help-28

[sharp vine]

so solutions would be 0, 90, 180, 270, 360, ...

according to the interval we have 3 solutions (be careful because it's < not <=)

[muted kayak]

how did you get 4theta = 360n?

im a bit confused where n comes from and why 360

[sharp vine]

ok so

[muted kayak]

Sorry, youve been really helpful so far. thanks a lot

[sharp vine]

you know if we do full rotation (or it's multiplicity) then result is same

right?

[muted kayak]

full rotation?

[sharp vine]

you can imagine unit circle

or in other words

if we add period's multiple then value won't change

360 degrees because period of cosine is 360 degrees

[muted kayak]

hmm alr

[sharp vine]

this is a periodic function that repeats every 360

[muted kayak]

sin also repeats 360 and tan periods at 180 correct?

[sharp vine]

exactly

and we use 360n where 'n' represents an integer because we want to have each solution, it's easier way than writing them down

each 'n' value corresponds to some solution

[muted kayak]

would you do something similar for -180?

instead of 0 at the start

[sharp vine]

n is an integer so it can produce negative solutions as well

[muted kayak]

so -180 <= x <= 180

[sharp vine]

but yes

yea, same process

[muted kayak]

hmm ok. ill make sure to study this more

thanks for all the help

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[versed ivy]

The sides of a square are doubled and the area is increased by 1200cm^2. What was the original length of the square. I know the answer is 20 using “(2x)^2 - x^2 = 1200, but I do not know where that equation comes from

[bold grotto]

Ok so initially you have a square

Let the side of the square be x

And the area will be x^2

Now

You double the sides of the square

So the sides become 2x

And the area becomes (2x)^2

And it is given in the question that the second square is 1200cm^2 bigger than the initial square

So area of second square - area of first square = 1200

[versed ivy]

Oooooh

Of course

That’s where the -x^2 comes from

Thanks a lot 🙂

[bold grotto]

Np

[versed ivy]

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[magic socket]

for the first question, I've factored by group. where would I go from there?

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[robust slate]

$=(x^2-9)(x-3)$ by factoring out $x-3$

[glossy valve]

enclave wangedrocht

[robust slate]

Do difference of squares to complete factorization

Roots are when a factor = 0, multiplicity is the power that factor is raised to

y intercept is when x=0

End behavior, just consider first term's coefficient and degree

Graph bring it all together

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[woeful oracle]

Can someone help me with this question please

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@woeful oracle Has your question been resolved?

[versed basalt]

@woeful oracle try drawing the graph described yourself

then compare your graph to Albert's

[woeful oracle]

Alr I’ll try thanks

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[terse creek]

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[stone galleon]

Not sure how to approach from here

calculus II partial fractions btw

[eager violet]

When you have a factor that's to a power, you don't just do A over that factor

For example, if one of the factors was (x - 1)³, you'd have to do A/(x - 1)³ + B/(x - 1)² + C/(x - 1)

[stone galleon]

so would i do A/(x^2) + B/(x) + C/(x^2+25)

[eager violet]

Be careful about the x² + 25 factor

You did it right in the problem, but not in that comment

[stone galleon]

oh so then how would i make b and c equal a value

i have A but im not sure on what to use for x for it to make b and c equal a value

[eager violet]

Wdym

Show what equation you have rn

[eager violet]

We already went over why thats incorrect

You're supposed to have A/x² + B/x + ...

[stone galleon]

wait but didnt u say i did right in the problem but not right in the comment (im assuming the comment is the reply i made)

[eager violet]

I was referring specifically to the x² + 25 term

It's not C/(x² + 25) like in your comment

It's (Cx + D)/(x² + 25)

[stone galleon]

oh i c mb

is the condition for letterx + letter that if its irreducible

[eager violet]

If it's an irreducible quadratic, yes

If it's cubic, then you'd do Ax² + Bx + C

And so on for higher powers

[stone galleon]

So from here not sure what to use for x

[eager violet]

1 isn't equal to that

The original fraction, 1/(x²(x² + 25)) is

[stone galleon]

ok got it

but it would be equal to one at the second line right?

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[torn jolt]

does anyone know how to make quadratic equation to standard for like

x²+6x+9=0

[hot herald]

in this context, by standard form do you mean vertex form?

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[buoyant pewter]

Quick question. Is 58% in one std dev and 100% within 2 std deviations close enough to normal to call it approximately or not

[buoyant pewter]

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[spiral badge]

If I have 2 5/6 x 1 2/9 how do i solve with a common denominator of 3

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[uncut tulip]

what do i do here

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[light jolt]

If contribution rate goes is X and sales price increases doesn’t CR go down?

[light jolt]

Formula is CR= CM/S

[tidal citrus]

please....try harder to explain what you are talking about.

at least define your variables because it's been a couple of years for some of us

[deft zodiac]

lol

we are old sry

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[light jolt]

Sec

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[formal tundra]

I managed to calculate the perimeter doing 72 s * 2,27 m/s = 164 m

But how do I calculate the sides?

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@formal tundra Has your question been resolved?

[formal tundra]

<@&286206848099549185>

can someone help me out?

<@&286206848099549185>

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@formal tundra Has your question been resolved?

[formal tundra]

<@&286206848099549185>

[formal tundra]

<@&286206848099549185>

[deft zodiac]

it isnt rly specified but im gonna guess its a circle

like myself

ok nvm ignore that

just find the surface area using 2 i guess

also dont ping helpers more than once

[formal tundra]

It’s a rectangle

[deft zodiac]

yea find the surface area using 2

[formal tundra]

After I found the surface area, how do I find the sides?

Because the problem asks for the size

[deft zodiac]

let l be one sides length

form an equ with area n perimeter with l

[formal tundra]

Sorry, I don’t get it

What should the equation look like?

[formal tundra]

?

Goddessssss 😭

[deft zodiac]

im bac

so

uve one length is l

what is the other length

how do u find that using the perimeter

[formal tundra]

That’s where I’m stuck

[formal tundra]

Is that an L?

[deft zodiac]

ye

so

uve perimeter is 164

let the sides be w and l

then u know that

2l+2w =164

right?

[formal tundra]

Yes

[deft zodiac]

so

write w in terms of l

[formal tundra]

w = (164 -2l)/2

Like this?

[deft zodiac]

yepp

fast

now simplify it a bit

[formal tundra]

w = 82- l

[deft zodiac]

yes

then what would the area be in terms of l

[formal tundra]

mhm

Area = l * w

[deft zodiac]

mhm and can u remove the w and write it in terms of just l

[formal tundra]

Area = l * (82-l)

[deft zodiac]

nice!

now

have u calculated the area using 2?

[formal tundra]

I haven't cause I'm stuck

Should I use the volume or what?

[deft zodiac]

yes

so normally volume is base * height right

[formal tundra]

4 000 000 liters are 4000 m^3

[deft zodiac]

and whats the base now?

[deft zodiac]

yepp

[formal tundra]

isn't it?

Now i have the volume

4 000 m^3

[deft zodiac]

yep

but u want to find the

base/surface area

using that and the height

so u use

area* height = vol

[formal tundra]

2,5 m is the depth

[deft zodiac]

yes

[formal tundra]

But I don't think the depth is useful?

[deft zodiac]

depth is

essentialy the same as height

the depth of the swimming pool is like

its height

for calculating volume anyway

[formal tundra]

area = volume/height

[deft zodiac]

mhm and that here is

[formal tundra]

area = 1600 m^2

right?

[deft zodiac]

yep

so now

[deft zodiac]

area = 1600 m^2

[deft zodiac]

Area = l * (82-l)

how would u find the sizes

of the pool

[formal tundra]

1600 = 82l - l^2

do you mean this?

[deft zodiac]

yess

[formal tundra]

it is a second grade equation 😓

[deft zodiac]

yes

use quad formula

[formal tundra]

but then i will get 2 solutions

and i need only 1?

Can you check the result

[deft zodiac]

ok

,w solve 1600=82l-l^2

[glossy valve]

Results provided by WolframAlpha

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

Requested by Circle#0025

[deft zodiac]

there

so

the 2 solns are just

the length

and

82-l

[formal tundra]

what do I do with 2 l?

Because I needed only one l

[formal tundra]

?

Where have you gone now 😭

[deft zodiac]

its

the same

try using 32

then try 50

like

one length is l

the other, w =82-l

if l=32

w=82-32=50

if l=50

w=82-50=32

so no diff

[formal tundra]

So I can invert them

[deft zodiac]

yes

[formal tundra]

what else now?

[deft zodiac]

thats all

[formal tundra]

I can say the size of the pool is 50x32x2,5?

[deft zodiac]

ye

use m

[formal tundra]

100% sure?

ok

[formal tundra]

thank you, goddess, i suppose you're female

[deft zodiac]

❤️

[formal tundra]

💙

my savior you are

[deft zodiac]

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[deft moss]

Hi, just want to clarify e^(i x theta)* is the conjugate of e^(i x theta) which means that it will give us the conjugate complex number of said complex number?

[next sail]

what's *

[deft moss]

i'm not sure my professor noted it as e^(i x theta)*

i'm assuming it is referring to a conjugate complex number

[gritty rose]

It's just wherever you see an i, you replace with -i

[deft moss]

b/c z* is a conjugate right

[gritty rose]

Yes

[deft moss]

okay got it so e^(i x theta) = cos (theta) + i sin (theta)

and e^(i x theta)* = cos (theta) - i sin (theta)

the conjugate

<@&286206848099549185>

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[daring tapir]

The equation is (7^2)^3*7/7^4, and the answer I ended up with was 7^2, but the answer book says its 7^3, so I must've made a mistake, can someone tell me where

[rapid rain]

could you show how you got there ?

[daring tapir]

Sure, one sec

[rapid rain]

ping me when you showed it

[daring tapir]

Got it

@rapid rain

bottom equation

[rapid rain]

7^6 x 7 = ...

[daring tapir]

I just assumed 7^6 * 7^1

oooooh

stupid

Sorry, should've spent more time checking before posting it here

Thanks for the help!

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[teal epoch]

I need help understanding the difference between the geometric random variable and the binomial random variable

[teal epoch]

currently i see the geometric is the exact probability of an event happening once in n trials

and binomial is the same thing but happening k times in n trials

however when i plug k=1 into the binomial theorem i do not get the geometric equation

and now im confused on how they are different

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[torn jolt]

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[normal dome]

English

A person rolls a die and continues until he first gets a result he has gotten before.
Let X denote the number of rolls and find the distribution of X

[normal dome]

can someone help me with this, i have no idea how to set this up

[clear lily]

sounds like a negative binomial

but a die is not a bernoulli so im not sure

[normal dome]

hmmm gotta be an easier way i have not learned about those things yet

[clear lily]

you have to find the distribution right?

[normal dome]

yeah

[twilit leaf]

what is the range of the distribution?

[rapid rain]

X can take values from 2 to ?.
P(X=2) = 1/6 (get the same roll twice).
P(X=3) = 5/6*1/3 (not get the same first two rolls, then get one of the same)
etc...

[twilit leaf]

can the number of rolls be 8 or more?

[normal dome]

yeah why not

[clear lily]

not with a die

youre guaranteed a duplicate if you roll a 6 sided die 8 times

[twilit leaf]

youre guaranteed at least 2 duplicates

so the max is 7

where there is guaranteed only one duplicate

[normal dome]

how are you guaranteed

ah wait yeah

lol

but do i just have to set this up as a table

[clear lily]

so the range is [2, 7]

you can do it by hand but that doesnt seem very educational

[normal dome]

yeah thats what i thought

like is there a way to set up like a function for this

[clear lily]

probably right

could you send the exact question?

like as written in the book

[normal dome]

what i sent is the whole question

A person rolls a die and continues until he first gets a result he has gotten before.
Let X denote the number of rolls and find the distribution of X

[clear lily]

oh alright

well what is the chance that you get a duplicate on roll x

thats the same as rolling x-1 uniques and then one youve had before

and the chance of rolling x-1 uniques is what

[normal dome]

wait let me process this lol

[clear lily]

its probably a good idea to consider 4 rolls of the dice first and then generalise

[normal dome]

wont this be correct?

[clear lily]

wait i gotta process this hahah

[normal dome]

wait here i assumed that x=1 is the second throw

here is with x=1 being the first throw

[clear lily]

alright the other didnt sound quite right

[normal dome]

this one?'

[clear lily]

this one didnt sound quite right to me

[clear lily]

well suppose x = 1 then you get 6!/(6*5!) = 1

thats not right

[normal dome]

but its the chance for uniques

so if i want chance for duplicate i just have to write 1-what i wrote

[clear lily]

oh i thought you meant the chance of getting a duplicate on roll x

smart

[clear lily]

i think you almost have it

[normal dome]

ah it doesnt give 0 after 7 throws

[clear lily]

i would say

[twilit leaf]

its not supposed to

[clear lily]

$P(x) = 1 - \frac{6!}{(6-x)!6^x}$

[glossy valve]

Gijs

[twilit leaf]

theres a rare chance you do get to the 7th throw

[normal dome]

how can i not get a duplicate after 7 throws

[clear lily]

also in probability theory its customary to write

[normal dome]

ah true

[clear lily]

that is the probability that random variable X takes on value k is ...

$P(X = k) = 1 - \frac{6!}{(6-k)!6^k}$

[glossy valve]

Gijs

[clear lily]

forgot to replace one x

this should be it

[normal dome]

yeah nice thanks a lot

really appriciate the help

[clear lily]

np

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[boreal solstice]

can someone help me with this

i got up to tan = 2

lore?

i think that helped

[hot herald]

you're missing the angle

tan**(t)** = 2

[boreal solstice]

i got up to that but what do i do after?

[hot herald]

then use inverse tan to get your angle t

[boreal solstice]

how do i get it to degrees?

[hot herald]

set you calc to degrees

and/or if your calc is in radians, convert it manually yourself

[boreal solstice]

oh gotcha so I just type tan^-1(2)

wait so I get that

but what angle are they even talking about

like the one across from the right angle?

[hot herald]

the angle between the part of the line above the x-axis and the positive x-axis

[boreal solstice]

ohhh I see tyty

also one more questoin

do you know how to do a.) of this

[hot herald]

!15m

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[boreal solstice]

oops myb

!close

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[dark niche]

<@&286206848099549185>

[deft zodiac]

!15min

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[deft zodiac]

what a way to start

[dark niche]

Hi are you able to hop into a vc?

i can screen share

[deft zodiac]

no

[dark niche]

so i am confused with grade 11 radicles

[deft zodiac]

[dark niche]

square root of 128

[light sonnet]

Google

[dark niche]

i tried

im just not getting it

[light sonnet]

"How to simplify radicals"

That simple

[dark niche]

yeah

[atomic blade]

Man goes in here, breaks a rule, and doesn't acknowledge the warning

[light sonnet]

Hundreds of videos exist

[torn jolt]

chill its just discord

[dark niche]

i just joined the server

chill

[light sonnet]

Then read #❓how-to-get-help

[atomic blade]

Heard this one before

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@dark niche Has your question been resolved?

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[twilit grove]

Let x_n be bounded a sequence of real numbers. Define s_n = sup{xP_k : k ≥ n} and
t_n = inf{x_k : k ≥ n}. (Recall that lim sup x_n = lim s_n and lim inf x_n = lim t_n.). I am given this

[twilit grove]

Prove that sn is decreasing and that tn is increasing, that is, prove that for all n ∈ N,
sn+1 ≤ sn, and for all n ∈ N, tn ≤ tn+1. I have to proof this.

Any help would be appreciated!

[native flicker]

you can show your progress if any

[twilit grove]

None yet. But I was thinking I have to show what s_n converges to and that is the lim of t_n

[native flicker]

let us assume that |xn| < M for all n E N

[native flicker]

so this is given
and now we observe that (sn)nEN is a sequence with decreasing Sn >= Sn+1

[twilit grove]

hmm...

[native flicker]

similarly for given t_n = inf{x_k : k ≥ n}
using the same argument we show taht (tn)nEN is an increasing sequence

hence we have (s_n) and t_n) as a monotone and bounded sequence

hence by the theorem of sequence that every bounded and monotone sequence is convergent, we can say that (s_n) and (t_n) is convergent

note that (s_n) is a decreasing sequence with |s_n| >= -M for all n

hence by theorem of sequence if (s_n) is convergent to s then :

[twilit grove]

Can you draw the graph containing both the sequences? Seeing the picutre would help i guess

sorry, I am little slow.

[native flicker]

here is your proof

[twilit grove]

can i ask you follow up questions ?

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[worldly comet]

The indicator on the oil tank in my home indicated that the tank was one-eighth full. After a truck delivered 240 gallons of oil, the indicator showed that the tank was half full. What is the capacity of the oil tank, in gallons?

[trim sandal]

What have you tried so far?

[worldly comet]

I fall asleep in class alot but i've tried doing 1/8 + 240 = 4/8 but i realized that wouldnt add out to a right answer i just dont really understand where to start

[trim sandal]

you're forgetting to represent the unknown quantity (the capacity of the tank) using a variable

[worldly comet]

would 1/8 be the where i put the variable?

[trim sandal]

anywhere the capacity of the tank is talked about

so X/8 + 240 = X/2, solve for X

[worldly comet]

would it be 240 divided by 3x/4 = 80 so x is 80?

[trim sandal]

no, and that would not be a sensible answer as that would mean the tank is smaller than the 240 gallon delivery

Are you unfamiliar with algebra entirely?

[worldly comet]

Kinda just started alg 2 after not doing alg 1 for 2ish years

[trim sandal]

In order to "solve for X" you must apply something to both sides of the equation a few times until you end up with X = insert answer here

take X/8 + 240 and X/2 and start by subtracting X/8 from both expressions

[worldly comet]

yea so would it be 240 = 3/8? because you x x/2 by 4 to get 4x/8 so you can do x/2 - 1/8?

[trim sandal]

not 240 = 3/8 but 240 = (3/8) * X

but the fraction is correct

so you have on one side a number, and on the other, the variable X multiplied by another number

an equation of the form a * X = b

if you divide both sides by a you will have solved for X at this point

[worldly comet]

what would be a?

[trim sandal]

3/8 in this case. Remember dividing by a fraction is the same as multiplying by its inverse

[worldly comet]

and what would be b/x?

wait would b be 240?

[trim sandal]

yes

[worldly comet]

would is be 240 divided by 3/8x to get 90?

actually wait nvm

that wouldnt make sense

because 90 x 3 = 270

[trim sandal]

there shouldnt be an x in that expression

240 / (3/8) like that

since dividing by a fraction is the same as multiplying by its inverse, 240 * (8/3) is equivalent, and may be easier for you to calculate

[worldly comet]

So 640?

= x

[trim sandal]

yes

the whole process looks like this

represent the problem as an equation, where a variable (x) is the unknown you are trying to find the value of:
x/8 + 240 = x/2

subtract x/8 from both sides:
240 = x/2 - x/8

simplify the right hand side:
240 = x * (3/8)

divide both sides by (3/8):
240 / (3/8) = x

simplify the left hand side:
240 * (8/3) = x

finish calculating the left hand side:
640 = x

[worldly comet]

Ok thank you so much

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[astral rapids]

Is there infinite number of equivalent angles (negative and positive) in all cases of directed angle? Or only when it's on the standard position? (Ik it's a dumb question but idk why I don't remember this at all)

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[astral rapids]

<@&286206848099549185>

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@astral rapids Has your question been resolved?

[night folio]

There are an infinite number of equivalent coterminal angles if that's what you want to know

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[snow depot]

https://cdn.discordapp.com/attachments/387447715312828416/1022556962614935693/image0.jpg

[snow depot]

What’s this asking exactly

[deft zodiac]

like

is it of the form

y=f(x)

where f is some function of solely x

[snow depot]

I assume without rearranging though right?

[deft zodiac]

yea

[snow depot]

So false

[deft zodiac]

ye

[topaz valley]

are you sure about that

[deft zodiac]

of course

[snow depot]

Yes

[topaz valley]

mmmmmmmm

sus

[deft zodiac]

way to kill the vibes dud

[snow depot]

🤔

[topaz valley]

well

unfortunately

cute little goddess has it wrong

[snow depot]

Oh no

[deft zodiac]

WOW

[snow depot]

Why does cute little goddess have it wrong

[topaz valley]

the question asks whether or not y is defined implicitly as a function of x

in other words

is there a unique y value for each value of x

[deft zodiac]

why does cute little goddess always get it wrong :c

[topaz valley]

because a function must have a unique output for each input value

[snow depot]

Mmm gotcha

I suppose that would make it true?

[topaz valley]

that would make it true

[deft zodiac]

wait

wouldnt x=+-1 map to the same y

[topaz valley]

y

[deft zodiac]

[topaz valley]

as a function of

x

y(x)

[deft zodiac]

ic

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[thin gale]

Hi there! I just did this question on a quiz, my answers being:
A) 312
B) 122.522
C)435

and it says I got them all wrong

is there something wrong with the quiz?

[scenic wren]

for the rectangular prism, notice that they only care about the visible areas

so the top area is discarded

[thin gale]

oh no

oh noo

what about the cylinder?

[scenic wren]

for B, I got the same

[thin gale]

well ok, I am sorry I am just really scared

my parents are going to be very upset

thanks you though

ok so what did you get for A?

[scenic wren]

we have two squares, so 2*6^2

then we have 3 rectangles, so 3 * 6 * 10

add them, we get

252

[thin gale]

I got 252

thanks

alright

I may have some more questions

can you help me solve this

A)40,326.89
B)1145.111

C)41,472

and this is the last one I need help with
A)4.4162
B)53

thanks again

<@&286206848099549185>

damn

[thin gale]

I think I figured this one out
A) 7.416
B) 89
correct me if I'm wrong

<@&286206848099549185> sorry for the double ping, I believe I am allowed to do this?

Not trying to rush, but i am still in need of some help haha

@scenic wren Hey sorry to bother you, I just have 1 more question

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[grim zealot]

hi could you please help solve this question?

[grim zealot]

i think it's parametric vector but i dont understand how to do that

[deft zodiac]

so try to reduce this to rref first

the above matrix

[grim zealot]

i just dont understand how since theres no numbers..

like its x1 instead of an actual number?

[deft zodiac]

u can just reduce it as per normal

just that with variables

u can think of it as a

[grim zealot]

ohh okay

[deft zodiac]

fill in the gaps with zeros

but even if its other variables u can just do rref as per normal

its like an augmented matrix

[grim zealot]

aa right, thank you

[deft zodiac]

np

❤️

[grim zealot]

after it's been reduced what would u do?

the formatting of +s[], is something ive never seen before

[deft zodiac]

how does the matrix look like

[deft zodiac]

so

[deft zodiac]

first we can let x4 be some dummy variable

in the Q its s

and uve to represent the other vars in terms of it

[grim zealot]

uve?

[deft zodiac]

u have

[grim zealot]

oh lol mb

what do u mean by represent the other variables

how would we represent them

[deft zodiac]

so like

[deft zodiac]

from here

observe each row

so

from first row weve

so we know that

x1 + x4 = -3

and x4=t

so its like

x1 + t = -3

or

x1 = -t -3

then u repeat that for all variables

[deft zodiac]

then throw it into this vector afterwards

[deft zodiac]

oh and i used t instead of s, just.. replace approriately

[grim zealot]

oh gotcha

thank youu

[deft zodiac]

np

❤️

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[mighty cypress]

How can we prove a triangle has no more than one right angle, without using the fact that all right angles are congruent? (Also the angle measure axiom is unavailable)

[next sail]

what's the angle measure axiom

[mighty cypress]

There is no means of measuring an angle or assigning a value to its measure

E.g. 90 degrees doesn't exist as a concept

And a right angle is defined simply as one which is congruent to its supplement

[next sail]

?

ok

i mean can we use the classic proof thingy

[mighty cypress]

What is the classic proof thingy

[next sail]

ok so this is the classic proof that the angles of a triangle sum to 180

because if you draw a line through the vertex of a triangle parallel to the opposite side

then you end up with these alternate angles that all sum to 180

[mighty cypress]

180 what

[next sail]

degrees

[mighty cypress]

Degrees don't exist

[next sail]

yeah i know

but we can adapt that here

clearly the sum of B and C is equal to the supplement of A

but if A is a right angle, the sum of B and C is also a right angle

[mighty cypress]

Can't sum things without values 😦

[next sail]

what

oh right

hmm

[mighty cypress]

Still making assumptions and using angle measure axiom

Yeah this was way harder than I was expecting too lol

[next sail]

ok how did euclid do it then

i bet his axioms were basic enough

well hmm

wait

still trying to adapt my proof

[mighty cypress]

As far as axioms we have 4: segment measure axiom, SSS, SAS, and angle noncongruence axiom

[next sail]

idk what any of that means

[mighty cypress]

Using these we've proven things about parallel lines and such

[next sail]

ok back to the diagram

so suppose B were a right angle as well

then A and B would be supplementary

since they're both right

so that would have to be a straight line, without including C

but it's clearly not

that's the best i've got lol

[mighty cypress]

Hold on 2 seconds then I'll read and reason through that

How do you know the second B is a right angle

Without being given that all right angles are congruent

Yeah I think you've got it ty

Because if one angle is a right angle and congruent to a second angle, we can prove the other is also a right angle

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[dense edge]

I can't seem to be able to write cube roots on WeBWork

Would this syntax also make sense, the way I have written it with ^1/3 power?

Makes sense the way I have written this answer? Or way and completely off? ..lol

[calm sphinx]

parentheses bru

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@dense edge Has your question been resolved?

[dense edge]

yeah I don't know how to get brackets working..

it removes them in WeBWork

Would this be fine?

Looks like it

.close

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[lime jackal]

Hi! This was a pre-calc question on the homework that we haven't learned yet,, can someone get me started?

[lime jackal]

how would i approach this problem

i know calculus and i still dont know how to do this sob

[next sail]

let x = the whole fraction

then x = 7 + 1/x

[lime jackal]

what do u mean the whole fraction?

[tame snow]

Set the expression = to x

[eager violet]

x = 7 + 1/(7 + 1/(7 + 1/(...)))

[tame snow]

Or any variable

[lime jackal]

ok and then i set x = 7 + 1/x?

what would i do next

?

[tame snow]

I can give you an example of a problem like that

[lime jackal]

that would be helpful :D

[tame snow]

x = 1/2 + 1/4 + 1/8 … (do you already know how to solve this)?

[lime jackal]

yes i think, it's a geometric sequence right

[deft zodiac]

oh these kinda Qs

[lime jackal]

u can see it as a + ar + ar^2 + ar^3 ... and so on and use the formula right?

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[gray dagger]

Have I done this right?

[gray dagger]

y=3x+2 btw

[toxic elbow]

seems good

[gray dagger]

ok ty

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[cyan wedge]

I got c = -6 which gives infinite solutions for the system of equations. Which means, the last row is all zeros and there's a free variable. I can't figure out part b though. I only know how the row of zeros applies to the column vectors (which is they don't span r3 since c must be -6 for there to be a solution)

[cyan wedge]

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[cyan wedge]

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[coarse seal]

I am wondering if I did the negation right :

[glossy valve]

Fluffy
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[brave basin]

p implies q is logically equivalent to ~p or q, so ~(~p or q) is equivalent to p and ~q

Talking about the last part, where you have written ~P(x)

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[modest sail]

So i started with y= 1+3x/5-2x

And then switched

X = 1+3y/5-2y

And here idk what to do

I can’t isolate the y now

It’s an inverse function question

There’s two y’s on one side so idk how to get rid of one

If u understand what i mean

[atomic blade]

Multiply both sides by (5-2y) first, and then distribute the x

@modest sail what do you get after that

[modest sail]

I got 5x-2xy = 1+3y

Still there’s two ‘y’s

[atomic blade]

Yeah

[atomic blade]

Subtract 1 from both sides, then add 2xy to both sides

Ayo @modest sail im gonna need you to not disappear every 2 minutes

[modest sail]

Oh sorry

Let me do that rn

Poh

Ooh

Its weird but worked

Oh interesting

[atomic blade]

You'll notice something a bit funny

Hold on lemme verify the formula though

[modest sail]

Ok

[atomic blade]

Here

[modest sail]

Wait my teacher got the same answer but with signs a bit different

[atomic blade]

Let $f(x) = \frac{a+bx}{c+dx}$, then $f^{-1}(x) = \frac{cx-a}{b-dx}$

So far, I only proved this works for a, b, c, d being nonzero real numbers

[glossy valve]

Umbraleviathan

[atomic blade]

Wdym

[modest sail]

Right one is my teachers

But it shouldn’t matter

[atomic blade]

They factored out -1 from both sides

[modest sail]

Right?

[atomic blade]

-1/-1 = 1

[modest sail]

Yeah

I noticed that

[atomic blade]

So it doesn't matter

[atomic blade]

Knowing this saves time though

[modest sail]

Oh boy

This is gonna bring me my ap history class vibe

The only reason i love math is i don’t “have to” memorize

Formulas

Except some mandatory ones

Such as the quadratic formula

But idk if that one’s necessary

A+bx/c+dx -> cx-a/b-dx

Ill try tho

Idk if it’s necessary

To memorize

But definitely thanks a lot for your help

😃

[modest sail]

Dx didn’t change before and after huh

Thats what i noticed

Other than that they all changed

X changed from second to the first

I mean the first x

Bx to cx

Interesting

Idk how ill get that in my head lol

At least i kinda brainstormed here

a first place to second place

Aight thank you @atomic blade !

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[torn jolt]

I defined function in mathematica 13.1 as f[x_] := x^2 and Solve[f[x]==36,x] but th output is {{}} instead of {{x -> -6}, {x -> 6}}, what did I do wrong??

[full forum]

@torn jolt Has your question been resolved?

[prisma sundial]

@torn jolt I checked my mathematica and it worked. Double check what you actually typed in.

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@torn jolt Has your question been resolved?

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[limber tiger]

x/2-2=4+1/4 iv been stuck not knowing where to start

[torn jolt]

bring all numbers to RHS

Right Hand Side**

[limber tiger]

elaborate im a bit confused

how do i do that

[stiff oak]

<@&286206848099549185>

[slate hemlock]

?

[stiff oak]

.

[slate hemlock]

You shouldn't ping for other people, and 15 minutes haven't passed anyways

[slate hemlock]

$\frac x2 -2=4+\frac 14$

[glossy valve]

Andrea276

[stiff oak]

Sorry

[slate hemlock]

This is what you have, right?

[slate hemlock]

Just to make sure

[limber tiger]

yes

[slate hemlock]

Ok, so you want to isolate the terms with x on one side and the terms without on the other

[slate hemlock]

Which are the terms with x here?

[limber tiger]

x/2

[slate hemlock]

Ok. All the others are without x, so you should try to move them to the right side, as peace said before

In this case you only need to move -2 since the others are already on the right side

[limber tiger]

so it become +2? right ?

[slate hemlock]

Yep

[limber tiger]

i see

[slate hemlock]

You now have $$\frac x2 =4+\frac 14 +2$$

[glossy valve]

Andrea276

[slate hemlock]

Do you know how to go on from there?

[limber tiger]

no not really my fundementals are pretty weak which is why im having trouble

[slate hemlock]

Now you just need to sum the terms on the right side

[limber tiger]

so i get 6 1/4 right

[slate hemlock]

Btw you might find this video helpful, you should watch it after we finish here to make sure you understood the procedure well: https://youtu.be/7DPWeBszNSM

[slate hemlock]

Yes

[limber tiger]

will do

[slate hemlock]

6 1/4 is 25/4, right?

[limber tiger]

yes

[slate hemlock]

I'd rather work with 25/4 than 6 1/4.
Anyways, now you have: $$\frac x2= \frac{25}{4}$$

[glossy valve]

Andrea276

[limber tiger]

no wait mb its 24

[slate hemlock]

Now you need to get rid of the 2 at the denominator

[slate hemlock]

No, it's 25/4
6 + 1/4 = 24/4 + 1/4 = 25/4

[slate hemlock]

Can you guess how to do that?

[limber tiger]

ohhh shi u right u right

[slate hemlock]

Don't worry

[slate hemlock]

Any ideas for that?

[limber tiger]

add it

[slate hemlock]

It's a fraction though, you can't get rid of it by adding

[limber tiger]

subtracting it ?

[slate hemlock]

That's basically the same as an addition, you can't really use it there

[limber tiger]

im lost tbh

[slate hemlock]

Since a fraction is a division, you should use a multiplication instead

[limber tiger]

ohhhhh

[slate hemlock]

$\frac x2 \times 2=x$, right?

[glossy valve]

Andrea276

[slate hemlock]

So you can multiply both sides by the denominator of the fraction, 2 in this case, to get rid of it

[limber tiger]

where did the 25/4 go tho

[slate hemlock]

I'm just showing you the denominator goes away, that's not the equation we were working with

[limber tiger]

oh wait nvm were gettinng rif of it mb carry on

[slate hemlock]

You would have to do:
$$\frac x2 \times 2= \frac{25}{4} \times 2$$

[glossy valve]

Andrea276

[slate hemlock]

Then simplify (if there's anything to simplify)

[slate hemlock]

That way the 2 at the denominator of x/2 will disappear, giving you x=some_value, which is your solution

[limber tiger]

i got 25/2 (12.5)on the right side

[slate hemlock]

Yes, that's correct

x=25/2, or 12.5

You just solved your equation 👍

[limber tiger]

tysm i appreciate
the help a lot

[slate hemlock]

Again, you should watch this, you'll also practice solving equations by doing it since there are many examples and you'll get used to it

[slate hemlock]

You're welcome

[limber tiger]

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[royal vapor]

hello i've been stuck on this one for a bit

[calm nacelle]

Hello! What have you tried so far

[royal vapor]

multiplying and adding

[calm nacelle]

Have you learned anything about finding the rule for a table

[royal vapor]

no because im sick at home so i wasn't able to go to school

[calm nacelle]

yea it involves multiplying and adding

okay first do you notice any patterns in the numbers

like how do you get from the first y value to the second one

[royal vapor]

adding tens on the y

[calm nacelle]

yea

so we know that when x=1, y=39; when x=2, y=39+10; when x=3, y=39+10+10

do you see why

[royal vapor]

uhh

it like

adds another ten as the x goes up

[calm nacelle]

yup so if you have x=4 then how many times do you have to add 10

[royal vapor]

3

[calm nacelle]

yup

how did you get that

[royal vapor]

oh ok i think i get it now

y = 10x + 29

[calm nacelle]

yep exactly :))

[royal vapor]

tyyy

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[calm nacelle]

np!

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[royal vapor]

sorry i kinda need help again because it suddenly gave me higher numbers lol

[royal vapor]

like i know the pattern for x and y is adding 30 but idk how to solve it from there

[buoyant tinsel]

It's a linear equation, find slope using any two points (rise over run) and use point slope

doesn't really matter what "pattern" x or y is since this isn't a sequence

[royal vapor]

oh ok

[torn jolt]

y=x+39

[royal vapor]

tyty

[torn jolt]

was that what you had?

[royal vapor]

not at all 😭

[torn jolt]

how did you do it?

[royal vapor]

i just did a bunch of random things

[torn jolt]

let me explain it to you.

[royal vapor]

using the y-y1 and x-x1 thing

[torn jolt]

first of all, a linear equation would satisfied all the points in the question since if you draw a linear equation, you get a line in a graph. All the values on the table are possible solutions.

y= mx+b

m= gradient or slope.

y= is the y coordinate or solution.

x= x coordinate or solution.

b is the y intercept, a point where the line crosses they y axis and x=0

now, the questions gives us the values, so we plugin them into the formula y=mx+b

y=-5, x=-44
-5=-44m+b
Now, there are two methods, you can either find a system of equation or find the gradient.

[royal vapor]

ok i get it now

it worked for the rest ty so much for explaining :)

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[lone pond]

5. Find the vertices of the region that contains the solutions to the system of linear inequalities .I have alredy placed my 3 inequalties ohn the grahp but im not sure how to get the vertices.

[neat bay]

The solutions are all the points which satisfy all three inequalities

In other words, they must lie inside every color’s shaded region

Wait wth

Are those the right inequalities? Your solution region is unbounded

[lone pond]

give me a sec im going to regraph them

yea i forgot 2

x>=0 andy>=0

so as a "vertex" i would do 0,0 , 0,5 and so on

<@&286206848099549185>

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