help-13

[idle urchin]

I already know that bisecting it gets us a smaller 90 degree right triangle

[lyric jungle]

You don't need to do that

Have you learnt the cosine rule for triangles yet?

[idle urchin]

Not with isosceles triangles no

[lyric jungle]

It doesn't matter what kind of triangle it is, it'll always apply

[idle urchin]

Oh

[lyric jungle]

The one that says cos(theta) = (a^2 + b^2 - c^2)/(2*a*b)

[idle urchin]

I think I remember

[lyric jungle]

So what is the length of the side you're not given?

[idle urchin]

24?

[lyric jungle]

Yes.

Now just immediately use the cosine rule.

[idle urchin]

I don’t think have the right answer

[lyric jungle]

What do you mean?

[idle urchin]

I got a large number for it

[lyric jungle]

You've put in the numbers wrong then

I got 2/3

[idle urchin]

What did you do?

[lyric jungle]

(32^2 + 24^2 - 24^2)/(2 * 32 * 24)

[idle urchin]

Oh I put 24^2 as A

So does the A = the base in this formula?

[lyric jungle]

c is always the side opposite the angle

[idle urchin]

I got the decimal 0.6667

[lyric jungle]

Which is 2/3

[idle urchin]

Oh i must have wrote it wrong sorry

If you don’t mind could we do one more?

[lyric jungle]

Go ahead

[idle urchin]

Btw it’s not a test it’s the Sat study guide

[lyric jungle]

Do you know what $\perp$ and $|$ mean?

[wraith dagger]

castroploiin

[idle urchin]

Yeah perpendicular and parallel

[lyric jungle]

Okay, so let's just deal with the angles here.

What can you say about DCE and ACB?

[idle urchin]

They’re opposites from each other?

[lyric jungle]

About their sizes, I mean.

[idle urchin]

Oh they’re similar to each other?

[lyric jungle]

The triangles are similar yes, so what do that tell us about the angles?

[idle urchin]

There angles must be similar sizes?

[lyric jungle]

Almooost.

They have to be the same.

DCE = ACB

[lyric jungle]

Also because they're "vertically opposite."

[idle urchin]

Oh

[lyric jungle]

So, if we know that in triangle DCE, angle DCE = angle ACB, line CE ~ line AC, and line CD ~ line CB.

What can you say about the angles CED, CAB, and angles ABC, CDE.

[idle urchin]

They’re similar I think

[lyric jungle]

In similar triangles, the corresponding sides are off by a scale factor, while the corresponding angles are identical.

So those sets of angles have to be equal.

Knowing that, we get a pair of alternating angles

If those 2 angles are equal, the angles are "alternating angles" and the bottom and top line have to be parallel

[crimson sedge]

c

[lyric jungle]

Meaning $\overline{AB} \ | \ \overline{DE}$

[wraith dagger]

castroploiin

[lyric jungle]

So the answer is c.

[idle urchin]

Oh

I get it now

Thx

Thanks for the help

.close

[cedar kiln]

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[cedar kiln]

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[willow lotus]

hi can someone help me? i got it that if n = k then i get δ_n,k = 1

[willow lotus]

but for n ≠ k it should be 0 but im not able to solve this...

[dull anchor]

Das sieht schwer aus, studierst du mathe?

[willow lotus]

zum glück nicht

informatik

aber mit mathematik der mathematiker

[dull anchor]

haha studiere e technik aber auch mit mathe der mathematiker

bin grad aber erst im 2ten Semester

[willow lotus]

ich nur im ersten

[dull anchor]

WAS

[willow lotus]

sommersemester angefangen

[crimson sedge]

????

English?

[willow lotus]

summer year began

[dull anchor]

????

[dull anchor]

maybe im helping him in german is that a problem?

[crimson sedge]

Idk

[dull anchor]

🤓

[crimson sedge]

https://tenor.com/view/iq-low-dumb-stupidity-stupid-gif-24161342

[willow lotus]

me right now

[crimson sedge]

Just kidding.

[cedar kiln]

@willow lotus Has your question been resolved?

[cedar kiln]

@willow lotus Has your question been resolved?

[cedar kiln]

@willow lotus Has your question been resolved?

[celest ledge]

You rewrite that sum as

$\sum (-1)^{n-l} \binom{n}{l} \binom{l}{k}=(-1)^{n-k} \binom{n}{k} \sum (-1)^{l-k} \binom{n-k}{l-k}= (-1)^{n-k} \binom{n}{k} (1-1)^{n-k}=0$

[wraith dagger]

Cogwheels of the mind

[cedar kiln]

@willow lotus Has your question been resolved?

[cedar kiln]

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[willow lotus]

.reopen

[cedar kiln]

✅

[willow lotus]

how did you calculate in order to get $(-1)^{n-k} \binom{n}{k} (1-1)^{n-k}=0$

[wraith dagger]

ඞDedekindඞ

[cedar kiln]

@willow lotus Has your question been resolved?

[ancient grotto]

if you mean the first step they just used the properties of C(n,r) (i.e. C(n,r)= n!/(r!(n-r)!) to get the C(n,k) term outside
if you are confused about (-1)^(l-k) you can basically use the fact that (-1)^(n+2x)=(-1)^n for all x in integers and use that to prove (-1)^(k-l) = (-1)^(l-k)
if you are confused about the second step they used the bionomial theorem

[cedar kiln]

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[median ermine]

Solve the equation 2^x=5 exactly and with 3 decimals with the help of
A) 10-logarithm
B) Natural logarithm

[violet flume]

sup

[median ermine]

im awake and ready

[violet flume]

all you need is

[tranquil plank]

isn't this just change of base?

[violet flume]

$\log \qty( a^x ) = x \log (a)$

[tranquil plank]

oh lol

[wraith dagger]

jan Niku (join us for @pomo)

[tranquil plank]

I overcomplicated it, yeah that works

[violet flume]

or change of base ye

but this works i think

[median ermine]

hmm

[violet flume]

then i think uhh

okay so you do this

you get to $x \log 2 = \log 5$

[wraith dagger]

jan Niku (join us for @pomo)

[violet flume]

this itself kinda looks like change of base-y

but this is probably just uneccessary simplification

[median ermine]

hmm ok i get it so far

[violet flume]

so

whats your exact solution

[median ermine]

im not sure

[violet flume]

[median ermine]

wym

[violet flume]

👀

solve the equation

[median ermine]

x 0.301=0.698

[violet flume]

thats one way

[median ermine]

what's the other

[violet flume]

exactly

[median ermine]

[median ermine]

i did all the questions in my book and still coulden't answe your question yesterday

[violet flume]

what question

[median ermine]

idk u were asking smth weird

[violet flume]

im trying to remember

i was not lol

it was something like uhh

so we were doing this

assume theres some number a

such that $b^a = 2$

[wraith dagger]

jan Niku (join us for @pomo)

[median ermine]

yes

[violet flume]

then i wanted you to use log form

so you would write $\log _b 2 = a$

[wraith dagger]

jan Niku (join us for @pomo)

[violet flume]

then you would exponentiate each side, by b

$b^{\log _b (2)} = b^a$

[wraith dagger]

jan Niku (join us for @pomo)

[violet flume]

but we know from our assumption that b^a = 2

[median ermine]

wtf i don't even know this

[violet flume]

$b ^{\log _b (2) } = 2$

[wraith dagger]

jan Niku (join us for @pomo)

[median ermine]

how was i supposed to answer lol

[violet flume]

its just log properties

i was trying to show you it doesnt matter what a is

$b^{ \log _b (n) } = n$

always

[wraith dagger]

jan Niku (join us for @pomo)

[median ermine]

ahhh

now i get it

what was the other way to solve this equation?

x 0.301=0.698

[violet flume]

naively

[median ermine]

[violet flume]

$x = \frac{ \log 2 }{ \log 5 }$

or whatever it was

[wraith dagger]

jan Niku (join us for @pomo)

[violet flume]

this is the exact answer

[median ermine]

tf

[violet flume]

might have 2 and 5 backwards

it dont matter

[median ermine]

wtf

[violet flume]

well it matters in context of the problem

but this is what i mean by exact

[median ermine]

why are we dividing

[violet flume]

$5x=2$

whats x

[wraith dagger]

jan Niku (join us for @pomo)

[median ermine]

0.4

[violet flume]

how did you get that

[median ermine]

just multiply with 0.4

[violet flume]

[median ermine]

[violet flume]

review algebra

5x=2

isolate x

divide both sides by 5

x = 2/5

[median ermine]

ahhhhhh

so with natural log

it's the same?

[violet flume]

it is nothing special

[median ermine]

ln2/ln5=x

[violet flume]

it is a number

so you can divide

[median ermine]

ok can we try a harder one

[violet flume]

👀

[median ermine]

solve the equation
A)ln x = 4 ln 2

[violet flume]

👀

[median ermine]

[violet flume]

use what i just told you

[violet flume]

this

[violet flume]

and these

[median ermine]

ln x = ln2^4

x=4

[violet flume]

use another property

this one is famouse

[median ermine]

wym

[violet flume]

well this one is

can be reasoned through

rarely you should memoriz something but

$\log x = \log y \longrightarrow x = y$

[wraith dagger]

jan Niku (join us for @pomo)

[violet flume]

this is nothing special about logs but more the kind of function a log is

[median ermine]

what are u tryna say lol

[violet flume]

well

you have $\ln x = \ln \qty (2^4)$

[wraith dagger]

jan Niku (join us for @pomo)

[violet flume]

what could this mean

[median ermine]

x=4

[violet flume]

[median ermine]

x=y

lnx=lny

[median ermine]

if i don't memorise the rules how am i supposed to apply them in the test

[violet flume]

you can memorize them

but you do have to be able to use them

[median ermine]

uhh

ok can we do the next subject

y=a^x

[violet flume]

aw

i wanted to do y=x^x

[median ermine]

y=2^x

[violet flume]

but you still didnt get the original question right

you used the rule incorrectly

[median ermine]

how

[violet flume]

.

[median ermine]

y= 2^4

[violet flume]

okay

[median ermine]

2^x * ln2

[violet flume]

huh?

[median ermine]

this

[violet flume]

yi dont understand

[median ermine]

derive y=2^x

2^x * ln2

[violet flume]

derive?

[median ermine]

idk the word in english

use derivative on y=2^x

[violet flume]

oh

differentiate

,w derivative 2^x

[wraith dagger]

Results provided by WolframAlpha

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

Requested by jan Niku#2119

[violet flume]

[median ermine]

😳

,w derivative 4*5^x

,w derivative 1,2 * e^0.5x

[wraith dagger]

Results provided by WolframAlpha

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

Requested by Skondarツ#6942

[median ermine]

[median ermine]

this was wrong btw

it was apparently

2^x ln2

not log

[violet flume]

this is dumb but a lot of the time if you see a log in the wild it means natural log

unless its in the context of like

axis scales

then it means log 10

[median ermine]

.close

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[brisk stratus]

can the value of students that didnt use anything be different in the venn diagram?

[cedar kiln]

Please don't occupy multiple help channels.

[obsidian coral]

Don't open multiple channels

[brisk stratus]

@obsidian coral soz

[obsidian coral]

Close this channel

[brisk stratus]

i cant find the other one

.ccoese

.close

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[ember drum]

what formula is being used here

And in the second line from the top is that completing the square ?

[cedar kiln]

@ember drum Has your question been resolved?

[ember drum]

<@&286206848099549185>

Pls help

[cedar kiln]

@ember drum Has your question been resolved?

[ember drum]

.close

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[ember drum]

How did the 5/2x+1/2 come aboutb

[cedar kiln]

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[glossy lintel]

don’t know where to find any formula sheet for this type of equation

[cedar kiln]

@glossy lintel Has your question been resolved?

[cedar kiln]

@glossy lintel Has your question been resolved?

[crimson sedge]

C

[cedar kiln]

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[marble lance]

A bird flies for 1600km at a speed of 600mph. One mile=8.5km. How many minutes will it take to fly the 1600km?

I started out with time= distance divided by speed and got 2.6 mins but I’m not sure where to go after that. What am I missing?

[cedar kiln]

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[marble lance]

.close

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[static hound]

Find the length of PQ, leaving "e" in your answer? P=(0,-e) and Q=(e/2 , 0). I put this in my calculator √[(e/2)-(0)]^2 + [(0)-(-e)]^2

[upper abyss]

No need to calculator it, since you can leave e in the answer

Simplify is the direction to go

[static hound]

and I get PQ= 3.0391314475 but the answer booklet says 1/2√5e

[static hound]

when i did it without the calculator i got √(5e^2 /4)

they're different answers aren't they?

[cedar kiln]

@static hound Has your question been resolved?

[static hound]

<@&286206848099549185>

[cedar kiln]

@static hound Has your question been resolved?

[narrow pulsar]

maybe recheck if u wrote P and Q right and and recheck the answer in the booklet
cuz if the P and Q u have written are right then yeh
ur answer is right
i dont see how it can be 1/2√5e

[static hound]

just checked the answer in the booklet is still 1/2√5e and P & Q respectively. could just be a typo

[narrow pulsar]

yeh probably

instead of √5e/2 they wrote 1/2√5e

[static hound]

yeah thought so

[cedar kiln]

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[autumn merlin]

can somoeone explain this to me?

like how the second line came to be

[tropic oxide]

they just swapped cos(cos(x)) and -sin(x)

if you want to get formal about it this is an example of the commutative law of multiplication

and nothing else

[ember python]

I think they mean going from the first line to the second

[ember python]

Chain rule

[autumn merlin]

ye but can you explain further

i understand they use chain rule but it seems weird

[tropic oxide]

why

the derivative of sin is cos, and that of cos is -sin

[ember python]

$(f(g(x))’ = f’(g(x))\cdot g’(x)$

[tropic oxide]

in fact the use of the chain rule couldn't be more transparent here than anything else - this function is literally written as a composition

[wraith dagger]

dk.dkn

[ember python]

It’s easy enough to prove if it makes you uncomfortable

You can even make a substitution for g(x), or clearly define both functions each time if it makes it easier

[autumn merlin]

whered the negative come from?

[ember python]

Derivative of cosine

[urban kestrel]

the derivative of cosx is -simx

[autumn merlin]

ahh

i see

[cedar kiln]

@autumn merlin Has your question been resolved?

[cedar kiln]

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[bright sorrel]

Using variations (discrete math): "Find how many teams had taken part in a basketball tournament if there were 21 matches with the system everyone against everyone"

[cedar kiln]

@bright sorrel Has your question been resolved?

[bright sorrel]

<@&286206848099549185>

[elfin hemlock]

what is "Using variations"? There is a formula for amount of matches in everyone vs everyone given amount of teams

so its just equation solving

[bright sorrel]

yeah but how do I get a model out of it?

Vx/21=2x?

I have no idea

[elfin hemlock]

what?

[bright sorrel]

i'm definitely wrong

a mathematical model

when you have a text problem => making a mathematical model

whether it would be an equation or w/e

[elfin hemlock]

using this formula if you know it

[bright sorrel]

Yeah Vk/n

I have n=21

but what's k?

[elfin hemlock]

what?

[bright sorrel]

that's the formula

idk how to write it otherwise

[elfin hemlock]

formula is n(n-1)/2

[bright sorrel]

with just keyboard

[elfin hemlock]

no idea what you are saying

[bright sorrel]

oh so I just replace?

21(21-1)/2?

[elfin hemlock]

if you actually knew the formula, yes

but no idea what it was you said

and no don't plug in n=21

n is amount of teams

[bright sorrel]

ight bet

.close

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[weak halo]

"The inequality $\frac{x^3-8}{x-1} \geq 0$has the same solutions as

[wraith dagger]

Help

[weak halo]

and then one of the options

Why is the answer a and not b?

[calm sierra]

ty

@sand ether pls dont delete modpings

[weak halo]

Since $\frac{x^3-8}{x-1}\cdot (x-1)^2 \geq 0 \iff (x^3-8)(x-1)$

[wraith dagger]

Help

[sand ether]

soz this is someones help channel and just wanted to clear it for him/her

[weak halo]

Appreciate it buddy!

[calm sierra]

i get it but it can confuse mods coming later

[sand ether]

ok noted

[elfin hemlock]

x=1 works in (b) but not in original

[weak halo]

ahh because division with 0 would occur

these questions are hard

@elfin hemlock Thank you mate

[crimson sedge]

keep getting the first answer, but not sure if im 100% correct here

[upper garnet]

get your own help channel

[cedar kiln]

@weak halo Has your question been resolved?

[dire geode]

not really a valid excuse, but the name "help" sure can confuse people

[cedar kiln]

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[topaz socket]

Hi! Does anyone know how to do box whisker plots?

[crystal raptor]

Does this picture help?

You just need to find those 5 things from your data

[tight herald]

Yes!

Do you know how to calculate Q1 and Q3?

[topaz socket]

I know how to, but not off the top of my head

[dire geode]

flashcards are great

[tight herald]

So is that a yes or no? lol

[topaz socket]

no sorry I’m not that smart

[tight herald]

You don't have to be smart, its just about learning

[topaz socket]

I know but this math is confusing

[tight herald]

So I assume you know how to calculate the median?

Because q1 and q3 use the same method :)
Q1 is just the median of the data in the set under the median
Q3 is the median of the data above the median

@dire geode Did I say something wrong

[dire geode]

not at all. i just never thought of it that way

[tight herald]

What????

[dire geode]

i just calculate percentiles like a pleb

[tight herald]

Lollll

I always have done it that way

I forgor how do you calculate for a lower or upper outlier

https://tenor.com/view/forgor-gif-22367783

[crystal raptor]

Yeah thats a nice way to think about it I'd never thought about lol

[tight herald]

You're welcome I guess lol

[cedar kiln]

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[cunning lance]

Just a quick question, to find x would that be to do x/17=20/13 or x/20=17/13.
Finding y is kinda hard too, how do I do that?

[cunning lance]

The triangles are "congruent?" Or smth

[crimson sedge]

This may help, these are the rules for Congruent Triangles. To my understanding, x should be directly proportional to (x+17), I believe the same scale will be used throughout.

[cunning lance]

Thanks man, what does SSS and ASA mean tho?

[crimson sedge]

SSS = Side Side Side

ASA = Angle Side Angle

In SSS the sides are directly proportional

[cunning lance]

Oh, I see. What about R?

[crimson sedge]

Right Hand Side I believe

You might wanna check me on that.

[cunning lance]

Alright, thanks. I'll come back to you

By the way, why is it x+17?

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[fallen vine]

How can I do #22 b)?

[fallen vine]

I get f'

but for f'' I'm not sure how I would solve something like that?

[cedar kiln]

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[fallen vine]

<@&286206848099549185>

[crystal raptor]

f'' tells you whether f is "concave up" or "concave down"

[fallen vine]

right

ohhhh

I see

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[crystal raptor]

Do you know how the graph of a function is related to the graph of its inverse?

[autumn cradle]

somewhat

[crystal raptor]

What do you think it is

[autumn cradle]

I think the answer is (4,4)

<@&286206848099549185>

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[hearty basalt]

s

[cursive cedar]

Rewrite the equations so that it says y equals to something. The left hand side is just y

[hearty basalt]

so like y = 1/2x - 5?

ok thanks!

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[final wyvern]

I need to state which of the two angles has a greater measure. I dont know how or where to start from this equation. can anyone please guide me on how do I answer this?

[cosmic steppe]

They should have the same angle measure

If ED is a straight line

And A is a bisector

That means RY is a transversal and thus mRAE and mDAY are equal

Well actually

Lemme see

Maybe they're slightly askew

[opal matrix]

I dont think they are equal

[cosmic steppe]

Yeah they're not

That means RY or ED aren't straight lines

Anyways

[cosmic steppe]

Which two segments are congruent?

[final wyvern]

RA and YA according from what I've calculated. the problem is if I dont know if its correct or not

and EA and AD are also congruent from what I've calculated aswell

RA is 6 and YA is 2x. 6/2 = 3 and 3 will be my "x". I use that x for my calculations for AD

I dont know if im going the right direction though.

[cosmic steppe]

Well to find the angle, just use your trigomentric ratios

[final wyvern]

oh ok

[cosmic steppe]

What would mRAE be

In terms of known sides

I'm going to assume taht there are 2 right triangles

Well, even if they weren't, you could use law of cosines

Are they two right triangles?

Or is that not given information

[final wyvern]

they are not given information

[cosmic steppe]

Alright then you have to use law of cosines

[final wyvern]

the question just told us to compare the 2 triangles and say who is greater from those 2

[cosmic steppe]

Let's look at triangle RAE

Well actually

Solve for x first

[final wyvern]

its 3 from what I've gathered.

[cosmic steppe]

Yeah

So now let's look at triangle RAE

You have sides 6, 5, and 3

To make things easier, I'm gonna assign 6, 5, and 3 to dummy variables, A, B, and C, respectively

Using law of cosines,

[final wyvern]

alright

[cosmic steppe]

$C^2 = A^2 + B^2 - 2AB\cos{\left(mRAE\right)}$

[wraith dagger]

Umbraleviathan

[cosmic steppe]

Plug in the appropriate values for A, B and C, and then solve for mRAE

I gtg but

You basically do this for both triangles

[final wyvern]

alright. thank you sir

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[livid shadow]

this is what i did

im not sure if this is right

<@&286206848099549185>

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[proud crane]

hi im kinda stuck with these 2 questions

[cedar kiln]

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[celest ledge]

I am considering let $a_{n},b_{n},c_{n},d_{n}$ be number of binary strings satisfying the condition, whose first two terms are 00,10,01,11 respectively. Then $\begin{pmatrix}a_{n}\b_{n}\c_{n}\d_{n}\end{pmatrix}=A \begin{pmatrix}a_{n-1}\b_{n-1}\c_{n-1}\d_{n-1}\end{pmatrix}$ where $A=\begin{pmatrix}1&0&1&0\1&0&1&0\0&1&0&1\0&1&0&0\end{pmatrix}=CDC^{-1}$ is diagonalizable. So you can solve $f_{n}=\begin{pmatrix}1&1&1&1\end{pmatrix} \begin{pmatrix}a_{n}\b_{n}\c_{n}\d_{n}\end{pmatrix}=\begin{pmatrix}1&1&1&1\end{pmatrix} CD^{n-3}C^{-1} \begin{pmatrix}a_{3}\b_{3}\c_{3}\d_{3}\end{pmatrix}$

[wraith dagger]

Cogwheels of the mind

[celest ledge]

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[deep patrol]

please help thank you

[crystal raptor]

What have you tried?

[deep patrol]

wait imma send my solution-

[crystal raptor]

Yep, very nicely done and laid out!

[deep patrol]

oh-

WAIT IT'S CORRECT?-

[crystal raptor]

It is yeah

[deep patrol]

i-

is there a way for me to like check if my answer is correct? do I just substitute it to the formula?

[crystal raptor]

Yep so plug your answer back into your first equation below pythagoras theorem

[deep patrol]

ohhhh okay okay thank you!

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[modest sand]

why is this wrong ?

i used the L'Hospital rule

after i did e^(ln16x)*(ln(7)+1)/ln(12x)+1

[upper abyss]

Good start. Then you want the limit of the exponent, and L'h can do that

[slim valley]

@modest sand move the 16x up, remember ln7+1 is a constant.

take the derivative of the top and bottom respectively and you should get that the bottom goes to zero, and when you take the limit you should get number/0 which is infinity

in the case of a limit, its infinity

remember that a linear function (of the same order) grows faster than a logarythmic one

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[modest sand]

thx i got it

[slim valley]

good job

[modest sand]

the answer was e^ln(7)+1

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[slim valley]

you sure its not infinity?

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[crimson sedge]

I cant figure out how he got H(X|Y=1) = 0 bits

When I use the formula I get values higher than 0

is my approach

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[crimson sedge]

Hi, i’ll send a picture and then my question.

[crimson sedge]

“m^2 = (2k^2) = 4k^2 = 6n^2” —> “n^2 is even, so n is even...”

I don’t get why you can make that jump to saying n^2 is even. Can’t n^2 be uneven as well, since 6 = 2k so 6n^2 is 2kn^2 so it doesnt matter if n is uneven or not?

Its about proving square root of 6 cant be rational, by the way

[celest ledge]

4k^2=6n^2 the power of 2 of the left hand side is at least 2, so n has to be even. Otherwise n is odd, the power of 2 of the right hand side is 1, contradiction

[crimson sedge]

‘The power of 2 ... is at least 2’ sorry, what exactly do you mean by ‘power of 2’? What exactly is at least 2?

[celest ledge]

I mean in the prime factorization of the left hand side, the index of 2 is greater than or equal to 2

[crimson sedge]

could someone explain that to me? Spell it out? I dont get the technical terms

[celest ledge]

? I didn’t say any term

[crimson sedge]

Prime factorisation.

[celest ledge]

Just 4 divides the left hand side right?

So 4 divides the right hand side

If n is odd, 4 doesn’t divide the right hand side, the right hand side would only be divisible by 2, contradiction

[crimson sedge]

Ahh, yes!

So 4 divides the left hand side, so it must also divide the right hand side. But it doesnt divide 6 so it must divide n^2, which is only possible if n is even.

[celest ledge]

No

4 | 6n^2

Therefore 2 | 3n^2

2 doesn’t divide 3, so 2|n

n is even

[crimson sedge]

Oh

Yeah it seems to make more sense now

Yeah

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[crimson sedge]

Thank you btw

[celest ledge]

Np

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[sweet ingot]

Hey

[sweet ingot]

The function is known to intersect the axis of y at y = 2.
Find a

[livid hound]

what have you tried

[sweet ingot]

I tried 2 = a/2x^2+5

[wanton glacier]

if it intersects the y axis, what is the x value?

[sweet ingot]

0

[reef venture]

now substitute that in this

[wanton glacier]

you didn't change your x here

[sweet ingot]

I did it before and I got that A = 7

I guess I did + instead of *

Alright, thanks!

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[wanton glacier]

did you add 5 instead of multiplying it?

[sweet ingot]

Yes, by a mistake

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[dense frigate]

how do I do this?

this is asking to integrate x^-1 correct?

nvm

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[sweet ingot]

Hey

[sweet ingot]

It is known that the functions intersect at the point where x = 0.8
A ) Find K
B ) Are the functions cut at another point besides the given point?
If so-find it.

So I have found K which is 0.48

How do I solve B?

I tried to equal between them and instead of K i wrote 0.48 and I don't get a good answer

[crystal raptor]

show us your working

[sweet ingot]

second

[sweet ingot]

This is what I did ^

The second picture is A

The first is B

[crystal raptor]

how did you get to that very last line?

in B

[sweet ingot]

Multiplication by crossover

[crystal raptor]

from $x^2-0.96 = x^4$ to $-0.96 = x^2$?

[wraith dagger]

iCaird

[sweet ingot]

ohg

i did x^4/x^2

[crystal raptor]

do you see what you did wrong?

[sweet ingot]

not really

[crystal raptor]

okay well you shouldnt be dividing by x^2 anyway, but if you were to divide both sides by x^2, you have to divide everything by x^2

you did not divide -0.96 by x^2

[sweet ingot]

so what the equation should look like

-0.96/x^2 = x^2?

and even then i still cannot really solve this

[crystal raptor]

it would look like $1 - \frac{0.96}{x^2} = x^2$

[wraith dagger]

iCaird

[crystal raptor]

but anyway what you're supposed to do is set $y = x^2$ and then you have a quadratic you can solve

[wraith dagger]

iCaird

[sweet ingot]

wait where did you add that 1 from

[crystal raptor]

$\frac{x^2}{x^2} = 1$

[wraith dagger]

iCaird

[sweet ingot]

oh

so like

i had to divide here everything by x^2

[sweet ingot]

and on which equation

f(x) or g(x)

[crystal raptor]

yeah you would have to divide everything

but we dont want to do that anyway, it doesnt make anything simlper

[sweet ingot]

so when you said y = x^2

how do we find out it's x^2 we want to replace y

[crystal raptor]

We solve for y

what will the equation look like if we set $y=x^2$?

[wraith dagger]

iCaird

[sweet ingot]

it we're talking about f(x) then x^2 = x/√x+0.48

[crystal raptor]

not f(x) no

the equation you just sent

we're trying to solve that to find intersection points

[sweet ingot]

this?

[crystal raptor]

yes

[sweet ingot]

i'm not sure then how it would look like

x^2-0.96=x^2

?

[crystal raptor]

not quite

we can replace x^2 with y right?

[sweet ingot]

ye but where's the y here

[crystal raptor]

its just what we're calling x^2

[sweet ingot]

oh oka

[crystal raptor]

itll make sense

in terms of y, what can we replace x^4 with?

[sweet ingot]

x^2

[crystal raptor]

in terms of y

if $x^2 = y$ then $x^4 = ???$

[wraith dagger]

iCaird

[sweet ingot]

y^2

[crystal raptor]

yep!

so whats our new equation in terms of y?

[sweet ingot]

x^2 = x - 0.48?

or the x^2-0.96 stays the way it is

[crystal raptor]

take $x^2-0.96 = x^4$

[wraith dagger]

iCaird

[crystal raptor]

and write it in terms of y

by how we just discussed

we dont want any x in there, just y

[sweet ingot]

y^2 = y - 0.96

[crystal raptor]

yes!

now this is a quadratic you know and love and can solve

so solve for y using the formula or however you want to

[sweet ingot]

how i had to think about that myself lol

what are the clues

like

[crystal raptor]

hm?

[sweet ingot]

if i have x^4

?

[sweet ingot]

i meant how can i know i need to understand that i need to divide it just for the exercise

what hints are there

[crystal raptor]

oh yeah basically if you only have even powers of x, you can let y = x^2 and get a polynomial that is half the degree you started with

if we had $x^4 + 5x^3 + 9x^2 + 3 = 0$

[wraith dagger]

iCaird

[crystal raptor]

we could not have done this trick

because of the x^3

[sweet ingot]

i see

by the way i got 2 points

-1.6 and 0.6

are they both valid?

or not because of the §

√

[crystal raptor]

,w solve y^2 = y - 0.96

[wraith dagger]

Results provided by WolframAlpha

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Requested by iCaird#6258

[crystal raptor]

whoops

what did i do

[sweet ingot]

√-1.6+0.48 gives a minus

so we can disqualify -1.6 right?

[crystal raptor]

,w (0.48)^2

[wraith dagger]

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Requested by iCaird#6258

[sweet ingot]

it was supposed to be like y^2 - y + 0.96

i think you can edit it

[crystal raptor]

your value for 0.48^2 is wrong

,w solve y^2 = y - 0.2304

[wraith dagger]

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Requested by iCaird#6258

[sweet ingot]

i check the answers actually, it's really k = 0.48

[crystal raptor]

yes but $(0.48)^2 \neq 0.96$

[wraith dagger]

iCaird

[sweet ingot]

oh

omg

lol

[crystal raptor]

ya timsed it by two haha

[sweet ingot]

idk why i wrote 0.96

so

i get 2 points 0.64 and 0.36

[crystal raptor]

keep them as fractions for now

remember what we just found, we found y

but we are looking for x

ages ago we set $y = x^2$

[wraith dagger]

iCaird

[crystal raptor]

so we have $x^2 = \frac{9}{25}$ or $x^2 = \frac{16}{25}$

[wraith dagger]

iCaird

[crystal raptor]

you know what to do

[sweet ingot]

for 16/25 I get 0.4096 and for 9/25 i get 0.1296

[crystal raptor]

what do you mean?

[sweet ingot]

oh i thought i should multiply

[crystal raptor]

multiply what?

we are trying to find x, look above at what we have, what should we do?

[sweet ingot]

ye, square root

[crystal raptor]

yep

what does that give us for x?

[sweet ingot]

so for √9/25 I get 0.6 and for √16/25 I get 0.8

[crystal raptor]

are you forgetting something when you square root?

[sweet ingot]

and we already got the point 0.8 so that leaves us with 0.6

brackets?

haha

[crystal raptor]

what is $(-0.6)^2$?

[wraith dagger]

iCaird

[sweet ingot]

9/25

[crystal raptor]

yep, so what is root(9/25)?

well

[sweet ingot]

-0.6

[crystal raptor]

what are the soltuions so x^2 = 9/25

[sweet ingot]

oh

+-0.6

[crystal raptor]

yep!

so finally we have $x = \pm 0.6$ or $x = \pm 0.8$

[sweet ingot]

so there are 2 more points

[wraith dagger]

iCaird

[crystal raptor]

like you said 0.8 was already given

[sweet ingot]

3*

[crystal raptor]

one last thing to check

we need to check whether when we plug these points into our functions, we actually get a value

because otherwise the lines cant intersect where atleast one of them isnt defined

it is $g(x)$ what may be undefined for some of these points, see if you can see why and for which points

[wraith dagger]

iCaird

[crystal raptor]

yeah try see which ones cause problems

[sweet ingot]

the ones with the minus'

[crystal raptor]

why?

[sweet ingot]

because a square roott cannot be negative

root*

[crystal raptor]

bingo

well done

[sweet ingot]

so that leaves me only with 0.6 now

[crystal raptor]

yep!

and just to show you all that work was correct

there they are in all their glory

[sweet ingot]

so the y that i got is 0.57

it makes sense with that illustration

[crystal raptor]

yep seems right

spot on

[sweet ingot]

thanks for not giving up on me caird

[crystal raptor]

thanks for giving up on you?

[sweet ingot]

not***

[crystal raptor]

oh hahaha

no worries! you got there in the end, be proud:)

[sweet ingot]

♥

[crystal raptor]

always learning

[sweet ingot]

ye

i try to study it as best as possible

[crystal raptor]

and thats all we can do, try our best

[sweet ingot]

ye, would you mind if i will pm you in case i face any problems

[crystal raptor]

sure, won't be around 24/7 but can get roudn to things when i have time

[sweet ingot]

oh no rush, just whenever you can

[crystal raptor]

cool, see ya around!

[sweet ingot]

i added you, thank you very much ♥

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[crimson sedge]

Hi

[crimson sedge]

Is this the help page?

I am so confused

[rustic siren]

read the bots message

[crimson sedge]

How on earth do I calculate this?

.close

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[jovial bough]

Q9

So i managed part a

Just stuck on part b

I am not sure what the final velocity must be for it to be a bearing of 45°

[cedar kiln]

@jovial bough Has your question been resolved?

[loud anchor]

,rccw

[jovial bough]

Guys?

<@&286206848099549185>

[cedar kiln]

@jovial bough Has your question been resolved?

[jovial bough]

u should be 6i-10j btw

[cedar kiln]

@jovial bough Has your question been resolved?

[cedar kiln]

@jovial bough Has your question been resolved?

[tidal nebula]

This seems like physics (kinematics)

The equation to be used is $a = \frac{v-u}{t}$

[wraith dagger]

24ayn5

[cedar kiln]

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[terse yoke]

https://cdn.discordapp.com/attachments/810625768685436968/968668998776553482/20220426_202359.jpg

[terse yoke]

The basic formula is the cubic root of x equals the function. For number 10, what is the new radical function

sorry for the vagueness, ping if anyone has questions

[cosmic steppe]

Cube root function @terse yoke ?

[terse yoke]

yes

[cosmic steppe]

Wdym radical functional though

The cube root itself?

Does that say $7 = Cl - \sqrt[3]{2}$?

[wraith dagger]

Umbraleviathan

[cosmic steppe]

What is Cl @leaden summit

[terse yoke]

cl is meant to be a, and there's no subtraction there, sorry for my writing

[cosmic steppe]

Oh

Alright

Yeah I was gonna say there should only be an amplifier

First of all, make sure you plugged your values correctly

Should be $-2 = a\sqrt[3]{7}$

[wraith dagger]

Umbraleviathan

[cosmic steppe]

And then a would just be $\frac{-2}{\sqrt[3]{7}}$

[wraith dagger]

Umbraleviathan

[cosmic steppe]

@terse yoke there

[terse yoke]

thanks a lot

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[sick river]

Hello everyone

[sick river]

Determine the extreme values ​​of the function f(x,y,z)=yx^2 + z in the region given by x^2 + y^2 + z^2 <= 1

Can someone help me pls ?

Idk how I will do this question

[cedar kiln]

@sick river Has your question been resolved?

[cedar kiln]

@sick river Has your question been resolved?

[cedar kiln]

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[ashen flower]

hello

[ashen flower]

<@&286206848099549185>

[celest ledge]

Ask a question or leave

[ashen flower]

A yearly package for adults who want to be physically fit is available in a fitness center. The package has a registration fee of ₱600 and a payment of ₱120 per access to the gym for workouts.

As one of the finance officers of the fitness center,
you need to present the graph that shows the current possible expenses of a member of your fitness center.
compare the current package to the proposed package below by answering the questions.

How much is the initial expense in each package without accessing the gym yet?
How much is the payment per access to the gym in each package?
How much will a member cost in a year if they access the gym twice a month in each package?
How many times can a member access the gym in a year if their budget for the fitness center is ₱3000 in each package?

[ashen flower]

A yearly package for adults who want to be physically fit is available in a fitness center. The package has a registration fee of ₱600 and a payment of ₱120 per access to the gym for workouts.

As one of the finance officers of the fitness center,
you need to present the graph that shows the current possible expenses of a member of your fitness center.
compare the current package to the proposed package below by answering the questions.

How much is the initial expense in each package without accessing the gym yet?
How much is the payment per access to the gym in each package?
How much will a member cost in a year if they access the gym twice a month in each package?
How many times can a member access the gym in a year if their budget for the fitness center is ₱3000 in each package?

anyone help

[celest ledge]

f(x)=600+120x
So
600
600+120x
600+(120)(24)=3480
600+120x<=3000,120x<=2400,x<=20. So 20

[ashen flower]

what question answers that?

[celest ledge]

Clearly f(-x,-y,-z)=-f(x,y,z) so f(x,y,z) is maximal iff f(-x,-y,-z) is minimal. We only need to find maximal value M, minimal value will be -M. So let’s find the maximal value, clearly x,y,z>=0 this case. For any f(rx,ry,rz)=(xy^2)r^3+zr is increasing in terms of r for any fixed x,y,z on the sphere x^2+y^2+z^2=1 so maximal when r=1, so we let x=sin(a),y=cos(a)cos(b),z=cos(a)sin(b) . Then yx^2+z=cos(a)[sin^2(a)cos(b)+sin(b)]<=cos(a)sqrt(1+sin^4(a))=sqrt(cos^6(a)-2cos^4(a)+2cos^2(a)) is monotonically increasing in terms of cos(a) so cos(a)=1, maximal value is 1

[ashen flower]

thanks for the help man

.close

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[upper vortex]

hey, i have a question about EDO by Power series, The equation is the following, and I expanded in Power series and I got the following expression.

I did take the a0 term, that was a0 = 1/6, but after that I got stuck and I could not find the recurrence formula no matter what

[cedar kiln]

@upper vortex Has your question been resolved?

[celest ledge]

General solutions of homogeneous part are {(e^(-2t))(ucos(sqrt(2)t)+vsin(sqrt(2)t): u,v from R} of course
For a particular solution y, you let y=w+z where w”+4w’+6w=1 and z”+4z’+6z=e^-t so w=1/6, z=(1/3)e^(-t) so the solutions are
{(e^(-2t))(ucos(sqrt(2)t)+vsin(sqrt(2)t)+(1/6)+(1/3)e^(-t) : u,v from R}

EDO is French version of ODE in English?

[upper vortex]

portuguese version

the problem is that I need to solve by power series

[celest ledge]

Then compare coefficients both side. Since Taylor expansion of e^-t is known

I see the dilemma here

That recurrence relation is hard to solve

Or you only need to find the recurrence relation not solve it? That case it is
$a_{n+2}(n+2)(n+1)+4a_{n+1}(n+1)+6a_{n}=\frac{(-1)^{n}}{n!}$ for $n \geq 2$

[wraith dagger]

Cogwheels of the mind

[upper vortex]

to solve it :/

[celest ledge]

…

Hard to understand why the professor gave this question insist on solving it using series… when it’s so easy to solve just normally 😂😂

[upper vortex]

indeed my friend, he likes to make the students suffer

[celest ledge]

Well, at least I can’t… that recurrence relation is not linear

Well, I just found out that

We can cheat:

Solving a_n since we can solve this ODE😂

a_n is linear combinations of b_n and c_n where

[upper vortex]

actually, my professor said that we need to solve this by variation of parameters, power series and laplace, all of thoose 3.

[celest ledge]

$b_{n}=\frac{(-2+\sqrt{2}i)^{n}}{n!}+\frac{(-1)^{n}}{3n!}$
$c_{n}=\frac{(-2-\sqrt{2}i)^{n}}{n!}+\frac{(-1)^{n}}{3n!}$

[wraith dagger]

Cogwheels of the mind

[celest ledge]

For n>=1

I mean all solutions {a_n} is {(x+yi)b_n +(x-yi)c_n: x,y from R}

[upper vortex]

alright, so the problem is that for us to find the recurrence we must have used the complex number right?

[celest ledge]

This is just cheating, solve a_n from the solutions of that ODE. But at least it’s solved. You can check b_n ,c_n form a basis of the linear space of sequences satisfying that recurrence relation

[upper vortex]

how, by the normal way of solving the ODE did you get to this?

cause I'll only use it to solve

[celest ledge]

Yeah, this is solutions of this ODE, I just took their Taylor expansion

I mean you don’t have to show it on paper

You secretly solved it normally

[upper vortex]

lmao

[celest ledge]

And used it to solve the recurrence relation on paper

[upper vortex]

yeah

[celest ledge]

Then act like you used this recurrence relation to solve ODE

A meaningless loop😂😂

[upper vortex]

yeah broh, my professor is a crazy man

[celest ledge]

Just to please this weirdo professor

[upper vortex]

maybe someday he will learn to have love in his heart

[celest ledge]

😂

[upper vortex]

thank broh, U helped a lot

[celest ledge]

Np

[upper vortex]

I'll use this recurrence

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