#help-39 | Mathematics | Page 223

elder narwhal Mar 20, 2025, 7:50 AM

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so i think my classmate was right!! they just did it an extended way

toxic lichen Mar 20, 2025, 7:51 AM

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you have your k and you have your probability table filled out

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the values of N that correspond to "less than advertised" are 48 and 49; you find the probabilities for each one, and add them

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you're honestly kind of overthinking it in here

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this is correct and the only thing that needs to be done

elder narwhal Mar 20, 2025, 7:53 AM

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toxic lichen you're honestly kind of overthinking it in here

haha yeah lol its a bit of a habit

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thank you very much though :D

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OH!!!! i have one more question if youd be willing to help

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this one is more logistic because i already solved it but i somehow got it wrong

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THIS ONEE has been tripping me up

elder narwhal Mar 20, 2025, 7:57 AM

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toxic lichen this is correct and the only thing that needs to be done

because i had the same process of what was on the answer key binomialpdf(144, 0.3, 5) but somehow different result??

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at first i had 0.670, and when i double checked i got another really long power to negative 16

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its probably something wrong with my calculator but you dont have to help if you dont feel like it!! ^^

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oh wait nvm i did the wrong thing lol

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wait

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nope still lost

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oh well ill open another room!! tyvm ^^

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.close

pearl pondBOT Mar 20, 2025, 8:00 AM

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Channel closed

Closed by @elder narwhal

Use .reopen if this was a mistake.

pearl pondBOT Mar 20, 2025, 8:15 AM

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Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

daring bay Mar 20, 2025, 10:19 AM

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if i have to check if the vectors are colinear

pearl pondBOT Mar 20, 2025, 10:19 AM

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daring bay if i have to check if the vectors are colinear

daring bay Mar 20, 2025, 10:19 AM

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do i have to do each a and b, a and c, b and c

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or is there an easier way

toxic lichen Mar 20, 2025, 10:21 AM

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,rccw

jolly parrotBOT Mar 20, 2025, 10:21 AM

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toxic lichen Mar 20, 2025, 10:21 AM

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daring bay do i have to do each a and b, a and c, b and c

in principle yes, but collinearity is transitive

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if you know a is collinear to b and b is collinear to c, then a will also automatically be collinear to c

pearl pondBOT Mar 20, 2025, 10:34 AM

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@daring bay Has your question been resolved?

pearl pondBOT Mar 20, 2025, 10:49 AM

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Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

daring bay Mar 20, 2025, 10:53 AM

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toxic lichen if you know a is collinear to b and b is collinear to c, then a will also automa...

oh

pearl pondBOT Mar 20, 2025, 10:53 AM

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daring bay oh

daring bay Mar 20, 2025, 10:53 AM

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okay okay

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thanks

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wait

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let me understand

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yeah ofc

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and lets say we have 4 vectors

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a b c d

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then we do a and b, b and c, c and d

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or how

toxic lichen Mar 20, 2025, 10:56 AM

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each vector needs to be involved in at least one comparison

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and then like

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you need to go based on info you already know

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idk what else to tell you there

daring bay Mar 20, 2025, 10:59 AM

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toxic lichen you need to go based on info you already know

i dont thinks thats gonna happen tho

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rither way doing all of them takes like a second

toxic lichen Mar 20, 2025, 10:59 AM

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then do them all lol

pearl pondBOT Mar 20, 2025, 11:14 AM

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@daring bay Has your question been resolved?

daring bay Mar 20, 2025, 11:16 AM

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toxic lichen then do them all lol

so

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i got another question

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if i have to check if 2 vectors are orthogonal

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perpendicular

toxic lichen Mar 20, 2025, 11:17 AM

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... what's your question tho

rough forge Mar 20, 2025, 11:17 AM

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where is the frage

daring bay Mar 20, 2025, 11:17 AM

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toxic lichen ... what's your question tho

do i use this formula

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or just do

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a • b

rough forge Mar 20, 2025, 11:18 AM

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daring bay a • b

= 0

toxic lichen Mar 20, 2025, 11:18 AM

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daring bay do i use this formula

if you like doing 200% more work than necessary then yes

daring bay Mar 20, 2025, 11:18 AM

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yep

toxic lichen Mar 20, 2025, 11:18 AM

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otherwise no, just find the dot product and check whether it is 0

daring bay Mar 20, 2025, 11:18 AM

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okay thanks

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man im so confident for my exam tomorrow

rough forge Mar 20, 2025, 11:18 AM

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tomorrow will be wild

daring bay Mar 20, 2025, 11:18 AM

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rough forge tomorrow will be wild

so

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can i ask another question

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its a hard one tho

rough forge Mar 20, 2025, 11:19 AM

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server limit is 2

toxic lichen Mar 20, 2025, 11:19 AM

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daring bay can i ask another question

you just did

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don't ask to ask

daring bay Mar 20, 2025, 11:19 AM

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you see this right

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the question is: For which real numbers a is x not expressible as a linear combination of the other given vectors?

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we have a parameter here

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this is so confusing ngl

rough forge Mar 20, 2025, 11:22 AM

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daring bay the question is: For which real numbers a is x not expressible as a linear combi...

If it's not expressible as a linear combination

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that means they are all linearly independent

daring bay Mar 20, 2025, 11:23 AM

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warte bro

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,rotate

jolly parrotBOT Mar 20, 2025, 11:24 AM

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daring bay Mar 20, 2025, 11:24 AM

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nicht darstellbar*

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also wie mache ich das rechnerisch

rough forge Mar 20, 2025, 11:25 AM

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jolly parrot

also x nicht als LK von a und b

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Ok

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In 2D ist das noch einfacher

daring bay Mar 20, 2025, 11:26 AM

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rough forge also **x** nicht als LK von **a** und **b**

Für welche zahlen a ist x nicht als linearkombination der übrigen gegebenen vektoren darstellbar

rough forge Mar 20, 2025, 11:26 AM

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a und b können eine Ebene aufspannen, und dann bilden sie eine Basis für R^2

daring bay Mar 20, 2025, 11:26 AM

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rough forge **a** und **b** können eine Ebene aufspannen, und dann bilden sie eine Basis für...

ja

rough forge Mar 20, 2025, 11:26 AM

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wo dann x natürlich drinnen wäre

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das heißt du musst das verhindern bro

daring bay Mar 20, 2025, 11:26 AM

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ja

rough forge Mar 20, 2025, 11:26 AM

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a und b dürfen nicht linear unabhängig sein

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Wie wählst du also a

daring bay Mar 20, 2025, 11:26 AM

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hä

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ich muss doch

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einfach ein LGS machen

rough forge Mar 20, 2025, 11:27 AM

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Nicht mal

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Mit Auge kann man hier

daring bay Mar 20, 2025, 11:27 AM

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und gucken für welche zahl a x ist nicht als linearkombination darstellbar ist

rough forge Mar 20, 2025, 11:27 AM

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Wann waren zwei Vektoren linear abhängig?

daring bay Mar 20, 2025, 11:27 AM

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rough forge Wann waren zwei Vektoren linear abhängig?

wenn sie kolinear sind

rough forge Mar 20, 2025, 11:27 AM

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bro

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sag einfach wenn sie vielfache sind

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von eiannder

daring bay Mar 20, 2025, 11:27 AM

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ja

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genau

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😭 😭 😭

rough forge Mar 20, 2025, 11:27 AM

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jetzt musst du a so wählen

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dass a und b Vielfache sind

daring bay Mar 20, 2025, 11:28 AM

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hä warte kurz

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also x soll nicht durch a und b darstellbar sein

rough forge Mar 20, 2025, 11:29 AM

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Ja

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das geht nur wenn a und b keine Ebene aufspannen

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das wiederum heißt a und b sind linear abhängig, also Vielfache

daring bay Mar 20, 2025, 11:30 AM

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dann müssen sie abhängig sein

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OMG

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ja

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das baut alles auf

rough forge Mar 20, 2025, 11:30 AM

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und die Wahl ist eindeutig

daring bay Mar 20, 2025, 11:30 AM

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mein wissen bildet sich

rough forge Mar 20, 2025, 11:30 AM

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für a

daring bay Mar 20, 2025, 11:31 AM

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also a ist

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also 6 durch 3 ist zwei

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4

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a isr 4

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A ist 4

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digga was ein krasser trick

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aber

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falls ich das rechnen solte

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während der klausur

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wie mache ich dass dann

rough forge Mar 20, 2025, 11:33 AM

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haha

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(a,6)k = (2,3)

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bzw. lass lieber mit k

daring bay Mar 20, 2025, 11:33 AM

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was

rough forge Mar 20, 2025, 11:33 AM

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du kannst schreiben

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ja also es soll ja ein k-faches vielfaches sein

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dann kriegst du

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ak = 2
6k = 3

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k = 1/2 also a/2 = 2 also a = 4

daring bay Mar 20, 2025, 11:35 AM

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rough forge k = 1/2 also a/2 = 2 also a = 4

warte mal

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ok ja

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mein lehrer

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ja ok jawoll

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ich hab mir heute freigenommen damit ich mathe behindert lernen kann

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also

rough forge Mar 20, 2025, 11:36 AM

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hahahahaha

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das nannte man damals bei mir schulschwänzen

daring bay Mar 20, 2025, 11:37 AM

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d

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die b)

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,rotate

jolly parrotBOT Mar 20, 2025, 11:37 AM

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daring bay Mar 20, 2025, 11:37 AM

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also

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man sieht a ist 1

rough forge Mar 20, 2025, 11:37 AM

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gleiche aufgaben stellung?

daring bay Mar 20, 2025, 11:37 AM

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rough forge gleiche aufgaben stellung?

ja

rough forge Mar 20, 2025, 11:37 AM

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mach den ansatz den ich dir gerade gezeigt habe

daring bay Mar 20, 2025, 11:37 AM

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ok

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aber

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hä

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hier gehts nt

rough forge Mar 20, 2025, 11:38 AM

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bei der a) wars halt echt offensichtlich, aber hier könnte es schief gehen

daring bay Mar 20, 2025, 11:38 AM

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ja also hier geht der ansatz nt

rough forge Mar 20, 2025, 11:39 AM

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daring bay man sieht a ist 1

wenn a = 1 ist dann (0,1) und (1,2) sind linear unabhängig

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nh

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also wie gesagt

daring bay Mar 20, 2025, 11:39 AM

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hä

rough forge Mar 20, 2025, 11:39 AM

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?

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mach mal den ansatz den ich dir gerade gezeigt habe

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überlass das raten/observieren den anderen

daring bay Mar 20, 2025, 11:40 AM

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also a und b müssen abhängig sein

rough forge Mar 20, 2025, 11:40 AM

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richtig

daring bay Mar 20, 2025, 11:40 AM

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aber

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hier ist nur a

rough forge Mar 20, 2025, 11:40 AM

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also ist a zum Beispiel ein vielfaches von b

daring bay Mar 20, 2025, 11:41 AM

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ja

rough forge Mar 20, 2025, 11:41 AM

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du definierst eine eigene variable k

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kann ja ein 2-faches, 3-faches, pi-faches vielfach sein

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das weiß du nicht

daring bay Mar 20, 2025, 11:41 AM

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rough forge du definierst eine eigene variable k

r

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ich will lieber r

rough forge Mar 20, 2025, 11:41 AM

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dann r

jolly parrotBOT Mar 20, 2025, 11:42 AM

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daring bay Mar 20, 2025, 11:42 AM

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ja

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also

rough forge Mar 20, 2025, 11:42 AM

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jetzt bildest du zwei gleichungen

daring bay Mar 20, 2025, 11:42 AM

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a ist 0

rough forge Mar 20, 2025, 11:43 AM

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ja

daring bay Mar 20, 2025, 11:43 AM

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ok warte

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ich schreibes kurz auf

rough forge Mar 20, 2025, 11:44 AM

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wenn du schon eigentlich weiß a = 0 muss sein, bist du durch

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musst halt noch begründen was mit a = 0 passiert

daring bay Mar 20, 2025, 11:45 AM

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rough forge musst halt noch begründen was mit a = 0 passiert

warte

rough forge Mar 20, 2025, 11:45 AM

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für jede andere wahl von a ist doch die 1. gleichung schon nicht erfüllt

daring bay Mar 20, 2025, 11:46 AM

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ja also

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digga omg

rough forge Mar 20, 2025, 11:46 AM

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außer r = 0 aber dann ist es eh trivial

daring bay Mar 20, 2025, 11:46 AM

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warte

daring bay Mar 20, 2025, 11:48 AM

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rough forge außer r = 0 aber dann ist es eh trivial

hä also

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a = r * 0

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a = 0

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0+1 = r * 1

ember cloak Mar 20, 2025, 11:48 AM

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rough forge das nannte man damals bei mir schulschwänzen

hahaha 😂

rough forge Mar 20, 2025, 11:49 AM

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du hast ar = 0 für r = 0 kommt eh der nullvektor also weg mit dem fall

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und für a = 0 sieh selbst

daring bay Mar 20, 2025, 11:49 AM

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die können garnicht linear abhängig sein

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hä

rough forge Mar 20, 2025, 11:50 AM

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💀

daring bay Mar 20, 2025, 11:50 AM

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weil

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guckmal

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wenn a 0 kst

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a + 1

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0 + 1

rough forge Mar 20, 2025, 11:50 AM

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scheiß mal auf b

daring bay Mar 20, 2025, 11:51 AM

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dann ist bei der eiben gleichung 0 und bei der anderen 1

rough forge Mar 20, 2025, 11:51 AM

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was ist dann a für ein Vektor

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wenn a = 0

daring bay Mar 20, 2025, 11:51 AM

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keine ahnung

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nullvektor

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?

rough forge Mar 20, 2025, 11:51 AM

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ja

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und jeder vektor ist linear abhängig zum nullvektor

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merk dir das

daring bay Mar 20, 2025, 11:52 AM

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ok

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achso

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ok krass

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also nochmal wenn etwas linear abhängig ist dann kann man eine linearkombination draus machen?

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ja

rough forge Mar 20, 2025, 11:57 AM

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daring bay also nochmal wenn etwas linear abhängig ist dann kann man eine linearkombination...

nein

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Für (1,1) und (2,2) wird niemals eine LK existieren für (1,2) zum Beispiel

daring bay Mar 20, 2025, 11:59 AM

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aber

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so bildich

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bildlich

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ich versuche es mit vorzustellen

rough forge Mar 20, 2025, 12:00 PM

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Angenommen (0,0) und (a,b) ist ein Vektor ungleich dem Nullvektor. Dann sind die immer linear abhängig weil (a,b)k = (0,0) mit k = 0 ein Vielfaches vom Nullvektor wäre

daring bay Mar 20, 2025, 12:02 PM

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ja

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also in einer ebene heißt dass sid

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sie

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ich check okay

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ja

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aber die können doch eine linearkombination darstellen ohne dass sie in einer ebene sind oder

rough forge Mar 20, 2025, 12:04 PM

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das geht nur noch mit einer geraden

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mit {(1,1)} kannst du jeden vektor auf der geraden y = x darstellen

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zum beispiel (2,2) = 2(1,1)

daring bay Mar 20, 2025, 12:06 PM

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ja

rough forge Mar 20, 2025, 12:06 PM

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ich glaube ich habe deine frage nicht verstanden

rough forge Mar 20, 2025, 12:06 PM

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daring bay aber die können doch eine linearkombination darstellen ohne dass sie in einer eb...

also das hier

daring bay Mar 20, 2025, 12:07 PM

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keine ebene = linear abhängig

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das merke ich mir einfach

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aber wie sieht das aus

rough forge Mar 20, 2025, 12:07 PM

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ich würde es mir anders merken

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also du meinst wenn zwei Vektoren keine Ebene aufspannen, oder?

daring bay Mar 20, 2025, 12:09 PM

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rough forge also du meinst wenn zwei Vektoren keine Ebene aufspannen, oder?

ja

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oder 3

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also allgemein

rough forge Mar 20, 2025, 12:10 PM

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3 Vektoren können im R^3 eine Ebene aufspannen, aber dann müssen sie abhängig sein

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zum Beispiel

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(1,0,0), (0,1,0) und (0,2,0)

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der (0,2,0) liegt dann drinnen zum Beispiel

daring bay Mar 20, 2025, 12:11 PM

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rough forge 3 Vektoren können im R^3 eine Ebene aufspannen, aber dann müssen sie abhängig se...

ist eine basis dasselbe wie ebene

daring bay Mar 20, 2025, 12:11 PM

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rough forge der (0,2,0) liegt dann drinnen zum Beispiel

achso

rough forge Mar 20, 2025, 12:12 PM

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daring bay ist eine basis dasselbe wie ebene

nein

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Ich würde mir es so merken, eine Gerade braucht genau einen Vektor, eine Ebene genau zwei Vektoren und ein Raum wie der R^3 braucht genau 3 Vektoren um erzeugt zu werden

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Weil dann weiß du sofort, wenns mehr sind, dann muss da lineare Abhängkeit sein

daring bay Mar 20, 2025, 12:16 PM

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rough forge Weil dann weiß du sofort, wenns mehr sind, dann muss da lineare Abhängkeit sein

hä also

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linearkombinationen sind eine kette nh

rough forge Mar 20, 2025, 12:16 PM

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ja

daring bay Mar 20, 2025, 12:17 PM

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ja also

daring bay Mar 20, 2025, 12:17 PM

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rough forge ja

x muss in dem fall eine vielfache von a und b sein nh

rough forge Mar 20, 2025, 12:18 PM

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wieso

daring bay Mar 20, 2025, 12:19 PM

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ne also

rough forge Mar 20, 2025, 12:19 PM

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die aufgabe war a so zu bestimmen, sodass keine LK von x mit a und b existiert

daring bay Mar 20, 2025, 12:19 PM

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also doch

daring bay Mar 20, 2025, 12:19 PM

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rough forge die aufgabe war a so zu bestimmen, sodass keine LK von **x** mit **a** und **b**...

achso

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srimmt ja

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das heißt das a und b linear abhängig sein müssen

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aber dann bildet sich doch eine kette wenn die abhängig sind

rough forge Mar 20, 2025, 12:20 PM

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hä

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du hast x = (1,3) und a = (0,0) und b = (0,1)

daring bay Mar 20, 2025, 12:21 PM

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ja

rough forge Mar 20, 2025, 12:21 PM

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du wirst niemals nach (1,3) kommen

daring bay Mar 20, 2025, 12:21 PM

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ja

rough forge Mar 20, 2025, 12:21 PM

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wenn du nur entlang (0,1) gehst

daring bay Mar 20, 2025, 12:21 PM

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ja

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also ich hab nicht genau verstanden

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warum a und b linear abhängig sein müssen damit x nicht eine linearkombination von a und b ist

rough forge Mar 20, 2025, 12:24 PM

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weil egal wie du a = (0,0) und b = (0,1) skalierst und addierst

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du wirst niemald zu (1,3) kommen

daring bay Mar 20, 2025, 12:24 PM

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ja

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stimmt

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wtf

rough forge Mar 20, 2025, 12:24 PM

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a und b liegen auf einer Geraden x = 0, die nicht (1,3) enthält

daring bay Mar 20, 2025, 12:25 PM

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also hier

rough forge Mar 20, 2025, 12:25 PM

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,rccw

jolly parrotBOT Mar 20, 2025, 12:25 PM

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rough forge Mar 20, 2025, 12:25 PM

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😭

daring bay Mar 20, 2025, 12:25 PM

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egal wie man

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(4,6) und (2,3) skaliert

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dann kommt man nie auf 0,9?

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ja stimmt eigentlich oder

rough forge Mar 20, 2025, 12:26 PM

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ja

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(4,6) und (2,3) liegen auf einer geraden

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y = (3/2)x

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und (0,9) ist nicht ein punkt auf der geraden

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der liegt auf der y-achse

daring bay Mar 20, 2025, 12:27 PM

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ja ok ok

rough forge Mar 20, 2025, 12:27 PM

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wenn du es malst

daring bay Mar 20, 2025, 12:27 PM

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du bist schon schlau ohne spaß

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soll ich dir was schweres zeigen

rough forge Mar 20, 2025, 12:28 PM

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ok

daring bay Mar 20, 2025, 12:28 PM

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die selbe aufgabe

rough forge Mar 20, 2025, 12:29 PM

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bei R^3 empfehle ich LGS

daring bay Mar 20, 2025, 12:29 PM

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ja

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ok ok

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hö aber

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ok warte

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ich schreib es auf

rough forge Mar 20, 2025, 12:31 PM

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,, r\vec{a}+s\vec{b}+t\vec{c} = \vec{x}_a

jolly parrotBOT Mar 20, 2025, 12:31 PM

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rough forge Mar 20, 2025, 12:32 PM

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Damit keine LK existiert müssen a,b und c linear abhängig sein

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sonst würden die eine Basis für den R^3 bilden

daring bay Mar 20, 2025, 12:32 PM

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rough forge sonst würden die eine Basis für den R^3 bilden

aber eine basis ist doch sozusagen keine linearkombination

rough forge Mar 20, 2025, 12:33 PM

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mit einer Basis kann man LKs bilden

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wenn du linear abhängige Vektoren hast, dann bilden die keine Basis mehr, und dann kannst du nicht immer eine LK finden

daring bay Mar 20, 2025, 12:35 PM

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rough forge wenn du linear abhängige Vektoren hast, dann bilden die keine Basis mehr, und da...

linear abhängig heißt das sie in einer ebene sind

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wie sieht das aus in 3d digga

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ich kann mir das nt vorstellen

rough forge Mar 20, 2025, 12:36 PM

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daring bay linear abhängig heißt das sie in einer ebene sind

alter

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gerade hatten wir doch gesagt, dass hängt davon ab wie viele vektoren duhast

rough forge Mar 20, 2025, 12:37 PM

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daring bay linear abhängig heißt das sie in einer ebene sind

wenn die drei vektoren in einer ebene liegen* dann kannst du folgern dass da eine lineare abhängigkeit existieren muss

rough forge Mar 20, 2025, 12:37 PM

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daring bay wie sieht das aus in 3d digga

halt einfach ein Blatt papier in dein Hand

daring bay Mar 20, 2025, 12:37 PM

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ja

rough forge Mar 20, 2025, 12:37 PM

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das ist deine Ebene im Raum

daring bay Mar 20, 2025, 12:38 PM

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accso

rough forge Mar 20, 2025, 12:38 PM

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und da können Vektoren drinnen liegen

daring bay Mar 20, 2025, 12:38 PM

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sowas

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ok icc chrkc

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ja ich check

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ok ich mach diesen lgs kurz

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also ich muss hier sozusagen

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einen widerspruch finden

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oder wie

rough forge Mar 20, 2025, 12:42 PM

#

ja

#

denn dann findest du nie r,s,t

daring bay Mar 20, 2025, 12:42 PM

#

hmm

rough forge Mar 20, 2025, 12:42 PM

#

hier ists am Besten das auf Zeilenstufenform zu bringen

daring bay Mar 20, 2025, 12:43 PM

#

haben wir nt gemacht

#

ich glaub sowas kommt dann auch nt dran

#

oder vllt

rough forge Mar 20, 2025, 12:43 PM

#

habt ihr nicht gelernt wie man lgs löst

daring bay Mar 20, 2025, 12:43 PM

#

keine ahnung

daring bay Mar 20, 2025, 12:43 PM

#

rough forge habt ihr nicht gelernt wie man lgs löst

ja aslo so wie ich es aufgeschrieben habe

rough forge Mar 20, 2025, 12:44 PM

#

doch

#

ihr hattet doch so themen in der kurzarbeit

#

wie hieß nochmal dieses eine theme

daring bay Mar 20, 2025, 12:44 PM

#

ja

#

lgs?

#

😭

#

was meinst du

rough forge Mar 20, 2025, 12:44 PM

#

braucht man da

#

wsa waren die themen nochma

daring bay Mar 20, 2025, 12:44 PM

#

substituieren und so

#

meinst du das

rough forge Mar 20, 2025, 12:44 PM

#

nein

#

mischprobleme

daring bay Mar 20, 2025, 12:46 PM

#

achso

#

ja aber

#

das sind doch textaufgaben

rough forge Mar 20, 2025, 12:47 PM

#

na wenn du meinst

daring bay Mar 20, 2025, 12:47 PM

#

daring bay

also

#

wie kommt hier ein widerspruch

#

ich checks nt

rough forge Mar 20, 2025, 12:48 PM

#

ja wird schwierig ohne ansatz

daring bay Mar 20, 2025, 12:49 PM

#

ja ich skip die aufgabe

#

wnen die kommt dann improvisiere ich

#

aber warte

#

#

,rotate

rough forge Mar 20, 2025, 12:50 PM

#

würde nicht

daring bay Mar 20, 2025, 12:50 PM

#

die geht ohne ansatz oder

jolly parrotBOT Mar 20, 2025, 12:50 PM

#

rough forge Mar 20, 2025, 12:50 PM

#

du musst einfach die umkreisenden loswerden in dem du vielfache von I und II drauf addierst

daring bay Mar 20, 2025, 12:51 PM

#

rough forge würde nicht

ja ich weiß wie man es llst

#

löst

#

ich kann das ez

rough forge Mar 20, 2025, 12:51 PM

#

ja dann mach doch

daring bay Mar 20, 2025, 12:51 PM

#

aber parameter

#

verwirrt mich

rough forge Mar 20, 2025, 12:51 PM

#

ja junge

#

das kommt am ende doch

daring bay Mar 20, 2025, 12:51 PM

#

ok ich mach

#

ok sorry bro

#

ich mach

rough forge Mar 20, 2025, 12:51 PM

#

du hast dann

daring bay Mar 20, 2025, 12:51 PM

#

spoiler

#

nt

rough forge Mar 20, 2025, 12:51 PM

#

ok

daring bay Mar 20, 2025, 12:51 PM

#

ich mach kurz selber

rough forge Mar 20, 2025, 12:51 PM

#

dein law

daring bay Mar 20, 2025, 12:57 PM

#

ich glaub etwas ist schief gelaufen bro

#

#

,rotate

jolly parrotBOT Mar 20, 2025, 12:58 PM

#

daring bay Mar 20, 2025, 12:58 PM

#

was mache ich falsch bro wtf

rough forge Mar 20, 2025, 1:02 PM

#

du hast ja jetzt

#

2 = r-s+t
2+a = 2r+4t
1-a = r

#

ja

#

jetzt set mal alles in I

#

2 = (1-a)-s+(3/4)a

daring bay Mar 20, 2025, 1:06 PM

#

rough forge 2 = (1-a)-s+(3/4)a

hö

#

also hab ich es richtig

#

digga da kam so 7/4 dann dachte ich mir ich lass es lieber aber ok

#

ich mach weiter

rough forge Mar 20, 2025, 1:08 PM

#

daring bay also hab ich es richtig

mir ist aufgefallen du hast falsch abgeschrieben

#

#

daring bay Mar 20, 2025, 1:09 PM

#

wtf

#

ich hab ne frage aber allgemein zu LGS

#

kann man die gleichungen die man bildet

#

dann auch weiter benutzen

rough forge Mar 20, 2025, 1:10 PM

#

da muss eine 5 btw

daring bay Mar 20, 2025, 1:10 PM

#

unendlich

daring bay Mar 20, 2025, 1:10 PM

#

rough forge da muss eine 5 btw

ja

rough forge Mar 20, 2025, 1:10 PM

#

daring bay dann auch weiter benutzen

ja solltest du auch

daring bay Mar 20, 2025, 1:10 PM

#

rough forge ja solltest du auch

ja

#

bruder MORGEN 15 punkte

rough forge Mar 20, 2025, 1:10 PM

#

du willst ja die solange variablen loswerden wie möglich

daring bay Mar 20, 2025, 1:12 PM

#

ok bro

#

ich muss mir noch geraden im raum anschauen

#

egal ich mach die aufgabe noch

#

ok ich fang jz an

daring bay Mar 20, 2025, 1:46 PM

#

ok also

#

@rough forge bist du da

#

#

ich hab die werte raus aber ka was ich jetzt mache

rough forge Mar 20, 2025, 1:54 PM

#

daring bay

was ein chaos

#

wenn ich dein lehrer wäre 0 punkte

daring bay Mar 20, 2025, 1:56 PM

#

rough forge was ein chaos

😭

#

also

rough forge Mar 20, 2025, 1:57 PM

#

ich muss mich anstrengen

#

IV und V, wenn du die gleichsetzt kriegst du 2-a = 2-4a also 4a = a

#

also existieren nur Lösungen für a = 0

#

das heißt es gibt keine Linearkombination, wenn a ungleich 0 ist

#

das sieht so aus für a = 0, der Vektor x liegt in der Ebene

#

und für jeden anderen Wert ist der weiße Vektor x draußen

daring bay Mar 20, 2025, 1:59 PM

#

DIGGA

#

ich checke ES

#

ICH sehe es

#

ich checks jz

#

aber

#

was hab ich falsch gemacht

rough forge Mar 20, 2025, 2:00 PM

#

sah bis jetzt richtig aus

#

die rechte Seite von IV und V ist gleich, also kannst du es gleichsetzten

#

und so auf a kommen

daring bay Mar 20, 2025, 2:02 PM

#

rough forge die rechte Seite von IV und V ist gleich, also kannst du es gleichsetzten

ah stimmt

#

also ziel ist es

#

nur a für sich zu lösen

rough forge Mar 20, 2025, 2:03 PM

#

sozusagen

daring bay Mar 20, 2025, 2:03 PM

#

ok ok ok

#

ok jetzt mache ich diese flächen und so

rough forge Mar 20, 2025, 2:03 PM

#

aber das ist so freestyle

daring bay Mar 20, 2025, 2:03 PM

#

danach geraden im raum

#

und dann schon fertig hö

rough forge Mar 20, 2025, 2:03 PM

#

ich hätte euch einen systematischen ansatz gezeigt

daring bay Mar 20, 2025, 2:03 PM

#

aber digga

#

erster teil ohne taschenrechner

#

fuck

rough forge Mar 20, 2025, 2:04 PM

#

weil man kann so ohne ein gespür zu haben viel falsch bzw. lost sein

daring bay Mar 20, 2025, 2:05 PM

#

rough forge weil man kann so ohne ein gespür zu haben viel falsch bzw. lost sein

meinst du mit unseren lgd

#

lgs

rough forge Mar 20, 2025, 2:05 PM

#

wie man lgs mit gauß algorithmus löst

#

und dann wie man guckt ob da lösungen existieren in abhängigkeit mit parameter

daring bay Mar 20, 2025, 2:07 PM

#

ja wir haben das nt gelernt

rough forge Mar 20, 2025, 2:07 PM

#

ja loser

#

also der lehrer

daring bay Mar 20, 2025, 2:07 PM

#

willst du die themen sehen

rough forge Mar 20, 2025, 2:07 PM

#

von mir aus

daring bay Mar 20, 2025, 2:07 PM

#

fuck es ist schon viel

#

rough forge Mar 20, 2025, 2:08 PM

#

er sagt es kommt kein lgs 🤣

daring bay Mar 20, 2025, 2:08 PM

#

WANN

#

ich meine mit parameter

rough forge Mar 20, 2025, 2:09 PM

#

daring bay ich glaub sowas kommt dann auch nt dran

.

daring bay Mar 20, 2025, 2:09 PM

#

so räudige parameter

daring bay Mar 20, 2025, 2:09 PM

#

rough forge .

ja das war schon eine harte aufgabe

rough forge Mar 20, 2025, 2:09 PM

#

wenn du glück hast werden es 3NP

daring bay Mar 20, 2025, 2:09 PM

#

oha

rough forge Mar 20, 2025, 2:09 PM

#

weil alle guten dinge sind drei

daring bay Mar 20, 2025, 2:09 PM

#

diese geraden im raum hab ich mir garnicht angeschaut

rough forge Mar 20, 2025, 2:09 PM

#

ich kann mich erinnern ich hab auch so mit vektorgeometrie gestrugglet

daring bay Mar 20, 2025, 2:09 PM

#

also ich weiß wi es geht

rough forge Mar 20, 2025, 2:10 PM

#

gab keine coolen leute dir mir das erklären konnten

daring bay Mar 20, 2025, 2:10 PM

#

aber noch nt geübt

#

digga unterbestimmte lgs

#

endgegner

#

und schrägbilder

#

DIGGA IN DER KURZarbeit

#

gab es so eine aufgabe

#

holy shit bro

rough forge Mar 20, 2025, 2:11 PM

#

was soll ich sagen

#

du hattest eig 2 wochen (auf kurzarbeit geschissen)

#

😂

#

schulleben ist eig. geil

#

immer diese dummen probleme hahahaha

daring bay Mar 20, 2025, 2:12 PM

#

rough forge Mar 20, 2025, 2:12 PM

#

,rccw

jolly parrotBOT Mar 20, 2025, 2:12 PM

#

daring bay Mar 20, 2025, 2:12 PM

#

ich hatte keine zeit mehr digga

rough forge Mar 20, 2025, 2:13 PM

#

er musste nur zeichnen

#

das mit den punkten haben wir geübt ja

#

er war auch so lieb und hat gesagt wie die angeordnet sind

daring bay Mar 20, 2025, 2:14 PM

#

rough forge er war auch so lieb und hat gesagt wie die angeordnet sind

😭 😭

rough forge Mar 20, 2025, 2:14 PM

#

daring bay

😭

daring bay Mar 20, 2025, 2:14 PM

#

ich muss es mir halt vorstellen können nh

rough forge Mar 20, 2025, 2:14 PM

#

so ein kaputter karton

daring bay Mar 20, 2025, 2:14 PM

#

wie ein quader aussieht

rough forge Mar 20, 2025, 2:14 PM

#

immer dieses vorstellen

#

noch nie eine box gesehen

daring bay Mar 20, 2025, 2:14 PM

#

doch

#

aber

rough forge Mar 20, 2025, 2:15 PM

#

selbst das mit den skizzen und zeichnen

daring bay Mar 20, 2025, 2:15 PM

#

so billig

#

geschenkte punkte eigentlich

#

digga ich schäme mich

rough forge Mar 20, 2025, 2:15 PM

#

war halt nicht weihnachten, nh

daring bay Mar 20, 2025, 2:15 PM

#

hä aber

#

kann man überhaupt so ein schrägbild davon zeichnen

rough forge Mar 20, 2025, 2:16 PM

#

ja klar

#

er hat sogar

#

koordinatensystem für euch gemacht

#

bester lehrer

#

damals musste man das selber machen

daring bay Mar 20, 2025, 2:16 PM

#

warte

rough forge Mar 20, 2025, 2:16 PM

#

wo viel zeit verloren ging

#

du musst doch einfach zählen können umd die punkte einzuzeichen

#

und die dann verbinden

#

dann ergibt sich das von selbst

daring bay Mar 20, 2025, 2:17 PM

#

das war die lösung

rough forge Mar 20, 2025, 2:17 PM

#

ja siehst du

#

wenn du die punkte malst, und sie verbindest

daring bay Mar 20, 2025, 2:17 PM

#

aber kann man auch nicht die längere version haben

daring bay Mar 20, 2025, 2:17 PM

#

rough forge wenn du die punkte malst, und sie verbindest

ja

#

stell dir vor pyramide kommt drsn

#

pyramide

rough forge Mar 20, 2025, 2:18 PM

#

längere?

#

du hast da iwas falsch gezählt und gezeichnet

daring bay Mar 20, 2025, 2:20 PM

#

rough forge du hast da iwas falsch gezählt und gezeichnet

so in der art

rough forge Mar 20, 2025, 2:20 PM

#

nein, ist schon eindeutig

daring bay Mar 20, 2025, 2:20 PM

#

ja stimmt

rough forge Mar 20, 2025, 2:20 PM

#

weil, wenn das so wäre

daring bay Mar 20, 2025, 2:20 PM

#

weil die punkte

rough forge Mar 20, 2025, 2:20 PM

#

dann wäre B viel weiter entlang y-achse

daring bay Mar 20, 2025, 2:20 PM

#

ja

#

genau

#

ich muss sowas üben

rough forge Mar 20, 2025, 2:21 PM

#

das kann nur so aussehen

#

wenn du die achsen skalierst

daring bay Mar 20, 2025, 2:22 PM

#

ja

rough forge Mar 20, 2025, 2:22 PM

#

zum beispiel eine längeneinheit auf y-achse sind 4 kasten und auf der x und z eine längeneinheit ein kasten

#

aber mach das nicht

daring bay Mar 20, 2025, 2:23 PM

#

ja

#

overthinkikg

#

also ich muss meine zeit strukturieren

#

ich mache jetzt fläche und so

#

dann volumen

rough forge Mar 20, 2025, 2:23 PM

#

hast ja nicht viel zeit

daring bay Mar 20, 2025, 2:23 PM

#

das ist aber bilig

daring bay Mar 20, 2025, 2:23 PM

#

rough forge hast ja nicht viel zeit

ja wtf

rough forge Mar 20, 2025, 2:23 PM

#

guck dass du schlaf genug kriegst

#

sonst war das umsonst

#

ich sags dir

daring bay Mar 20, 2025, 2:23 PM

#

kopfrechen teil wird mich FICKEN

#

ich hab vergessen wie man brüche addiert subtrahiert multipliziert

#

ich schau mir das alles nochmal an

#

egal

daring bay Mar 20, 2025, 2:46 PM

#

so wichtig das ich mir es anschaue

daring bay Mar 20, 2025, 2:46 PM

#

rough forge sonst war das umsonst

#

,rotate

jolly parrotBOT Mar 20, 2025, 2:46 PM

#

daring bay Mar 20, 2025, 2:47 PM

#

also

#

sind die klammern falsch

#

oder

#

ohne klammern nh

rough forge Mar 20, 2025, 2:47 PM

#

jolly parrot

genau so

#

wehe ohne

daring bay Mar 20, 2025, 2:48 PM

#

hä

#

wtf

#

aber im taschenrechner

#

ok ich check

#

ich check

#

dieses vektorprodukt

#

hurensohn produkt digga

pearl pondBOT Mar 20, 2025, 3:20 PM

#

@daring bay Has your question been resolved?

daring bay Mar 20, 2025, 3:24 PM

#

@rough forge was würdest du erwarten in dem kopfrechnen teil

#

also was glaubst du würde dran kommen

rough forge Mar 20, 2025, 3:27 PM

#

ich denke rechenaufgaben, die man im kopf lösen kann

daring bay Mar 20, 2025, 3:35 PM

#

rough forge ich denke rechenaufgaben, die man im kopf lösen kann

ja

#

und

#

wenn ich zb die fläche von einem dreieck berechnen muss

rough forge Mar 20, 2025, 3:49 PM

#

daring bay wenn ich zb die fläche von einem dreieck berechnen muss

Ein Dreieck ist die Hälfte eines Parallelogramm

#

Parallelogram war die Länge des Normalenvektors

#

Solche Sachen kann googlen und auf einen Merkzettel schreiben

daring bay Mar 20, 2025, 3:57 PM

#

rough forge Ein Dreieck ist die Hälfte eines Parallelogramm

ja aber wenn nur das dreieck gegeben ist

#

nvm

#

ich weis es

rough forge Mar 20, 2025, 3:58 PM

#

grade sag ich es noch

daring bay Mar 20, 2025, 4:04 PM

#

rough forge grade sag ich es noch

hä

#

ok ok

daring bay Mar 20, 2025, 4:38 PM

#

what did i do wrong here

#

A(0/0/1) B(-4/-1/-1) C(-2/6/1) D(-6/5/1) S(-3/-1/-4)

#

bruh

#

wait i did something wrong

plush bramble Mar 20, 2025, 4:44 PM

#

,rotate

jolly parrotBOT Mar 20, 2025, 4:44 PM

#

daring bay Mar 20, 2025, 4:45 PM

#

plush bramble ,rotate

wait

#

wait

#

i had the wrong point

plush bramble Mar 20, 2025, 4:46 PM

#

?

daring bay Mar 20, 2025, 4:46 PM

#

its D(-6/5/-1)

#

but i still got it wrong, wait

#

#

,rotate

jolly parrotBOT Mar 20, 2025, 4:46 PM

#

daring bay Mar 20, 2025, 4:46 PM

#

yeah so what did i do wrong here

fallow jay Mar 20, 2025, 4:51 PM

#

wait can you tell me the points once again?

daring bay Mar 20, 2025, 4:55 PM

#

fallow jay wait can you tell me the points once again?

A(0/0/1) B(-4/-1/-1) C(-2/6/1) D(-6/5/-1) S(-3/-1/-4)

fallow jay Mar 20, 2025, 4:59 PM

#

are you sure these are the points?

#

cause if it is your answer is right

daring bay Mar 20, 2025, 5:04 PM

#

fallow jay cause if it is your answer is right

the website says 30

fallow jay Mar 20, 2025, 5:04 PM

#

maybe the website's wrong?

daring bay Mar 20, 2025, 5:04 PM

#

fallow jay maybe the website's wrong?

yeah

fallow jay Mar 20, 2025, 5:05 PM

#

cause your answer's right if youve copied the question right

fallow jay Mar 20, 2025, 5:05 PM

#

daring bay yeah

just check it on any other website to confirm tho

daring bay Mar 20, 2025, 5:06 PM

#

,rotate

daring bay Mar 20, 2025, 5:06 PM

#

fallow jay just check it on any other website to confirm tho

i did the b) here

jolly parrotBOT Mar 20, 2025, 5:06 PM

#

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

daring bay Mar 20, 2025, 5:06 PM

#

i copied correctly

fallow jay Mar 20, 2025, 5:07 PM

#

yeah so the answer might be wrong

daring bay Mar 20, 2025, 5:08 PM

#

fallow jay yeah so the answer might be wrong

also i have another question

fallow jay Mar 20, 2025, 5:09 PM

#

yeah?

daring bay Mar 20, 2025, 5:10 PM

#

#

so

#

are the vectors always pointed like this in a parallelogram

fallow jay Mar 20, 2025, 5:11 PM

#

no not necessary

#

you could make one with cdba

daring bay Mar 20, 2025, 5:12 PM

#

okay hm

#

my teacher so weird

#

anyways

fallow jay Mar 20, 2025, 5:12 PM

#

why what happened?

daring bay Mar 20, 2025, 5:12 PM

#

if i had to give the diagonals in a and b

daring bay Mar 20, 2025, 5:12 PM

#

fallow jay why what happened?

he said its always A B D C

fallow jay Mar 20, 2025, 5:12 PM

#

no lol

daring bay Mar 20, 2025, 5:13 PM

#

daring bay if i had to give the diagonals in a and b

AD would be = a + b and BC would be - a + b

#

but how do i know this without

#

having

fallow jay Mar 20, 2025, 5:13 PM

#

wait is this the diagram of the question you asked me??

daring bay Mar 20, 2025, 5:14 PM

#

wdym

#

no it isnt i just drawed it rq

fallow jay Mar 20, 2025, 5:14 PM

#

daring bay

is this llelogram the one you asked in the question??

dense sluice Mar 20, 2025, 5:14 PM

#

Can someone help me on help 28

daring bay Mar 20, 2025, 5:14 PM

#

fallow jay is this llelogram the one you asked in the question??

yeah

fallow jay Mar 20, 2025, 5:14 PM

#

daring bay no it isnt i just drawed it rq

oh okay

#

the vectors should be connected head to head or tail to tail

#

so in your parallelogram, AD would be b-a

daring bay Mar 20, 2025, 5:16 PM

#

yes yes

#

okay uhh

#

i got another question

fallow jay Mar 20, 2025, 5:16 PM

#

yeah?

fallow jay Mar 20, 2025, 5:17 PM

#

dense sluice Can someone help me on help 28

ill help you in a while

dense sluice Mar 20, 2025, 5:17 PM

#

Alright

daring bay Mar 20, 2025, 5:18 PM

#

so the opposite vectors are always pointed in the other direction

fallow jay Mar 20, 2025, 5:18 PM

#

daring bay he said its always A B D C

wait wait, i think you have drawn it wrong. it should be abcd but you made it abdc

daring bay Mar 20, 2025, 5:19 PM

#

fallow jay wait wait, i think you have drawn it wrong. it should be abcd but you made it ab...

yh cuz my teacher said its like this

fallow jay Mar 20, 2025, 5:19 PM

#

oh lol

daring bay Mar 20, 2025, 5:19 PM

#

or wair

#

maybe it is

#

ill ask tomorrow

fallow jay Mar 20, 2025, 5:20 PM

#

okay

dense sluice Mar 20, 2025, 5:21 PM

#

#

Do yk the slope

fallow jay Mar 20, 2025, 5:21 PM

#

daring bay so the opposite vectors are always pointed in the other direction

wait ill send a pic

fallow jay Mar 20, 2025, 5:21 PM

#

dense sluice Do yk the slope

are the point (200,3) and (300, 4.5)?

dense sluice Mar 20, 2025, 5:22 PM

#

yes

fallow jay Mar 20, 2025, 5:22 PM

#

the slope would be (y1-y2)/(x1-x2)

#

so here it would be 100/1.5

#

or 200/3

dense sluice Mar 20, 2025, 5:23 PM

#

So

#

The slope is

#

3 ?

fallow jay Mar 20, 2025, 5:24 PM

#

daring bay ill ask tomorrow

i think youve made the figure wrong

fallow jay Mar 20, 2025, 5:24 PM

#

dense sluice The slope is

no the slope is 200/3

dense sluice Mar 20, 2025, 5:24 PM

#

Ohhhh

#

Mb I’m kinda slow

fallow jay Mar 20, 2025, 5:24 PM

#

slope=(200-100)/(4.5-3)

dense sluice Mar 20, 2025, 5:25 PM

#

Like this

#

fallow jay Mar 20, 2025, 5:26 PM

#

yeah

dense sluice Mar 20, 2025, 5:26 PM

#

It’s wrong

fallow jay Mar 20, 2025, 5:26 PM

#

what's the answer?

dense sluice Mar 20, 2025, 5:26 PM

#

It don’t show until I get the right answer

fallow jay Mar 20, 2025, 5:27 PM

#

oh mb sorry

dense sluice Mar 20, 2025, 5:27 PM

#

Don’t worry

fallow jay Mar 20, 2025, 5:27 PM

#

it would be 3/200

#

i took the points wrong lol

dense sluice Mar 20, 2025, 5:28 PM

#

It’s okay lol

fallow jay Mar 20, 2025, 5:28 PM

#

i took y2 as 200 instead of 4.5 and y1 as 100 instead of 3

#

haha

#

@daring bay is your doubt clear??

dense sluice Mar 20, 2025, 5:29 PM

#

fallow jay i took y2 as 200 instead of 4.5 and y1 as 100 instead of 3

Haha

#

But ur so smart

fallow jay Mar 20, 2025, 5:29 PM

#

oh thank you!

dense sluice Mar 20, 2025, 5:29 PM

#

I wish I could be smart like you

fallow jay Mar 20, 2025, 5:29 PM

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is the answer correct tho?

dense sluice Mar 20, 2025, 5:29 PM

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Yes

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It was

fallow jay Mar 20, 2025, 5:29 PM

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nice

dense sluice Mar 20, 2025, 5:29 PM

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Yes

#

fallow jay Mar 20, 2025, 5:29 PM

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yeah its same thing

dense sluice Mar 20, 2025, 5:30 PM

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The same answer

fallow jay Mar 20, 2025, 5:30 PM

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i just told you the formula, you have to use it here too

fallow jay Mar 20, 2025, 5:30 PM

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dense sluice The same answer

no it wont be

dense sluice Mar 20, 2025, 5:30 PM

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I think the answer is 0.015

fallow jay Mar 20, 2025, 5:30 PM

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the points are different

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no no

dense sluice Mar 20, 2025, 5:30 PM

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Yeah I just realized that

fallow jay Mar 20, 2025, 5:30 PM

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haha yeah

dense sluice Mar 20, 2025, 5:31 PM

#

How long it take u for learn math

#

?

fallow jay Mar 20, 2025, 5:31 PM

#

well i am in grade 12 rn so like 12 years ig? lmao

dense sluice Mar 20, 2025, 5:31 PM

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Dayum

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That’s to much

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I’m 9 grade

fallow jay Mar 20, 2025, 5:32 PM

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oh cool cool

dense sluice Mar 20, 2025, 5:32 PM

#

Do u think I would pass the grade I only have 3 F

daring bay Mar 20, 2025, 5:32 PM

#

fallow jay <@419857529032409088> is your doubt clear??

yeah i guess

fallow jay Mar 20, 2025, 5:32 PM

#

uhh idk cause here we dont use grading system so i dont know how that works

dense sluice Mar 20, 2025, 5:33 PM

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fallow jay uhh idk cause here we dont use grading system so i dont know how that works

Oh

fallow jay Mar 20, 2025, 5:33 PM

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daring bay yeah i guess

okay, ill send you the pic of the vector one

flat cedar Mar 20, 2025, 5:33 PM

#

fallow jay wait ill send a pic

While adding or subtracting vectors, both arrows (heads) should be together or both ends (tails) should be together

rn_image_picker_lib_temp_59873f2e-bc60-4f17-8af2-c78b887b1ce0.jpg

fallow jay Mar 20, 2025, 5:33 PM

#

@daring bay this might help!

daring bay Mar 20, 2025, 5:34 PM

#

flat cedar While adding or subtracting vectors, both arrows (heads) should be together or b...

ok yeah

rough forge Mar 20, 2025, 6:25 PM

#

fallow jay <@419857529032409088> is your doubt clear??

hahaha

fallow jay Mar 20, 2025, 6:29 PM

#

??

daring bay Mar 20, 2025, 6:41 PM

#

@rough forge bei geraden im raum muss ich nur die gerade haben um sie einzuzeichnen zu können oder

rough forge Mar 20, 2025, 6:48 PM

#

daring bay <@587715738484211754> bei geraden im raum muss ich nur die gerade haben um sie e...

wenn die gerade schon da ist, musst du sie nicht mehr zeichnen

#

oder was meinst du

daring bay Mar 20, 2025, 6:50 PM

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rough forge wenn die gerade schon da ist, musst du sie nicht mehr zeichnen

also

#

#

sowas

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dann brauche ich nur die beiden vektore

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um die zu zeichnen kh

#

nh*

rough forge Mar 20, 2025, 6:51 PM

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such dir zwei punkte aus und verbinde sie

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zum beispiel

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x(0) und x(1)

daring bay Mar 20, 2025, 7:13 PM

#

rough forge such dir zwei punkte aus und verbinde sie

okay okay okay

#

bro es geht

#

ich muss nur noch die lücken zu machen

#

dann hab ich es drauf

rough forge Mar 20, 2025, 7:33 PM

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daring bay ich muss nur noch die lücken zu machen

hoffentlich machst du sie mit beton und nicht zahnpasta zu+

daring bay Mar 20, 2025, 7:34 PM

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rough forge hoffentlich machst du sie mit beton und nicht zahnpasta zu+

OMG

#

#

also wenn a und b linear unabhängig sind dann ja

#

wenn abhängig dann nein

rough forge Mar 20, 2025, 8:12 PM

#

ja genau

pearl pondBOT Mar 20, 2025, 8:38 PM

#

@daring bay Has your question been resolved?

daring bay Mar 20, 2025, 8:38 PM

#

rough forge ja genau

fuck bro

#

es ist schon viel ngl

daring bay Mar 20, 2025, 10:16 PM

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bro letzte frage

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bevor ich schlafe

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@rough forge

#

also BCDE ist ja klar

#

aber wie bekomme ich FGH

pearl pondBOT Mar 21, 2025, 12:16 AM

#

@daring bay Has your question been resolved?

pearl pondBOT Mar 21, 2025, 2:16 AM

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rain vessel Mar 21, 2025, 2:23 AM

#

I was taught that for a differential equation of the form ay''+by'+cy = 0 where a and b are constants that the general solution will be y = ke^(rx) + nxe^(rx)

pearl pondBOT Mar 21, 2025, 2:23 AM

#

rain vessel I was taught that for a differential equation of the form ay''+by'+cy = 0 where ...

rain vessel Mar 21, 2025, 2:23 AM

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k and n are constants

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r is the root of the polynomial am^2+bm+c=0 with multiplicity two

#

I'm having a bit of trouble extending this to high order differentials

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I'm not sure what to do if we have a root repeated more than two times

plush bramble Mar 21, 2025, 2:29 AM

#

rain vessel I'm not sure what to do if we have a root repeated more than two times

Take a guess

pearl pondBOT Mar 21, 2025, 2:41 AM

#

@rain vessel Has your question been resolved?

rain vessel Mar 21, 2025, 2:41 AM

#

no idea

#

@plush bramble sorry for responding so late

#

My best guess is that for say a polynomial with root r of multiplicity 3

#

y = ae^(rt)+bte^(rt)+ct^(2)e^(rt)

#

just because the main difference that stood out to mean between repeated roots and different roots was that extra t

#

I have no idea how these things are working under the hood, I only really know what to do but not why

plush bramble Mar 21, 2025, 2:48 AM

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rain vessel y = ae^(rt)+bte^(rt)+ct^(2)e^(rt)

This looks right

plush bramble Mar 21, 2025, 2:48 AM

#

rain vessel I have no idea how these things are working under the hood, I only really know w...

What is your actual question then

plush bramble Mar 21, 2025, 2:50 AM

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rain vessel y = ae^(rt)+bte^(rt)+ct^(2)e^(rt)

But also you should plug into the original DE to verify

rain vessel Mar 21, 2025, 2:54 AM

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plush bramble This looks right

damn, I literally purely guessed that

#

I think I should review the supplemental videos my teacher gave me a bit more

#

I literally have no clue what's going on

#

anyways thanks

#

.cloes

#

.close

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young dagger Mar 21, 2025, 3:11 AM

#

pearl pondBOT Mar 21, 2025, 3:11 AM

#

young dagger

young dagger Mar 21, 2025, 3:11 AM

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Help

#

I got x=0.5 and x=8

#

But the solution says 0.5≤x≤8

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Idk what I did wrong

feral sedge Mar 21, 2025, 3:20 AM

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Nothing

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Oh wait

#

I'd substitute y = log_2(x)

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I can't tell you what you did wrong since I don't know what you did though

plush bramble Mar 21, 2025, 3:28 AM

#

young dagger But the solution says 0.5≤x≤8

Show

young dagger Mar 21, 2025, 3:30 AM

#

plush bramble Show

This is the solution

#

I dont get it

plush bramble Mar 21, 2025, 3:31 AM

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That is quite surprising

plush bramble Mar 21, 2025, 3:33 AM

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young dagger This is the solution

Which line in the solution confuses you first

young dagger Mar 21, 2025, 3:34 AM

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Oh wait

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On second thought

#

I think i kinda understand

#

Thanks

#

.close

pearl pondBOT Mar 21, 2025, 3:35 AM

#

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sly void Mar 21, 2025, 4:17 AM

#

Q. If a,b,c,d are positive integers such that a = bcd, b = cda, c = dab, d = abc, then find the value of [(a+b+c+d)^4]/[(ab+bc+cd+da)^2]

pearl pondBOT Mar 21, 2025, 4:17 AM

#

sly void Q. If a,b,c,d are positive integers such that a = bcd, b = cda, c = dab, d = abc...

cinder thistle Mar 21, 2025, 4:19 AM

#

multiply the 4 eqn given to you

sly void Mar 21, 2025, 4:19 AM

#

then

#

it's (abcd)^3

#

then

cinder thistle Mar 21, 2025, 4:22 AM

#

so
abcd = (abcd)^3

#

as you know that a,b,c,d are positive inetgers, you can solve the value of a,b,c,d

compact ridge Mar 21, 2025, 4:22 AM

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lmao the everything is 1 moment

cinder thistle Mar 21, 2025, 4:25 AM

#

compact ridge lmao the everything is 1 moment

shh

compact ridge Mar 21, 2025, 4:25 AM

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it's pretty obvious after this step anyways

cinder thistle Mar 21, 2025, 4:25 AM

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||it was obvious from the beginning tho||

compact ridge Mar 21, 2025, 4:25 AM

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no but I wouldn't spoil it if I did not already see (abcd) = (abcd)^3

#

cause (abcd) = (abcd)^3 is legitimately basic algebra

cinder thistle Mar 21, 2025, 4:26 AM

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true

pearl pondBOT Mar 21, 2025, 4:35 AM

#

@sly void Has your question been resolved?

pearl pondBOT Mar 21, 2025, 4:50 AM

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lucid moth Mar 21, 2025, 11:12 AM

#

pearl pondBOT Mar 21, 2025, 11:12 AM

#

lucid moth

lucid moth Mar 21, 2025, 11:13 AM

#

I dont understand how he did these steps

#

The question was to integrate cos^2(x)

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cos^2(x)

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(1/2)(1 + cos(2x))

tardy reef Mar 21, 2025, 11:13 AM

#

You know the double angle trig formulae right?

#

Whats cos(2x)?

gloomy cliff Mar 21, 2025, 11:14 AM

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it's the double angle formula

#

cos(2x) = cos²x - sin²x

compact ridge Mar 21, 2025, 11:14 AM

#

gloomy cliff cos(2x) = cos²x - sin²x

shhhhhh they probably know this already or some variation

lucid moth Mar 21, 2025, 11:14 AM

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But im confused about the steps he took in this image

#

He put the 1/2 in front of the integral

#

Is that allowed?

gloomy cliff Mar 21, 2025, 11:15 AM

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constants can be pulled out of the integral buddy

tardy reef Mar 21, 2025, 11:15 AM

#

Well, if you write the full formula, youd get cos(2x) = 1 - 2cos^2(x)

#

and you rearrange the terms to get the cos^2(x)

#

it becomes exactly what the person wrote

gloomy cliff Mar 21, 2025, 11:16 AM

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tardy reef Well, if you write the full formula, youd get cos(2x) = 1 - 2cos^2(x)

you can get this from cos(2x) = cos²x - sin²x by simply substituting sin²x = 1 - cos²x

lucid moth Mar 21, 2025, 11:16 AM

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tardy reef it becomes exactly what the person wrote

Ah I see

#

Alright thanks!

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❤️

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.close

gloomy cliff Mar 21, 2025, 11:16 AM

#

it's just more simplification

pearl pondBOT Mar 21, 2025, 11:16 AM

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mental magnet Mar 21, 2025, 12:19 PM

#

Do I use product rule or chain rule for this question?

pearl pondBOT Mar 21, 2025, 12:19 PM

#

mental magnet Do I use product rule or chain rule for this question?

supple mural Mar 21, 2025, 12:20 PM

#

mental magnet Do I use product rule or chain rule for this question?

both

spark parcel Mar 21, 2025, 12:22 PM

#

mental magnet Do I use product rule or chain rule for this question?

oh ive done this question

pearl pondBOT Mar 21, 2025, 12:34 PM

#

@mental magnet Has your question been resolved?

pearl pondBOT Mar 21, 2025, 12:49 PM

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versed remnant Mar 21, 2025, 1:48 PM

#

Just a random question but can someone recommend me a good guide video on MATLAB?

pearl pondBOT Mar 21, 2025, 1:48 PM

#

versed remnant Just a random question but can someone recommend me a good guide video on MATLAB...

frail temple Mar 21, 2025, 1:52 PM

#

matlab onramp

#

https://matlabacademy.mathworks.com/details/matlab-onramp/gettingstarted

MATLAB Onramp | Self-Paced Online Courses - MATLAB & Simulink

Learn the basics of MATLAB through this introductory tutorial on commonly used features and workflows. Get started with the MATLAB language and environment so that you can analyze science and engineering data.

compact ridge Mar 21, 2025, 1:53 PM

#

versed remnant Just a random question but can someone recommend me a good guide video on MATLAB...

not a video but

#

highly recommend "MATLAB : a practical introduction to programming and problem solving" by Attaway

#

to use as a book

compact ridge Mar 21, 2025, 1:54 PM

#

versed remnant Just a random question but can someone recommend me a good guide video on MATLAB...

also can't stress practicing problems enough

#

and MATLAB's Cody is super super nice to get the basics covered and practiced

versed remnant Mar 21, 2025, 1:54 PM

#

Okie Ty , first time using this xD

#

Gotta check it out before I can do any problems lol

#

.close

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torn sky Mar 21, 2025, 2:16 PM

#

Given an equation for a plane, and a line that goes through it, how can you check at what points the line goes through the plane?

pearl pondBOT Mar 21, 2025, 2:16 PM

#

torn sky Given an equation for a plane, and a line that goes through it, how can you chec...

spare lark Mar 21, 2025, 2:17 PM

#

so the line isn't include in the plane it just go through at a certain point ?

#

in that case you have a parametric form of your line or cartesian ?

torn sky Mar 21, 2025, 2:18 PM

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I'll pull up the task, 2 sec

#

So given the parametric equation

x = 1+2t+s
y=-1+t-s
z=2+t+2s

and equation for the plane

3x-3y-3z = 0

A line l is given by the parametric equation

x = 1+t
y = 2t
z = -2-t

Find the coordinates where l crosses the plane

spare lark Mar 21, 2025, 2:21 PM

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ok

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so you know that the point has to verified l equation and plane equation

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so you take the values of x,y,z in line equation

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and you plug in the plane equation and solve for t

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and so to get the coordinates you take the line equation back and you plug in t value

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to get x,y,z

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that will be your point

torn sky Mar 21, 2025, 2:23 PM

#

i see, I'll go try that

torn sky Mar 21, 2025, 2:24 PM

#

spare lark that will be your point

I ended up with 9 = 0.

I assume this implies that it doesn't cross?

spare lark Mar 21, 2025, 2:28 PM

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it is

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and there is a reason for it

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the directionnal vector of l

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is (1, 2, -1)

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where the orthogonal vector of the plane is (3, -3, -3)

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doing dot product gives you 0

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so they're orthogonals

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which means

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the line is either parralel or included in the plane

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to make the difference

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take a point on l

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let t = 0 for example

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so you have that the point P_0 = (1, 0, -2) lies on l

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lets check if its on the plane, if yes, it is included, else its parallel

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by a rapid check with the equation of the plane we see that its clearly not in it

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so they're parralel

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which means no intersect

torn sky Mar 21, 2025, 2:35 PM

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ah alright, thats pretty interesting

pearl pondBOT Mar 21, 2025, 2:50 PM

#

@torn sky Has your question been resolved?

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lament knot Mar 21, 2025, 8:58 PM

#

Yo

pearl pondBOT Mar 21, 2025, 8:58 PM

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lament knot Yo

lament knot Mar 21, 2025, 8:58 PM

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Kinda wanna continue my previous help convo

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Are we sure the halting problem function is not computable?

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There seem to be conflicting sources

#

My understanding rn seems to be that if for some elements in the domain, a function is not computable

#

But for some it is computable

#

The function is still called computable

#

. close

#

.close

pearl pondBOT Mar 21, 2025, 9:12 PM

#

Channel closed

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meager trellis Mar 21, 2025, 9:16 PM

#

.reopen

pearl pondBOT Mar 21, 2025, 9:16 PM

#

✅

meager trellis Mar 21, 2025, 9:16 PM

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lament knot Are we sure the halting problem function is not computable?

to answer this: yes, we are

lament knot Mar 21, 2025, 9:16 PM

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Oh really

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So the set of partial recursive functions does not include the halting problem?

meager trellis Mar 21, 2025, 9:17 PM

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meager trellis to answer this: yes, we are

the sources you found that you think are conflicting are either wrong or you're misunderstanding what they're saying

meager trellis Mar 21, 2025, 9:18 PM

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lament knot So the set of partial recursive functions does not include the halting problem?

yes, in the sense that e.g. no recursive function takes a turing machine and outputs 1 if the machine does halt and 0 if it doesn't

#

if what you mean by "the halting problem is not computable" is not equivalent to this statement then that might be part of why you're confused

#

in particular bringing up partial recursive functions seems kind of odd given that the halting problem is not a partial function, every turing machine either halts or doesn't

lament knot Mar 21, 2025, 9:23 PM

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Umm

meager trellis Mar 21, 2025, 9:23 PM

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lament knot My understanding rn seems to be that if for some elements in the domain, a funct...

also this is the wrong way to think about it

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a function isn't computable or not for some element of its domain, the whole function is either computable or not

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for any particular program there is a program that solves the halting problem for that program: either the program that outputs "yes it halts" or the program that outputs "no it doesn't halt"

lament knot Mar 21, 2025, 9:26 PM

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meager trellis a function isn't computable or not for *some element of its domain*, the whole f...

We define halting problem over the set of all possible {turing machine, input tape} pairs to {0,1} halts or not halts right?

meager trellis Mar 21, 2025, 9:26 PM

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yes

lament knot Mar 21, 2025, 9:27 PM

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For some pairs it's computable, but for some pairs it's not computable

meager trellis Mar 21, 2025, 9:27 PM

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no

#

each pair, individually, is computable, by one of these two programs:

"output 0"
"output 1"

lament knot Mar 21, 2025, 9:28 PM

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Ah

#

I see what u mean

#

So we can only talk computability for the entire function

meager trellis Mar 21, 2025, 9:29 PM

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yes

lament knot Mar 21, 2025, 9:29 PM

#

So the set of partial recursive functions does not include the RE languages?

#

It's only the R languages ?

meager trellis Mar 21, 2025, 9:32 PM

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well the recursive functions are basically just the languages in R
partial recursive functions are something a bit more complicated

lament knot Mar 21, 2025, 9:33 PM

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meager trellis well the *recursive* functions are basically just the languages in R partial rec...

Recursive meaning total recursive?

meager trellis Mar 21, 2025, 9:34 PM

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yes

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...i'm not sure if there's an easy characterisation of partial recursive functions in terms of total recursive functions/things like R and RE, actually

lament knot Mar 21, 2025, 9:35 PM

#

meager trellis Mar 21, 2025, 9:35 PM

#

but the halting problem is still definitely not partial recursive given that it's not even a partial function lol

lament knot Mar 21, 2025, 9:36 PM

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meager trellis but the halting problem is still definitely not partial recursive given that it'...

Makes sense

lament knot Mar 21, 2025, 9:37 PM

#

lament knot

Did you take a look at this

meager trellis Mar 21, 2025, 9:37 PM

#

yes

lament knot Mar 21, 2025, 9:38 PM

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Turing machines that do not terminate for certain inputs, in case of halting problem would be inputs which make it loop forever

lament knot Mar 21, 2025, 9:38 PM

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lament knot

This bit makes me think it's suggesting that HP is a partial function

#

As it's drawing a connect between "non termination" and "undefined result for given input"

#

If non termination implies undefined result then the hp function is undefined for certain elements in its domain, meaning its a partial function

meager trellis Mar 21, 2025, 9:40 PM

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...no, that's completely wrong

lament knot Mar 21, 2025, 9:40 PM

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Oh

#

How so

meager trellis Mar 21, 2025, 9:40 PM

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the halting problem is a total function, it always outputs either "halts" or "doesn't halt"

lament knot Mar 21, 2025, 9:42 PM

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Hmm

meager trellis Mar 21, 2025, 9:42 PM

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the article is very specific about what "connection" it's talking about between non-termination and not being defined and it very clearly has nothing to do with the halting problem

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specifically it's saying that you can define that a partial function is computable iff there's some turing machine which, when ran on an input where the function is not defined, runs forever, and when ran on anything else, halts and outputs the value of the function at that input

lament knot Mar 21, 2025, 9:44 PM

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I see

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But such a function which lists out all {tm, input tape} pairs such that tm will halt on input tape is computable

meager trellis Mar 21, 2025, 9:52 PM

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yes, but that's not the same as computing the halting problem

lament knot Mar 21, 2025, 9:52 PM

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Yea

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Ok so

meager trellis Mar 21, 2025, 9:53 PM

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meager trellis yes, but that's not the same as computing the halting problem

in fact the sets for which you can computably list their elements like that are exactly the RE sets

lament knot Mar 21, 2025, 9:54 PM

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Partial recursive functions are all the functions which can be computed by a Turing machine, meaning the machine halts on all inputs, so they are the decidable language membership functions right?

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And partial recurisve functions given as they generalise total recursive functions are the exact same set as computable functions

meager trellis Mar 21, 2025, 9:55 PM

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lament knot Partial recursive functions are all the functions which can be computed by a Tur...

no, a machine computing a partial recursive function usually won't halt on all inputs, it halts on exactly the inputs where the function is defined

lament knot Mar 21, 2025, 9:56 PM

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This is whats bothering me

meager trellis Mar 21, 2025, 9:56 PM

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the partial recursive functions that are also total functions are the same as total recursive functions

lament knot Mar 21, 2025, 9:57 PM

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Why do we care about inputs where the function is not defined

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In the first place

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Is it because the domain is not defined