help-10

[fast cliff]

1.5% interest

then which account was x

[cinder quest]

First account

[fast cliff]

yea and the question asks about the second account

so we know the first account has 1050

so how much does the second account have

[cinder quest]

1510?

[fast cliff]

yea exactly

that should be the answer

[cinder quest]

R u sure

[fast cliff]

reasonably sure

[cinder quest]

hey do u mind if we re do it

i just wanna study it im sorry lol

[fast cliff]

ok

do you want to try and do it

[cinder quest]

yea

[fast cliff]

ok go ahead

[cinder quest]

hey

i have (0.15)x+(0.045)y=83.75

[fast cliff]

yea thats good

do you know what x and y represent exactly

[cinder quest]

mhmm

[fast cliff]

so what other information does the question gibe

give

[cinder quest]

i have 15+45(x+460)=83.75

is that wrong

[fast cliff]

yea thats wrong

[cinder quest]

15+45(x+460)=83750

[fast cliff]

going from this to that you want to multiply to get rid of decimals

you need to multiply both sides thats what you did wrong

[fast cliff]

yea good

[cinder quest]

im on x=1050

what do i do no

[fast cliff]

you need another equation

[fast cliff]

you already made an equation using interest rates and total interest earned

[cinder quest]

83.75=.045(1050)

[fast cliff]

nope

what does the question tell you about

how much more one bank account have

[cinder quest]

o

h

I add 460

but how do i put that on a equation

[fast cliff]

so

if x is the amount of money in the smaller account

y is the amount of money in the bigger account

theres a difference of 460

can you figure that out

dont overthink it, its way easier

[cinder quest]

idk lool

im sorry

1050 is X

and 1050+460 is the second

ok i see

[fast cliff]

sorta but

[fast cliff]

you wouldnt have gotten this unless you had

y = x + 460

thats the other equation

[cinder quest]

ok

i got it

TYSM

ILYYY

.close

[obtuse pebble]

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[teal smelt]

how would i start this

would i get rid of the sqrt3?

also how would i get rid of the sqrt3

[final thunder]

You can convert this into harmonic form or whatever it’s called

Say LHS = Rsin(x+α) and work out what what R and α are

[teal smelt]

the question is asking to evaluate and solve for x over the domain of [0,2pi]

[final thunder]

Yeah ik

[teal smelt]

im not familiar with the harmonic form

is it an identity?

[final thunder]

Look it up

[teal smelt]

i havent learned this

im in grade 12 adv functions

@final thunder

[final thunder]

Yeah my bad

Just move cos to the other side

Then divide by cos(x)

[teal smelt]

do you think you can show me

@final thunder

because someone is telling me this

but i dont know where they got the denominator of 2 from

[final thunder]

$\sqrt 3 \sin{(x)} = -\cos{(x)} \implies \tan{(x)} = - \ \frac{1}{\sqrt 3}$

[warm shale]

Pure

[teal smelt]

so with sin having the sqrt

how did you remove it when converting to tan

because

tan = sin/cos

im confused

@final thunder

[final thunder]

Divide both sides by root 3

[teal smelt]

so that would be sin = -cos/sqrt3

and then to make tan it would be sin(x)/cos(x)/sqrt3

?

@final thunder

[final thunder]

$\sin{(x)} \times {\color{red}\frac{1}{cos{(x)}}} = \frac{\cos{(x)}}{\sqrt 3} \times {\color{red}\frac{1}{cos{(x)}}}$

[warm shale]

Pure

[teal smelt]

so tanx = -sqrt3/3

is equal to 11pi/6

as well as 5pi/6

so is that the answer?

@final thunder

x1 and x2

[final thunder]

Yeah

[teal smelt]

my classmate did it a different way

is it still correct?

@final thunder

.close

[obtuse pebble]

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[final thunder]

Correct

[final thunder]

I started this wrong I thought it was +cos(x)

[obtuse pebble]

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[smoky zenith]

i dont know how they got this, im putting it in my calculator as 15tan-1(1/37.36) but i get 22.99 so i tried 15tan-1(37.36) but i get 0.401

[smoky zenith]

so i thought make it makes i should look for radians and nof for degrees, but nothing comes back as 19.55

[obtuse pebble]

@smoky zenith Has your question been resolved?

[obtuse pebble]

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[ember cairn]

hi

[ember cairn]

i’m stuck on part b any help

is appreciated

[marble junco]

how did you verify a?

[ember cairn]

sin2(t) + cos2(t) = 1

[marble junco]

ok and?

[ember cairn]

oh should i show it? sorry

[marble junco]

yeah

[marble junco]

ok that's pretty accurate

since we know it's in Q4, we also can figure out what t value is

try to get it

[ember cairn]

am i able to use arcsin?

[marble junco]

ok, actually we don't need to get the value of t

do you know what sin(a+b) =?

[ember cairn]

yes the sum of angles formula

sin(a)cos(b)+cos(a)sin(b)

[marble junco]

ok use it

$sin(t+pi/4)$

[ember cairn]

a = -0.64 and b would be pi/4 right?

just double checking

[marble junco]

no need to replace -0.64

just use t and expand

[ember cairn]

so -3/5 as a and pi/4 as b

[marble junco]

no, t as a

[ember cairn]

i’m sorry i’m not getting the desired answer

[marble junco]

again, don't plug in -0.64

we already know the values of sint and cost from the condition and a)

[ember cairn]

would it be (-3/5)cos(pi/4)+(4/5)sin(pi/4)

[marble junco]

maybe

and we know the values of cos(pi/4) and sin

[marble junco]

sin(pi/4) is 0.014 yes, but it can be written more accurately

using sqrt

[marble junco]

ok so that's your answer

you can simplify it a bit

[ember cairn]

thank you so much 🙏

.close

[obtuse pebble]

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[spark field]

i pluged in -6 and 7
into the equation and got
f(-6) = -103
f(7) = -506
then i did the mean value theorm
f'(c) = -506-103/7-6
= -403
then i pluged that into the dervative of the equation
6x^2-18x-108=-403
and i got x = 1.5, 6.849

[obtuse pebble]

Please don't occupy multiple help channels.

[spark field]

can sm1 help me understand what im doing wrong

[marble junco]

everything is correct except

7 - (-6) is not 1

[spark field]

omg

bruh

[marble junco]

yeah

[spark field]

i got 5.383

is that correct?

for b

and A) i got -31

[marble junco]

formulas are correct. making sure numbers are correct is your responsibility 🙂

[spark field]

f'(c) = -506-103/7-6 = -31

6x^2-18x-108=-31

x = 5.3,-2.3

[marble junco]

seems right

[obtuse pebble]

@spark field Has your question been resolved?

[smoky zenith]

Dang fancy Calc

[ember cairn]

aha yea

[obtuse pebble]

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[obtuse pebble]

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[bleak eagle]

Do I have the right idea for number 9 or am I misunderstanding?

[obtuse pebble]

@bleak eagle Has your question been resolved?

[bleak eagle]

<@&286206848099549185>

[tardy epoch]

Where your work

[bleak eagle]

The second picture there is my work that I have. It seemed like a problem you could just interpret based off the questions without showing additional work

[obtuse pebble]

@bleak eagle Has your question been resolved?

[obtuse pebble]

@bleak eagle Has your question been resolved?

[obtuse pebble]

@bleak eagle Has your question been resolved?

[obtuse pebble]

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[sage dagger]

for part b i am just wondering what equation im supposed to use

[sage dagger]

for sum of roots and prod of roots

[obtuse pebble]

@sage dagger Has your question been resolved?

[obtuse pebble]

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[dark crag]

find element of a 22 i need help

[wet drift]

First multipley the mattix three times

[wet drift]

You will see some pattren will follow

[dark crag]

okey

thank u

i see

[obtuse pebble]

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[kind snow]

anyone know how to make this kind of graph using rational functions?

[obtuse pebble]

@kind snow Has your question been resolved?

[tawny fog]

Y=k/x

K ≠0

If you are talking about hyperbola

[tardy gull]

I think that you can't make this kind of graph using rational functions because, on the graph you drew, pretty much all the elements on the x-axis are assigned to 2 elements on the y-axis

[autumn adder]

This graph basically fails for the vertical line test for "functions"

[obtuse pebble]

@kind snow Has your question been resolved?

[obtuse pebble]

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[tight forge]

for this question i split it into two sequences:
1 + 3 + 5 + 7 + ...
and
2 + 4 + 6 + 8 + ...

so i said the difference of these two sequences needs to be more than or equal to 100

so the sum of the first series would be 2n^2 and the second series n^2 + 1

[wet drift]

Can you break -2 like -1-1 , -4 like -1-3 , -6 like -1-5 may be this can help

[tight forge]

if u subtract these two, u get (n-1)^2 >= 101

what do you mean?

[wet drift]

That will cancle out all the positive part

Hope we can get some hint from there

[tight forge]

hmm ok i sort of get where ur going

thanks ill try it out

ohh so it would be 199

i see

so i understand this way, but why didnt my way work? i got n = 11 before

[wet drift]

Can you explain how you got 2(n-1)

[tight forge]

The sequence was 1 + 2(n-1) so isn’t the series (n/2)(2 + 2(n-1))?

Like the LHS sequence

[obtuse pebble]

@tight forge Has your question been resolved?

[obtuse pebble]

@tight forge Has your question been resolved?

[obtuse pebble]

@tight forge Has your question been resolved?

[obtuse pebble]

@tight forge Has your question been resolved?

[hexed agate]

Can't you make it a reverse geometric series?

Like (-1)^(n+1) n + (-1)^n n + (-1)^(n-1) n +.... +3-2+1

[obtuse pebble]

@tight forge Has your question been resolved?

[obtuse pebble]

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[median forum]

Hey

[median forum]

I am looking for help understanding this question

I do not know why he start with condition 3 I get the solutions

but am not sure how we get them like the logic behind it

[marble junco]

so you need to make a(x) and b(x) which both converge to 0 when x->3, but also make the 3rd function converge to -3

[crisp tendon]

is anybody willing to help ?

[small thicket]

#❓how-to-get-help

[marble junco]

i am

he asked for help understanding the question

[crisp tendon]

is there something wrong with this polynomial expression?

i have to turn it into factored form

[small thicket]

sigh go to an empty channel. like #help-0

[obtuse pebble]

@median forum Has your question been resolved?

[obtuse pebble]

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[placid forge]

Need help with a)

[placid forge]

For the top question, not lower one

[violet sentinel]

hi there! So it's utilizing the distance formula. Do you know what that formula?

[placid forge]

Here’s what I did, can’t find what I did wrong

The left side

Is wrong

Bottom left

Should be 3 according to answer sheet

[violet sentinel]

$(-2)^2 \neq -4$

[unique spear]

You arrange the coordinates wrong

the first step of your working

should not be 5 - a and be a - 5

[placid forge]

Wait

[proven zephyr]

it's the same no?

[violet sentinel]

well that shouldn't matter tbh. $(1-3)^2 = (3-1)^2$

[warm shale]

MellowDramaLlama

[proven zephyr]

yeah

[unique spear]

oh right

[placid forge]

Isn’t is 5-a still tho?

[proven zephyr]

but yeah (-2)^2 is not -4 (said MellowDramaLlama)

[unique spear]

i just have a fear that it will mess up my entire working so I usually just arrange them in order

[placid forge]

Oh

[placid forge]

My calculator said it is tho??

[unique spear]

You forgot the brackets

[proven zephyr]

-2^2 is not (-2)^2
in the first one, you're only squaring the 2 and not the -

[unique spear]

You must’ve typed -2 ^ 2

[placid forge]

Ohhhhh

So it’s 4 then

[unique spear]

yes

-2 ^ 2 just makes the calculator think you’re multiply 2 two times and after that add a negative sign

[placid forge]

So apart from that, everything else is fine

[proven zephyr]

how did the 12 turn into +-12

[placid forge]

That’s the formula?

a is equal to two things in this equation

[proven zephyr]

ok i'm confused

[violet sentinel]

this part confuses me

[proven zephyr]

yeah exactly

me too

[violet sentinel]

what is the open patronesses?

[placid forge]

On the right side?

[violet sentinel]

yeah

[placid forge]

It’s just ‘Im not gonna do the stuff in there’

My maths teacher says it’s fine

[violet sentinel]

yeah I would get rid of that habit. Otherwise you'll make algebra errors

[placid forge]

As long as it’s included in the end

[proven zephyr]

how did it become 5 - a^2 -_-

and then becomes 5 - a

[placid forge]

Square root is gone

[violet sentinel]

$(5 -a)^2 \neq 5 - a^2$

[warm shale]

MellowDramaLlama

[placid forge]

Well… I’m following the textbook for that part lol

[proven zephyr]

uhh they square rooted both sides...

square root 1 is +- 1

you can't just put +- behind the 12

[violet sentinel]

no if you notice that in the book it's still of the form $(9 - a)^2$, not $9 - a^2$

[warm shale]

MellowDramaLlama

[placid forge]

Oh right

Then what do I do?

[violet sentinel]

$8 = 4 + (5 - a)^2$ Now you can follow what your book was doing 🙂

[warm shale]

MellowDramaLlama

[violet sentinel]

so subtract 4 by both sides, take the square root, then solve for a

[placid forge]

But wouldn’t a have two different answers?

Text book says it should be 3 and 7

[violet sentinel]

yes

[violet sentinel]

look here. a = 8 and a = 10

[violet sentinel]

and yes that's correct

[placid forge]

Yep

[violet sentinel]

so a has two answers there 🙂

its' the same steps

just diff values

[placid forge]

So how do I get a?

[violet sentinel]

same steps. different values

[violet sentinel]

here

[placid forge]

Alr

So do I use the +_ or no?

[proven zephyr]

note: they didn't just put a +- behind the 1, they square rooted the 1 to get +- 1

[violet sentinel]

yes. Since you are square rooting it comes with a +/- unless specified otherwise (or some other higher level math stuff)

[placid forge]

Ok thanks

[violet sentinel]

but for your case yes it will have +/-

[placid forge]

Wait I’m not understanding the end bit too much

+-4=5-a

Yes?

Then what do I do..

[violet sentinel]

solve for a. You add +/- 4 to both sides (which is still +/-4) and then add a to both sides

[violet sentinel]

it's not +/- 4

[placid forge]

Oh?

[violet sentinel]

$\sqrt{4} \neq \pm 4$

[warm shale]

MellowDramaLlama

[placid forge]

It was 8, then I minused 4

[violet sentinel]

yes

then you take the square root

of both sides

[placid forge]

Then what?

[violet sentinel]

$8 = 4 + (5 - a)^2\$
$4 = (5-a)^2\$
$\sqrt{4} = \sqrt{(5-a)^2}\$
$\pm 2 = 5 - a\$
$a = 5 \pm 2$

[warm shale]

MellowDramaLlama

[placid forge]

Oh so I root the left number

[violet sentinel]

you're square rooting both sides

[placid forge]

Ohhh

Gotcha

[violet sentinel]

fwiw you can also solve it this way

$\sqrt{8} = \sqrt{(1-3)^2 + (5-a)^2}\$
$8 = (1-3)^2 + (5-a)^2\$
$8 = (-2)^2 + 25 - 10a + a^2\$
$0 = a^2 - 10a +21\$
$0 = (a - 7)(a - 3)\$
$a = 3, 7$

[warm shale]

MellowDramaLlama

[violet sentinel]

this is another approach

if factoring is more comfortable for you

[placid forge]

Do I switch it from negative to positive?

5-a=-2
a=-7

So would I change that to 7?

[violet sentinel]

well remember you're doing both -2 and +2. Those just flip signs

[placid forge]

I don’t get that?

.close

[obtuse pebble]

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[trail tiger]

would this be a good place to ask for programming questions ?

[trail tiger]

it's related to prime numbers

and sieving primes up to like a few millions

well, you never know until you try : here's the problem

In Python, I have one bitwise segmented sieve, and one normal sieve

the normal one is ~3x faster than the bitwise one, which is not supposed to happen

and they are both pure copies of each other, but list operations were replaced with bit operations

Does anyone know if list operations are faster than bit operations or something ... ?

[violet sentinel]

well faster is subjective with CS. There are a lot of factors that can effect runtime. What is the runtime between them?

theoretically bitwise should theoretically be faster

[obtuse pebble]

@trail tiger Has your question been resolved?

[trail tiger]

normal sieve found 664579 primes less than 10000000
bitwise sieve found 664579 primes less than 10000000

normal sieve : took 2.14882675 seconds
bitwise sieve : took 8.024978542 seconds

[dark mango]

could you share the code

[trail tiger]

def sieve(upper_bound):
    increment = int(upper_bound ** 0.5)
    primelist = list()

    #Sieve the lowest segment normally
    array = [True for _ in range(increment)]
    for index in range(2, increment):
        if array[index]:
            primelist.append(index)
            for nonprime in range(index, increment, index):
                array[nonprime] = False

    #Sieve by incrementing with root of n
    for low in range(increment, upper_bound - increment, increment):
        array = [True for _ in range(increment)]
        for prime in primelist:
            # check prime < root of max value in segment
            if prime ** 2 <= low + increment:
                # start at smallest multiple bigger than low
                for nonprime in range(bool(low % prime) * prime - low % prime, increment, prime):
                    array[nonprime] = False
            else:
                # This line makes the code run 50x faster
                break
        
        #Harvest the primes
        for index in range(increment):
            if array[index]:
                primelist.append(low + index)

    #Sieve the remaining numbers
    low += increment
    array = [True for _ in range(upper_bound - low)]
    for prime in primelist:
        if prime ** 2 <= upper_bound:
            for nonprime in range(bool(low % prime) * prime - low % prime, upper_bound - low, prime):
                array[nonprime] = False
        else:
            break
    #Harvest the last primes
    for index in range(upper_bound - low):
        if array[index]:
            primelist.append(low + index)
    
    return len(primelist)

[violet sentinel]

okay yeah if it's that much of a dff there's something wonky going on

[trail tiger]

normal ^^^

def bitsieve(upper_bound):
    increment = int(upper_bound ** 0.5)
    primelist = list()
    print(f'increment = {increment}')

    #Sieve the lowest segment normally
    array = (1 << increment) - 1
    for number in range(2, increment):
        if array & (1 << number):
            primelist.append(number)
            for composite in range(number, increment, number):
                array &= ~(1 << composite)

    #Sieve by incrementing by root n
    for low in range(increment, upper_bound - increment, increment):
        array = (1 << increment) - 1
        for prime in primelist:
            # Check if prime < root of segment's max value
            if prime * prime <= low + increment:
                # This just finds the smallest multiple of the prime above low
                for composite in range(bool(low % prime) * prime - low % prime, increment, prime):
                    array &= ~(1 << composite)
            else:
                # This one break makes the code 50x faster
                break
        
        # Harvest the primes
        for prime in range(increment):
            if array & (1 << prime):
                primelist.append(prime + low)

    # Sieve whatever remains
    low += increment
    array = (1 << (upper_bound - low)) - 1
    for prime in primelist:
        if prime * prime <= upper_bound:
            for composite in range(bool(low % prime) * prime - low % prime, upper_bound - low, prime):
                array &= ~(1 << composite)
        else:
            break

    # Harvest
    for prime in range(upper_bound - low):
        if array & (1 << prime):
            primelist.append(prime + low)


    return len(primelist)

bitwise ^^^

if I had to guess it's because i'm doing this thing :
array &= ~(1 << composite)

it looks like it's 3 operations at the same time

but honestly I just don't know

also that one break statement is so important; without it my code just can't complete the sieve up to 10 000 000

feels like something went terrible wrong

[obtuse pebble]

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[placid forge]

i need help.

[placid forge]

.close

[obtuse pebble]

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[winged salmon]

https://www.desmos.com/calculator/9xqfpeqdec

[winged salmon]

i need help

lining these two lines up in desmos

[hollow fjord]

need help with a funtion problem, as of right now I'm not too sure on how to do this and I dont know the formula

[winged salmon]

like the gradient

needs to be the same

[winged salmon]

pretty sure it would be f(x)=mod 2x-9

or 2x-7

[royal basin]

@hollow fjord channel busy please move

[winged salmon]

anyone here?

https://www.desmos.com/calculator/9xqfpeqdec

i need help making the gradients of those 2 lines the same

anyone...

someone...

gettin kinda lonely here

[obtuse pebble]

@winged salmon Has your question been resolved?

[obtuse pebble]

@winged salmon Has your question been resolved?

[obtuse pebble]

@winged salmon Has your question been resolved?

[obtuse pebble]

@winged salmon Has your question been resolved?

[twilit ivy]

hey so i found your thing interesting and tried myself
and i think you should reduce the number of different curves you are plotting
bcz i just used 2 to get a good enough result

that way you dont have to worry about lining up those two dots

[winged salmon]

I need 4-7 tho

[twilit ivy]

oh do you want to like make a perfect fin?

[winged salmon]

ah i made it perfect

took a while

but i derived an ellipse

lowkey proud of myself :3

but um

can u help me explain this funtion

[obtuse pebble]

@winged salmon Has your question been resolved?

[obtuse pebble]

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[teal turret]

Use product rule to find f’(x)

[timid silo]

chain rule for the square root

[rose otter]

k ty

.close

[obtuse pebble]

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[strong nebula]

Im not sure how to set up c

[strong nebula]

I dont really know what to do with 0 = 6t^2-14+4

[teal turret]

Solve it

It’s a quadratic

Then once u get the first positive instant when v(t) = 0, plug that t-value into a(t) to find the acceleration at that moment

[strong nebula]

i got 72 for the positive and plugged that into a(t) and got 850 but it says it is wrong.

(14+sqrt((-14)^2(4(6)(4)))/12

[teal turret]

t=72?

Try again

[strong nebula]

alright

[strong nebula]

is it 12

12(12)-14 = 130?

[teal turret]

No

How are u solving it

[strong nebula]

@teal turret using the quadratic formula
a=6
b=-14
c=4
and I get
14 + sqrt((-14)^2 - 4(6)(4)) / 2(6)

(14+sqrt(100))/12 = 24/12

I would also like to solve this problem
the derivative of that would be
-6^-4+16
and the 2nd derivative
24x^-5+16
and i plug in 6
so 24(6)^-5+16 = 16.003 but it says its wrong

[obtuse pebble]

@strong nebula Has your question been resolved?

[strong nebula]

<@&286206848099549185>

[carmine bear]

Well this is right and theres a 2nd solution if you use the - bit of the quadratic equation so i mean you basically have the answer

since it's just putting t back in to the function for acceleration

[strong nebula]

i got 12(2)-14=10 but it says it is wrong

[carmine bear]

have you tried using the second solution for t

[strong nebula]

no, i thought it wanted the first positive instant

[carmine bear]

well the other root is positive and smaller than 2

[strong nebula]

i got 0.3 from the quad. formula
and plugged it into a(t) and got -10.4 and it is still wrong

[carmine bear]

use 1/3

[strong nebula]

ok the correct answer is -10. but how do I decide what solution from the quad formula to use?

[carmine bear]

well here it said the first positive instant when the velocity is 0. So the first time the curve crosses the x axis (v = 0) when t > 0 (since t is positive)

since 1/3 is smaller than 2 thats the solution you use

[strong nebula]

ok gotcha

Could you also help me out with the other problem above the Bot comment

[carmine bear]

sorry about this, I dont 100% understand what you're asking here

[strong nebula]

Im suppose to find the 2nd derivative of the problem given right, and plug in -6

[carmine bear]

ah ok i somehow didnt see that

[strong nebula]

The 2nd derivative is 24x^-5+16

[carmine bear]

yeah

[strong nebula]

Aftering plugging in -6 for x I get 15.9969

[carmine bear]

x is 6 in that question

but even then the answers are similar

[strong nebula]

yeah, 16 if 6

[carmine bear]

did you try putting in 16 instead of 16.003 or 15.9969

[strong nebula]

Yeah it says its wrong

[carmine bear]

that's weird

the answers must be wrong or there might be a typo in the question

[strong nebula]

ok thats fine, Im just glad i was able to find the derivative of 2*1/3^3 correctly

ill email the teacher about it

.close

[obtuse pebble]

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[ember cairn]

for question b i am stuck on finding what the value of x is

[hardy widget]

If it helps, think of setting the argument equal to u

and finding what u has to be for sin = 1/2

then substitute back to find x

[ember cairn]

i’m sorry?

[hardy widget]

solve sin u = 1/2

then set 3x - pi/2 = u

and find x

[ember cairn]

ok i’ll try that ty

.close

[obtuse pebble]

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[stoic merlin]

Quick question but for a) would i count 7 in the combination

or do i use 6

[desert sinew]

Guess so. I think you should do $$P(X = 7 \cup X = 8)$$

[warm shale]

jimmy1234

[stoic merlin]

Thank you!

.close

[obtuse pebble]

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[lean tendon]

hello

[lean tendon]

im working on finding the absolute max and min of a multivariable problem

how do i express this boundary as an inequality?

[twin sapphire]

its essentially three lines

[brisk matrix]

you can write it as just 2 inequalities

[lean tendon]

hm

$-2 \leqslant x \leqslant 10$

$-1 \leqslant y \leqslant 6$

[lean tendon]

like this?

[brisk matrix]

no

[warm shale]

Ziggy

[brisk matrix]

if you start with $-2\le x\le 10$, your second inequality should be for y and should have an x in there as well

[warm shale]

maximo

[lean tendon]

oh woops

my bad

[warm shale]

Ziggy

[lean tendon]

these two right?

[brisk matrix]

its closer

[lean tendon]

hm

[brisk matrix]

but that region is a square

[lean tendon]

yes i have a triangle

[brisk matrix]

you want something like $-1\le y\le f(x)$

[warm shale]

maximo

[lean tendon]

f(x) ?

[brisk matrix]

yes, some function of x

[lean tendon]

like the line

the hypotenuse?

[brisk matrix]

yes

[lean tendon]

this is a right triangle

[brisk matrix]

yes

[lean tendon]

$-1 \le y \le \frac{58 - 7x}{12}$

[warm shale]

Ziggy

[lean tendon]

thats the equation of this inperticular line

[brisk matrix]

yes that's your second inequality

[lean tendon]

and my first one is correcT?

[brisk matrix]

yes

[lean tendon]

okay im trying to figure out how to do my absolute max and min in mathematica

but this is a constraint on the domain of the function

correcT?

[brisk matrix]

that's one way to look at it

[lean tendon]

whats another way

[brisk matrix]

you're just looking for an max/min in that region, doesn't necessarily mean the function is restricted to this domain

[lean tendon]

yes

its a constraint then

a domain constraint

[brisk matrix]

not necessarily

f:R->R by f(x) = x has a local maximum in the region 0 <= x <= 3 at x=3
doesn't mean we restricted f to [0, 3]

[brisk matrix]

ah i missed this

i dont know about mathematica, i think that may work

[lean tendon]

we're just looking in that region

but why did is the y value restricted by a function?

because we need to determine the location of the hypotenuse of the triangle?

[brisk matrix]

i mean you can always think of the y value as restricted by a function

$-1 \le y\le 0$ has functions $f(x) = -1, g(x) = 0$ such that $f(x) \le y \le g(x)$

[warm shale]

maximo

[lean tendon]

hm

[brisk matrix]

obviously this is overkill notation, but it shouldn't surprise you that we can bound it like this

[lean tendon]

yes it's the output of the function

f(x) = y

f(x,y) = z

f(x,y,z) = k

[brisk matrix]

what are you saying

[lean tendon]

y is dependent on a function?

thats why its bounded by a function

[brisk matrix]

the region we constructed has y depending on a function, but we could have restricted y to -1 <= y <=6

and then used f^-1 to restrict x by a function instead

[lean tendon]

hm

[brisk matrix]

the reason we use functions to restrict the variables is that it's no longer just horizontal or vertical lines

[lean tendon]

hmhm

OKAY

whoops sorry for caps

[lean tendon]

thank you for your help

im going to bed

goodnight boss

🙂

.close

[obtuse pebble]

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• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[austere umbra]

Hey

[austere umbra]

I’m back

Need help with Properties of triangles congruency geometry stuff

[obtuse pebble]

@austere umbra Has your question been resolved?

[austere umbra]

<@&286206848099549185>

Anyone?

@twin sapphire

All unit

[teal turret]

u need to send a specific question

[austere umbra]

Entire unit**

Uh?

No specific question

[austere umbra]

This and proofs relating to that including cpctc sss sas asa aas

I have lot of questions on the entire unit

[fierce lagoon]

You need a specific question

We're not gonna give you a whole ass unit

[austere umbra]

I need like somebody to teach me bruh

[fierce lagoon]

We don't teach here

We assist

[austere umbra]

Some people did :/

[fierce lagoon]

Unless you have a specific question, we can't do much

And that would be a waste of my time

To go through a whole unit

[austere umbra]

Yes

[austere umbra]

Yes

[fierce lagoon]

Have you tried going on Khan Academy

[austere umbra]

Yea but

I can’t exact stuff I need like

[plain tulip]

There are a lot of resources online why would you wanna waste someone's time here

[austere umbra]

I can’t ask question online

Bruh

[plain tulip]

You can tho

[austere umbra]

If I don’t understand something then I can ask

[plain tulip]

That's what this channel is for, actually

[true granite]

yeah, but this channel is meant for specific questions

[fierce lagoon]

Yeah then you need to ask for a specific something

[true granite]

not an entire unit

[fierce lagoon]

Not a whole unit

[plain tulip]

IF you don't get a specific thing, ask it here

[austere umbra]

but I have too many specific questions

[fierce lagoon]

That's different than trying to get us to teach you a whole unit Bruhh

[austere umbra]

Like how exactly does sas asa sss etc work

[true granite]

ask them one at a time
we’re here to help, but we don’t work for you

[fierce lagoon]

SAS is when you have two congruent sides and a congruent angle between those sides

ASA is two congruent angles with a congruent side between them

SSS is self explanatory

HL is for right triangles only

[austere umbra]

Some my thought was

I know I need like examples

The one they gave us I don’t understand

Like

[fierce lagoon]

Send a picture of what they gave you

[austere umbra]

When they are connnected I don’t get it

Also based on how many questions I have I feel like Vc would be faster

Can we do Vc?

[austere umbra]

Wtb aas

Angle angle side?? I don’t see how that works

[fierce lagoon]

AAS is not the same as ASA

AAS doesn't prove anything

[austere umbra]

Yea I know

[fierce lagoon]

As far as I'm concerned

I mean

[austere umbra]

Wdym

[fierce lagoon]

No it does I was thinking about SSA

AAS is for congruency

It's when you have angle, angle, and then a side

That's because there's only certain lengths of two other sides that will form an opposite angle to the congruent side

[austere umbra]

Ok

Can we Vc ?

[fierce lagoon]

No

[austere umbra]

Y?

It would be faster

[fierce lagoon]

My mic broke

[austere umbra]

oh

[fierce lagoon]

Too much raging at video games

[austere umbra]

Wel ima do Vc u can text response

[fierce lagoon]

Just text here

[austere umbra]

Bruh

Ok

What is different ebwteeen like aas Asa

Look here

12 and 16

They are like the same thing

17 y is it not SSS cause it’s like 11

[fierce lagoon]

12, you have to use vertical angle properties

[austere umbra]

What are those

[fierce lagoon]

Angle XKF is congruent to angle LKJ

[austere umbra]

Oh yes those yea

[fierce lagoon]

You're gonna need to review your terminologies, I'm gonna be using them a lot

[austere umbra]

What were corresponding angles again

Wait I can Google

[fierce lagoon]

Oh it's a stupid png

[austere umbra]

I got it

I fucked my last tests prooof

We aren’t gonna have 27 step proofs on this test tho

It’s like under 10 steps but still

@fierce lagoon

U there??

[fierce lagoon]

I'm looking at it

The diagram is cut off

Show the whole page, but btw don't show your name

[austere umbra]

ok

[fierce lagoon]

Imma go in like 19 minutes btw

[fierce lagoon]

That's a lotta givens

So is the green marks the stuff you got wrong @austere umbra

[austere umbra]

it’s actually harder

Cause

There are different way of solving it

Like u can do different steps

[fierce lagoon]

Yeah but you have to fill in the blanks

[austere umbra]

But they gave u stuff so u have to use logic and find what they did

[fierce lagoon]

Obviously you have to build on each case

[austere umbra]

It’s harder trust me

Like my entire class ….

[fierce lagoon]

2 left column should be GHK is an isosceles triangle

[austere umbra]

Felt the same way

[fierce lagoon]

Otherwise you can't use 3R

3 right box

Imma shorthand which column

[austere umbra]

I mostly don’t get 23 to 26

[fierce lagoon]

So 2L is GHK is an isosceles triangle

For whatever reason

[austere umbra]

How do I get 22

[fierce lagoon]

Angle BAE is congruent to angle JKG

But like

You need to get those by looking at the previous cases

They're all sequential

[austere umbra]

They how would u simplify it

[fierce lagoon]

Well there's a shit ton of information

[austere umbra]

How would simplify this

[austere umbra]

.

Or substitute I mean

Not simlify

[fierce lagoon]

I mean that's just alternate interior angle stuff

[austere umbra]

Bro I know

[fierce lagoon]

Wdym substitute

[austere umbra]

How would u substitute it

The next step

What would u do for that

Oh they give us joy

Jkg

Nvm got it

[fierce lagoon]

ABE + BAE + AEB = 180

Something like that

They could really like

[austere umbra]

No

[fierce lagoon]

Condense these into bigger steps

[austere umbra]

Step 23

[fierce lagoon]

Yeah you can combine 22 and 23

As substitution into the postulate

[austere umbra]

Ur not suppose to Buh

I kno I can

[fierce lagoon]

You have a dumbass teacher then

[austere umbra]

The whole point is show our entire thinking

Also this is last unit

This unit

It’s something different

Like

I need help with how u use

Asa aas sas sss in proofs

[fierce lagoon]

It's based on the pairs of congruent parts that you have as well as their sequential positions

[fierce lagoon]

This whole problem's scuffed af lol

6R is substitution because of isosceles triangle properties. You shouldn't wrong though still

2 is that GHK is an isosceles triangles because of isosceles triangle properties

11 is JFM = JFD + DFM

[austere umbra]

I know

I have answer key

[fierce lagoon]

Absolute ass question

[austere umbra]

lol

[fierce lagoon]

This is critical though

For SSS, AAS, etc

Especially the sequential positions

[austere umbra]

Ok

[austere umbra]

We don’t need hl he said

[fierce lagoon]

Well I mean the diagram is over killing HL anyways lol

Everything else is fine

But note their sequential positions of congruent parts

Where the congruent angles or sides are

SAS really means as "side then angle then side", not "side and angle and side"

[austere umbra]

ok

I need

Like how u use that it in proofs

[true granite]

I mean, HL is just a specific case of SAS, so that’s a good point

[austere umbra]

I need help with it in proofs

Like

What is cpctc

And how does that fit into proofs

Examples of our proofs btw

[fierce lagoon]

Congruent parts (of) congruent triangles (are) congruent

It should really be CPCPC for "congruent polygons" but triangles kinda govern the world so

[austere umbra]

So how do u use that in prof

Like for my reason

Is this right

What do I put in question marks ?

[fierce lagoon]

I mean let's say that triangle ABC is congruent to triangle DEF, then AB is congruent to DE because CPCTC

[austere umbra]

Substitute?

[austere umbra]

Can u show pic?

I confusion

[fierce lagoon]

Order of how I named the triangle

"I mean let's say that triangle ABCis congruent to triangle DEF then AB is congruent to DE because CPCTC

[austere umbra]

ok

Oh

[austere umbra]

do u like complete proof

[fierce lagoon]

My NFT

[austere umbra]

With that included

[fierce lagoon]

You might

[austere umbra]

Dang

[fierce lagoon]

Usually you do

Sometimes you don't

[austere umbra]

Oh

[fierce lagoon]

Like what if an isosceles triangle was formed from sides from congruent triangles

[austere umbra]

Huh

[fierce lagoon]

Then CPCTC can be used to prove that two sides of the isosceles triangle are congruent, thus that triangle is an isosceles triangle

Like if you knew that the pink triangles were congruent

And such that those sides are congruent (because of CPCTC) then boom, white triangle is isosceles

[austere umbra]

Wasn’t it corresponding parts of congruent triangles

[fierce lagoon]

Yeah

Are congruent

[fierce lagoon]

I mean up here, I'm giving you that those lines are congruent because of CPCTC

That's just me giving you a fact for example's sake

Point is, CPCTC doesn't always end a proof

[austere umbra]

ok

We have 1 example

And I wanna do that later when I figure out everything

As practice

I think I got it

I need this now

Weird

[fierce lagoon]

$$180(n-2), n\in\bN\geq 3$$

[warm shale]

Umbraleviathan

[fierce lagoon]

That's the formula for the interior angle sum of a n-gon @austere umbra

And then if it's a regular n-gon

Each angle is $180 - \frac{360}{n}$ (only for regular n-gon).

[warm shale]

Umbraleviathan

[austere umbra]

ok

I don’t under stand the proofs with it?

How do u do proofs with that

[fierce lagoon]

Ngl, I haven't done anything that requires me to use that

It's more like "haha here's a pentagon what's the total sum of the angles"

[austere umbra]

Ok wtb this

What is this

Sss Asa aas sas

[fierce lagoon]

What are you triangle to prove

What's the end goal

[austere umbra]

Jam congruent to ime

[fierce lagoon]

Prove that AE = ME?

Oh

JAM to IME

Haha jam

[austere umbra]

Yes

[fierce lagoon]

What's given

[austere umbra]

Lol

[fierce lagoon]

Oh I see

Okay so you know what the reflexive property is right

[austere umbra]

Yes like a=a

[fierce lagoon]

Yeah

[austere umbra]

What’s the point of that

[fierce lagoon]

Because you get to use SAS

[austere umbra]

Huh

I don’t get it

[fierce lagoon]

The sum of congruent parts will be equal. Aka:

If JE + ME = JM, and IE + EA = IA,
Because JE = IE and ME = EA,
JM = IA (one side) [SIDE]

MEA is an isosceles triangle (because ME = EA, so iso triangle properties). Therefore EMA = EAM [ANGLE (BETWEEN AM AND JM, MA AND IA)]

AM = AM because funny reflexive property (yay) [SIDE]

Forogt what the theorem is called

But if:
A+B = C
D + E = F
D = A
B = E
Then C = F

It's through substitution, really

[austere umbra]

Couldn’t u have done like 2 Colum

This is confusing

Je plus me equals jm is segment addition postulate

Right?

I didn’t understand a single thing

Like 0

Like I don’t know what to do

After given where do I start

For other examples how would I figure it out

@fierce lagoon

Can’t it be sss

Bro

[fierce lagoon]

Uff I'm trying to do stuff ndjshfirhejdhdhd

[austere umbra]

oh

[fierce lagoon]

Wdym other examples

Which ones

Choose like one more because I gotta sleep

[austere umbra]

I got this for now

In given in says

2 corresponding sides are congruent

I used cpctc

[fierce lagoon]

What question is this

The same one?

[austere umbra]

To say the 3rd is congruent too

Yes

[fierce lagoon]

Just turn this into a two column proof

[austere umbra]

All 3 corresponding sides are congruent

[fierce lagoon]

You can assume straight lines so

[austere umbra]

So sss

[fierce lagoon]

I mean yeah, that's also one way to do it

That example is very nice because you can do multiple ways

[austere umbra]

.

So I can sss?

[fierce lagoon]

You can use SSS

Yeah

[austere umbra]

o

[fierce lagoon]

You'd have to prove that JEA = IME due to SAS though

[austere umbra]

Ok pls wait leme try to work this one problem

[austere umbra]

Why?

I proved they are sss

So they are sss and sas??????

[fierce lagoon]

That's to use CPCTC

You wanted to show that JA = IM to use SSS

To prove JA = IM you have to use CPCTC

To use CPCTC you need to prove that JEA = IME first

Sequential ordering

[austere umbra]

I can only use cpctc after I say sss sas aas etc ?

[fierce lagoon]

You can only use CPCTC after those, or you're straight out given that two triangles are congruent

CPCTC will always entail a proof that those triangles are congruent

[austere umbra]

ok

[fierce lagoon]

Always

You cannot just say it randomly

Your teacher, and I, will get mad

Now I have to sleep so feel free to ping a helper

[austere umbra]

Noooo

Can we finish problem :/

Oof no way u sleep lol

That is abnormal for teen/ adult

Usually they are sleep deprived

[solemn osprey]

Why yes I am thank you for asking

[solemn osprey]

What do you need help with?

Ping me when you read

[austere umbra]

ok

Number

2

Like how would I do that

@solemn osprey

In picture

Like how would I do a. Similar problem

[solemn osprey]

,rotate

[warm shale]

Couldn't find an attached image in the last 10 messages.

Oct 30, 2022, 5:09 AM

[solemn osprey]

,rotate

[warm shale]

Couldn't find an attached image in the last 10 messages.

Oct 30, 2022, 5:09 AM

[solemn osprey]

Dang

Can u repost all the relevant photos

?

@austere umbra

[obtuse pebble]

@austere umbra Has your question been resolved?

[obtuse pebble]

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[spark field]

Hey I just wanted to ask

[spark field]

if this was correct

$\frac{\frac{1}{:2}\left(x-1\right)^{-\frac{1}{2}}\cdot \left(2x+4\right)-\left(x-1\right)^{\frac{1}{2}}\cdot 2}{\left(2x+4\right)^2:}$

[warm shale]

Hoesmad

[spark field]

i simplifed that to

$\frac{\left(2x+4\right)-2\sqrt{x-1}}{2\left(\sqrt{x-1}\right)\left(2x+4\right)^2}$

[warm shale]

Hoesmad

[spark field]

Is this correct?

[timid silo]

why are you ghost pinging helpers?

[spark field]

the one time they actually respond 💀

[spark field]

so u not gonna help?

[timid silo]

nope

🙂

[spark field]

i hope u stub ur toe

[timid silo]

that won't help you get better at math

[spark field]

this is not about math anymore

[tardy epoch]

don't be rude

[spark field]

joke

.close

[obtuse pebble]

Channel closed

Closed by @spark field

Use .reopen if this was a mistake.