal grail]

nono

just apply the formula

like this

it's completely doable

and simple

it's just

(-3+-sqrt of 9+16)/-2

[pine tulip]

yes

[tribal grail]

(-3+-5)/-2

[pine tulip]

i wanted to make a simpler example

[tribal grail]

which is

-1 or 4

how do you write -1 and 4

as 0s

[pine tulip]

Wait, so you didnt change the sign of -x^2 before doing this right?

[tribal grail]

yes

[pine tulip]

okay, so we are left with x<-1 and x>4

[tribal grail]

or are we

don't jump to conclusions!

let's write the brackets that make up the quadratic first

[pine tulip]

(x+1)(x-4)>0

[tribal grail]

nice!

now what do we have in the denominator

[pine tulip]

x-4

[tribal grail]

now we need to see

when the top of the division line

and the bottom of the divison line

or rather

the numerator and the denominator

have the same

sign

• or -

[pine tulip]

they do

[tribal grail]

but when

[pine tulip]

now they do have the same sign

(x+1)(x-4)/(x-4)>0

[tribal grail]

nonono

when do they have th same sign

for an example they won't have the same sign

for the value of -2

[pine tulip]

ah you meant for which values of x

we would have different signs

[tribal grail]

yes!

btw if you would like to vc

so I can explain it better

I'm down for that too

[pine tulip]

I can only listen tho

is that fine?

[tribal grail]

good enough

x<-1 U x>4

[pine tulip]

Okay i feel like i finally understood what you meant with all this

thanks 🙂

[tribal grail]

no problem!

if u have any questions feel free to dm

[pine tulip]

I appreciate it 🙂

.close

[safe radish]

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[safe radish]

Please don't occupy multiple help channels.

[hollow torrent]

I got all 4 answers on this grid wrong

I'm confused why though

[ionic brook]

this is the graph of its derivative not the graph of f(x)

[hollow torrent]

so how do i find the info of f(x)

[ionic brook]

write those statements in terms of f'(x), for example, if f has a relative min at x, then f'(x)=0 and f' should be negative just before x and positive just after x

[hollow torrent]

ok so i get how to do the first 3

i dont get the 4th thpough

[ionic brook]

if the graph is concave up, that means that f'(x) is increasing, hence, f''(x)>0

[hollow torrent]

kk

cool ty

[hollow torrent]

please check help ticket 26, that other problem still says im getting something wrong

.close

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[lean otter]

Please explain me 2.33.

[tame charm]

What is the random variable described in 2.3

[tame charm]

What

That doesn’t make sense

Where did they get the 1/4 and 3/4 stuff?

Is that what you’re asking?

[lean otter]

Yes 🙂

Also 0 and 1.

[tame charm]

Maybe it’s a typo

Did they give an example of a random variable before this section?

[lean otter]

Yes, let me share that with you.

[lean otter]

Here you go.

[tame charm]

Ok there’s the random variable from 2.3

[lean otter]

Here is the book: https://drive.google.com/drive/u/0/folders/1jVs0o_ZTc9URrTR6mPiszFzLrEvgQPMj

[tame charm]

Ah i understand

There’s a 1/4 probability that X is between 0 and 1, i.e. (T, T)

There’s a 3/4 probability that X is between 0 and 2, ie (T,T),(H,T),(T,H)

Are you following

[lean otter]

Sorry, I am not.

[tame charm]

Do you know what X is

It’s a function that takes in a flipping of coins, eg (H, T), and spits out how many Heads there are

Basically it counts how many heads you flipped

Following?

[lean otter]

Yes 🙂

[tame charm]

Ok cool

When you flip a coin twice, there are only 4 possible outcomes

TT, HT, TH, and HH

[lean otter]

Yes

[tame charm]

So the probability of X (the number of heads) being 0 is 1/4

And the probability of X being 1 is 2/4

And P(X = 2) = 1/4

Still got it?

[lean otter]

Yes 🙂

[tame charm]

so we can create a function that associates each outcome of X with its associated probability

This is called a probability distribution

So our probability distribution would look like
P(0) = 1/4
P(1) = 1/2
P(2) = 1/4

A cumulative distribution function associated each outcome of X with the probability that X is at most that value

For our example, it would be like F(x) is the probability that the number of heads is at most x, ie
F(0) = P(0) = 1/4
F(1) = P(0) + P(1) = 1/4 + 1/2 = 3/4
F(2) = P(0) + P(1) + P(2) = 1

[lean otter]

Ohh. Gotcha 👍

[tame charm]

If you’re familiar with calculus, it is simply the integral of the probability distribution

But yeah that’s that

[lean otter]

I see 😀

Thank you for being patient with me and helping me out 🙂

.close

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[pliant zephyr]

hello

[pliant zephyr]

Hello i am doing some summer practice to make sure im ready for calculus next year i was wondering if my answers are correct as the guide says im wrong and idk what i did wrong

so the questions and answers i have are as followed and i think the guide says im all wrong on them so please le me know if my thinking is incorrect
lim x -> -5^- / f(x) =-9
lim x -> -5^+ / f(x) =-6
lim x -> -5^ / f(x) = undef
f(x) = -6

[lean otter]

"-/"? That's not an operation, is it?

[pliant zephyr]

the "/" is a seperator between question and answer sorry

[lean otter]

Oh

[pliant zephyr]

sorry

[lean otter]

lim x -> -5^-
But then this just doesn't make sense

[nova creek]

I think it's the limit of -5 from the left

[pliant zephyr]

its like limit of x as it approaches -5 from the left

[lean otter]

Oh

...So, what's the question?

Oh

[pliant zephyr]

am i right cause it says im wrong and i dont understand how i could be if i am

[lean otter]

The limit as x approaches 5 from the left of f(x) is -9?

Is that what you're saying?

[pliant zephyr]

cause unl;ess i got the directions of the approaching left and right wrong idk how its wrong

[pliant zephyr]

yes

[lean otter]

If it's approaching from the left, it can't be -9

Perhaps you mixed up left and right

[pliant zephyr]

would it be -6 then?

[lean otter]

But approaching from the left, that would be -6

[pliant zephyr]

yeah i mixed them up

i thought it ment like going lower in value for going left

[lean otter]

Yeah, you thought it meant going to the left, didn't you?

But nah

[pliant zephyr]

yeah

would lim as it approaches -5 it would be undefined since theres a jump

[lean otter]

I'm actually not sure about this one

[pliant zephyr]

for sure and then i have a question about constants in a piecewise function if you could help

[lean otter]

Sure

I would like to say it's like limx->0(1/x), but 1/0 is undefined, unlike f(-5)

[pliant zephyr]

yeah i get you

[lean otter]

Well, I really just don't know. If your guide says it's wrong, then it's probably -6

But that's only a guess

[pliant zephyr]

bet thanks

so my constant question you have to solve for "a" to make the graph constant but when i solve it it cancels out so idk what to do

[nova creek]

The lim as x->-5 should be the same from both sides. Since it's not the same, it doesn't exist

[pliant zephyr]

-2x^2-x-4a as x < -4
ax-3 as x > -4

[pliant zephyr]

bet thats what i was thinking but wasnt sure

[pliant zephyr]

so when i do this i set them equal to one another and plug in "x" but when I solve for a they both cancel as both sides have a "-4a"

[lean otter]

What's the question?

That's just a function

[pliant zephyr]

so they want me get a constant "a" to make the two functions continous as they are apart of a piecewise function but when i try to solve for "a" the equations just cancel out

cause i get

-2(-4)^2+4-4a = -4a-3

so the -4a cancels

[lean otter]

Uhm

[pastel abyss]

f(x) = - 6 if x=6
f(x) = equation if - 5<x<10
f(x)= equation if x<-5

https://youtu.be/NQ1QtXHpvaQ

This is how you should write it

Piecewise function

[nova creek]

If the as cancel, either all values of a make it continuous, or no values of a do

To find out which, continue simplifying and see if you get a contradiction or not

[pastel abyss]

He was given a graph though

[nova creek]

He's asking about an unrelated question

[pastel abyss]

Oh i just looked at the first message srry

[lean otter]

Are you sure the question is to find a constant "a" such that the piecewise function is continuous? It seems that it would never be continuous anyway

[pastel abyss]

Take the limit as x approaches - 4 of both equations

It has no solutions

-28-4a=-4a - 3

[safe radish]

@pliant zephyr Has your question been resolved?

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[lean otter]

Tangent Line from Substituting the values x 1 , y 1 gives the slope of the tangent line: 2 + y ' = 1 - y ', so y ' = -1/2

[lean chasm]

no?

how did you get -1/2?

[lean otter]

Someone says I have to solve it and gave me that message

That would be the full one lol

Tangent Line from Substituting the values x 1 , y 1 gives the slope of the tangent line: 2 + y ' = 1 - y ', so y ' = -1/2

[lean chasm]

wat

what is your question

[lean otter]

Wait

[lost forum]

This doen't make much sense

[lean otter]

Release date of 1.5 is: Given to you if you solve the Tangent Line from Substituting the values x 1 , y 1 gives the slope of the tangent line: 2 + y ' = 1 - y ', so y ' = -1/2

That’s the full message

[lean chasm]

the full message of what

[lean otter]

That I got

I don’t even understand what I should do

[lost forum]

Ok ok so, first step would be to identify the full question. We can get some hints from the message but the full think would be better. Anyhow, lets try with what we got

[lean otter]

Ok thank you

[lost forum]

One idea is that they are asking you to find the equation of the tangent line of a function y = f(x) at some x where f'(x) = y' and y' is given by the equation you provided. Does that sound like something the person who sent the message would expect?

[lean otter]

I think so

I could ask some more things what’s the exact problem?

[lost forum]

That would be helpfull yes.

[lean otter]

What information is missing?

[lost forum]

Ohhh, I read your reply as: I could ask some more things (such as) what is the exact problem

For the problem I proposed (the one with the tangent line) we have all the information

Would you like us to try and solve the tangent line problem or would you like to ask the person to send you the full problem?

[lean otter]

We should try it

[lost forum]

ok good

are you interested in understanding the problem and its solution or do you just want a solution?

[lean otter]

Hmm I would like to understand the way to solve it but I don’t have much time

So the Solution first.

Thx

[lost forum]

If the point on the curve $y = f(x)$ where $y' = -1/2$, is $y_0 = f(x_0)$, then the equation for the tangent line to $(x,f(x))$, passing through $(x_0,f(x_0))$ would be: $y = 1/2(x_0-x)+f(x_0)$

hold up

latex issues

[lean otter]

Ok

[flat frigate]

Stamatis

[lost forum]

In the notation of your problem

you can write

[lean otter]

Um

[lost forum]

$y = 1/2(x_1-x) + y_1$

sorry hold up again

[flat frigate]

Stamatis

[lost forum]

There you go

[lean otter]

Thats the solution?

[lost forum]

Yes

[lean otter]

Oh ok thx

[lost forum]

Would you like to understand it?

[lean otter]

It it like f(x)=m(x)+b ?

Is it*

[lost forum]

Yes! But let me start with a question

[lean otter]

Ok

[lost forum]

Stop me if you get confused

[lean otter]

Ok

[lost forum]

We are looking for the tangent line to a function y = f(x)

at some point x_1

[lean otter]

Ok i got that

[lost forum]

more precesily at (x_1,y_1) where y_1 = f(x_1)

[lean otter]

And now?

[lost forum]

So, what do you know about tangent lines?

Roughly

[lean otter]

Not much lol

[lost forum]

Anything that comes to mind

[lean otter]

Hmm

[lost forum]

I does not have to be technical

[lean otter]

That we have to get x in ()

[lost forum]

What does this mean?

I am insisting because you might have confused some earlier concepts

[lean otter]

Yeah um idk

[lost forum]

and without straightening those out you will only be getting more and more confused

Well lets go back to the question

We are looking for a tangent line

and one thing we know is that it is a line!

what do you know about lines?

[lean otter]

Nothing right now

That confuses me

[lost forum]

Ok good, we are getting closer to the problem I think

Btw if you do not have time for this let me know

[lean otter]

I have some Minutes left

[lost forum]

what do you know about something that looks like this: $y = f(x)$

[flat frigate]

Stamatis

[lean otter]

Ik we had some task like that

y=3+2x
-y=3+2x

[lost forum]

Ok, so lets take an easier example

when you see y = x+1

[lean otter]

Yes

[lost forum]

what do think you can do with this, what is it? If someone asked what is y = x+1 what would you say?

[lean otter]

A Tanget Line?

[lost forum]

Any answer, ie answering "it is a task I was given" is ok

[lean otter]

Lol

[lost forum]

Well, let me start answering some questions and not only asking

so when you are given y=x+1, you are given a factory. Think of y as the output and x as the input

[lean otter]

Wait that guy has send a message

Im gonna send screenshots

He got that

Now I’m very confused

He cut the first part?

[lost forum]

ok, I see

for a start, if you replace $x_1=1,y_1=1$ in our solution you get the correct solution

[flat frigate]

Stamatis

[lost forum]

I had misunderstood you input. But now the question is full

So we get the same answer as he does

[lean otter]

Oh

[lost forum]

Now, I unfortunately have to go, but I will leave you with a series of questions you can ask your teacher so that he can help you straighten your understanding out

[lean otter]

Ok thank you

[lost forum]

1. I do not understand how I use functions y = f(x)
2. How is the graph of y = f(x) and the explanation you gave in my first question connected? For example, how can I construct the graph of y=f(x)

More coming hold on

[lean otter]

Yea

[lost forum]

3. How can I draw a tangent line to y=f(x)?
4. How can I find some function y = g(x), such that the tangent line we just drew is its graph
5. If instead of y=f(x) I have something like g(y) = f(x), can you give me examples to understand what this means?
6. Ok, so how do I draw the graph of something like g(y) = f(x)
7. How do I find the tangent line to something like g(y) = f(x) at some point?

That's all

[lean otter]

Ok

[lost forum]

If you show your teacher these questions I think they will be able to help you. If you understand the answer to all these 7 questions you will be able to solve problems like the one you sent here on your own

[lean otter]

I hope so 😄

[lost forum]

Actually replace question 2 with:
2) what is the graph of y=f(x) and how do I make it

thats better

[lean otter]

K

[lost forum]

anyway, goodbye!

[lean otter]

Bye!

[safe radish]

@lean otter Has your question been resolved?

[safe radish]

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[tender mountain]

What algebra factoring strategies (methods) do I need to learn to factor most polynomials?

[tender mountain]

Only the absolute necessary ones

[worthy hemlock]

All of them

Quadratic formula, completing the square, factor by grouping, etc

All of them would be used, depending on the problem

[tender mountain]

how long will that take

[worthy hemlock]

Depends on how long it takes you to learn

[tender mountain]

can you recommend a video?

[worthy hemlock]

No

But you can Google them

Or look them on youtube

It's not hard to do that yourself

[tender mountain]

ok

so what are the strategies?

[worthy hemlock]

Just learn them, it's not that hard to Google the concepts and learn them

[tender mountain]

ok

https://tenor.com/view/thanks-thank-you-gif-20870035

bye

.close

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[pliant inlet]

No clue how to approach this

[pliant inlet]

Perhaps PAP^1 = D with PBP^1 = D or something alone those lines

just no idea

[stoic dune]

I've been drinking a little and might be saying something dumb. Do diagonal matricies commute?

[pliant inlet]

Commute?

If AB = C then BA = C

you mean?

[stoic dune]

Two matrices X and Y commute if XY = YX

Yeah, I think diagonal matricies commute. What does that mean for you?

[pliant inlet]

PAP^-1 * PBP^-1 = PBP^-1 * PAP^-1?

[stoic dune]

Yeah that's the statement. Does it simplify?

[pliant inlet]

uh

P^1*P =I

so PABP^-1 = PB AP^-1

hm

[stoic dune]

Good! Then we can multiply both sides by P⁻¹ on the left

And multiply both sides by P on the right

[pliant inlet]

AB^p1 = BAP^-1

and then

AB = BA

huh

makes sense

OK yeah

i understand

.close

[safe radish]

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[humble grove]

I have to calculate the area of this triangle

[lean otter]

herons formula

[humble grove]

Could you please elaborate

[lean otter]

given the sides of a triangle herons formula tells u the area basically

$s=\frac{a+b+c}{2}$

area=$\sqrt{s(s-a) (s-b)(s-c)}$

[flat frigate]

sqrt(-1) is approx -30

[lean otter]

where a, b, c are the sides

[humble grove]

Yes

[lean otter]

so where u are also told the sides

so u can just plug it in

[humble grove]

ah i add them then divide

then plug it in

okay let me try

[lean otter]

yes

[humble grove]

I've actually never learnt this before

[lean otter]

perimetee/2

[thin bridge]

the problem is overspecified

[lean otter]

it's a grade 7 topic

atleast I learned it in grade 7

first

[humble grove]

It didn't work in the triangle

[humble grove]

I believe there's a different method

[thin bridge]

using the 6m and 14m side and the angle between them, the third side is actually supposed to be around 18.347m,
not 18m exactly

[humble grove]

Perimetere divided by 2 is 20

[lean otter]

true

[humble grove]

just ignore it

[humble grove]

now i plugged it in

the answer i got is 643

[thin bridge]

if you take the diagram at face value, you'll get slightly different results depending on which formula you apply

[lean otter]

,w sqrt(20(20-18)(20-14)(20-8))

[flat frigate]

Results provided by WolframAlpha

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

Requested by pi=e#5213

[humble grove]

huh

how

wtf

my calculator gives me 643

i plugged it in exactly

[lean otter]

[humble grove]

ah its because it wont square root the entire formula

[thin bridge]

can you show what you're plugging into your calc

[humble grove]

only the 20

smh this shitty calc

[thin bridge]

wdym by only the 20

[humble grove]

should i just find the anwer then square root?

[lean otter]

like sqrt(20) * (20-18) * (20-14) *(20-8)

[humble grove]

okay there we go

im gonna have to get used to thsi on the exam

smh

[thin bridge]

put sufficient parentheses around the whole thing you're taking the square root of

were you asked to round the answer?

[humble grove]

How do I do the second question?

[thin bridge]

based on the solution and that there was an angle given in the question

it seems like they wanted you to apply the trig formula for area of a triangle

[humble grove]

so what could have I used

[humble grove]

does herons formula work for any triangle?

[thin bridge]

exactly what I said

[humble grove]

does herons formula work for any triangle?

.

[thin bridge]

herons formula can be used to calculate the area of a triangle given its 3 sides

but based on the given solution it's not what's intended here

[humble grove]

Ah okay

[thin bridge]

since this is part of a trig course

[humble grove]

Btw when it says ACB angle what does that mean

which angle is it

Is it this one?

[thin bridge]

the angle formed between segments AC and CB

yes

[humble grove]

Ah okay

[humble grove]

looking at this triangle how do ik which is the adj and which is the hyp

[thin bridge]

those terms aren't applicable to non-right triangles

[humble grove]

so how do i find the angle?

[thin bridge]

sine rule

[humble grove]

ok ima try my best and u correct me pls

[thin bridge]

as for part i), and mentioned earlier

if you take the diagram at face value, you'll get slightly different results depending on which formula you apply
using herons gets a value that'd round to 52.7
so it seems like they want you to apply
Area = 1/2 ab sin(C)

[humble grove]

I tried the basic formula

but i have to find the angle C before i do it

[thin bridge]

you have angle C

[humble grove]

110 isnt angle A?

[thin bridge]

I mean you have the angle represented by C in the formula

just because the formula uses
a,b,C
does not mean that you need the sides and angles labelled a,b,C in your diagram
and also does NOT mean that if your diagram is labelled with other letters that the formula wouldn't be able to be applied

[humble grove]

oh

that makes way more sense

but even if i plugged in the formula

i get 110

[thin bridge]

Area = 1/2 ab sin(C),
from labelling convention,
a,b are two of the sides and C is the angle between them

here your given angle is the 110°

which two sides form that angle between them

[humble grove]

8 and 14

ohh i was using 18

[thin bridge]

8 and 14
yes

[humble grove]

ah yes

i got 52 once again

once again

thank u so much

that formula is way easier than herons rule

[thin bridge]

onvr shsin?

[humble grove]

can you help me use the sine rule

[humble grove]

what

[thin bridge]

.

[humble grove]

typo lol

[thin bridge]

the stuff after the dp is important

,W calc 56sin(110deg)

[flat frigate]

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Requested by ramonov#0921

[thin bridge]

note that this rounds to the given intended answer

now moving on, can you show your attempt at applying the sine rule

[humble grove]

x14/sin110 = 18/sin110

[humble grove]

Correct?

[thin bridge]

no

you didn't apply the sine rule correctly

[humble grove]

could you please show me how to

[humble grove]

i think i messed up w the second one right

[thin bridge]

can you repost the diagram so I don't have to scroll

[thin bridge]

18/sin(110°),
would be part of it

[humble grove]

oh yeah

u said

my bad sorry

[thin bridge]

it's the thing on the left side of the equation that's problematic

[humble grove]

oh

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what's the angle you're trying to find,
what's the side opposite that

[humble grove]

8 is opp

so im supposed to use 14 and 18

huh

im confused now

[humble grove]

.

[thin bridge]

which was <ACB .
you'd just leave <ACB in the sine rule like that since it's currently unknown

[humble grove]

okay

[thin bridge]

the side opposite that angle has length 8(m)

and use that to set up your equation

[humble grove]

can you please do it for me so i can takenotes

im genuinely not sure at all about this

[thin bridge]

well you're looking at the sine rule in front of you right?

[humble grove]

yeah

8/sin110=x14/sin110

[thin bridge]

from labelling convention, the lowercase letter represents the side opposite the respective capital letter angle/vertex

the 110° angle and side opposite that is given (18m)

<ACB is unknown and the side opposite that is given 8m)

plug those into the sine rule

[humble grove]

dude

i dont know

i tried

my brains gonna explode

[thin bridge]

and seeing as all pairs are covered in the full sine rule

[humble grove]

The angle is 24.7

[thin bridge]

just simply plug in what's known

[humble grove]

i dont know how to

[thin bridge]

and you can even follow the labelling in your diagram

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i told u i dont know how to

[thin bridge]

try not to overthink this

[humble grove]

can you just write out the formula

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what's is C

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or equation

ill take notes that way

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you said you're looking at a copy of the sine rule right?

[humble grove]

and understand it better

yes

[thin bridge]

do not overthink this

[humble grove]

please hurry up

i gtg in 7 mins

[thin bridge]

you are given an angle to be 110°

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im gonna kill myself

okay butwhere do i plug in the 110

[thin bridge]

which letter in your diagram represents that angle

do not overthink this

[humble grove]

A?

[thin bridge]

yes

[humble grove]

okay

now what

[thin bridge]

so you'd plug in 110° for A in the sine rule