#help-17
1 messages · Page 44 of 1
yes
that’s not correct
1 0 0 | alpha -1
0 1 alpha-2 | 0
0 0 2a-4 | -2b-2
if i am trying to finagle this form i get alpha = 2 for no solutions if b-1 is not equal to 1
why no solutions
why is it independent of b
i typed too slow, it is dependant on b unequal 1
if alpha is 2 and b is 1 then i have exactly one solution right?
no
the last row can be three options. either 0=0, 0 = nonzero or nonzero=nonzero. to which of 0, 1 or infinite solutions does each correspond
whoops, forgot nonzero=0
0=0 is infinte solutions
0 = nonzero is no solution
nonzero = nonzero is one solution
right?
yes
so alpha = 2 beta -1 = all solutions
alpha = 2 beta not -1 no solution
none of the above one solution?
yes
@uneven jay Has your question been resolved?
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Guys I need help really really bad I feel so upset
Ping me when you are here
What's up?
might be a good idea to calm yourself down if you want us to help you w/ math
also just post your question
How do you flip the picture?
Oh
Okay
Is the method for this add them toghether divide them by 2
?
"the" method 
what's wrong is the unspoken assumption that every single problem, or at least this one in particular, has one and only one correct way of solving it, all others declared wrong outright
Oh
but anyway, yeah, sure.
let b, s and t be the heights of the table, soccerball and tennis ball as shown here
then you are told b + s - t = 100 and b + t - s = 60
Yes
adding these together does indeed give 2b = 160
would you like us to address your state of upsetness?
What do you mean?
you came here complaining that you are upset over this problem
do you or do you not want us to address that
,rccw
ok, so you are upset over something else. still, do you want us to address that?
I'm confused on that; what do you mean?
Yes
Oh yeah
Math questions
Making me upset
Wait
Let me try to solve it first
Does anyone have an idea on how to do this
Why are you here?
Go to help
@paper depot ok I will try to solve it
Help-21
Is the answer 15, Ann?
Is that correct
yes, it is
Yay
And then
Unfortunately
The questions got harder
Let me post it
I'm not very good at factorials
Is there a method?
So am I right the last number will be 0
Because anything after 6! Is 0 at the end?
,rccw
So am I right the last number will be 0
Because anything after 6! Is 0 at the end?
this wording is slippery
Ok, let me rephrase
After the factorial 6!, The last digit of every factorial after 6 factorial is 0 as the last digit
Am I correct?
clunky, but yes.
are you on a deadline?
Yes
:x
ok
keep in mind that you want the last 3 digits and not only the last 1 digit, by the way.
This is difficult for me
Yes I know
I am only able to figure out one
But I don't know the method for the other 2
finding the last 3 digits of a number is the same as taking its remainder on division by 1000
in the same way as the one last digit is the remainder on division by 10
I'm a bit confused what you mean
let me edit the word "m*d" out.
Okay
do you know what division with remainder is?
the remainder of a number on division by 10 is equal to the number's last digit.
the fact that you're phrasing this as a decimal is a little bit sus.
i am talking about division with remainder.
Can you give me an example?
which would be stated in words as "Dividing 283 by 10 gives quotient 28 and remainder 3."
and with an equation as 283 = 28 * 10 + 3.
yes, the asterisk means multiplication.
Oh ok
would be stated in words as "Dividing 283 by 10 gives quotient 28 and remainder 3."
and with an equation as 283 = 28 * 10 + 3.
this is the example i gave
Oh ok
i took your example and rephrased it properly.
anyway
at this point you could simply
write out what the last 3 digits of all the factorials are
you don't need to.
What do I do then?
5! ends in 120
6! ends in 120 * 6 = 720
7! ends in 720 * 7 = 5040 = 040
8! ends in 040 * 8 = 320
9! ends in 320 * 9 = 2880 = 880
10! ends in 880 * 10 = 8800 = 800
11! ends in 800 * 11 = 8800 = 800
12! ends in 800 * 12 = 9600 = 600
and so on...
i omitted 2! and 4! as they are obvious
So what's the relation?
Oh wait
If I add long numbers over and over again
Would it be 000?
we are not adding, we are multiplying.
and yes, the last 3 digits of the factorials do eventually end up zeroing out.
,rccw
,rrtw
it's ,rccw, for rotate counter-clockwise.
So what's the hint in the bottom for
i do not see any hint at the bottom.
that's not a hint, that's a reminder of the definition of factorial.
Oh ok
no, it is not.
you are confusing the sum with its last term
What do you mean?
considering the factorials individually, you will get that they end in 000 starting from a certain one onwards.
this alone doesn't mean the entire sum is equal to zero.
Oh wait
You are right
Listen
Tell me if correct
So
2 + 6 + 24
Oh wait
It's even factorials
So
2 + 24
So it should end with 026 right?
no
there's a lot of terms that aren't zero but which you have thrown out.
why did you throw out 6!, which is 720, and doesn't end with three zeros?
sure does, but you're interested in the last three digits.
no, this is not an impossible question.
Oh
the terms DO zero out at some point -- it just isn't at 6 factorial!
Oh ok
i worked out the last 3 digits of a bunch of factorials for you here.
you should keep going here and find the first factorial that zeros out.
Ok watch me
600 * 13 = 7800
5! ends in 120
6! ends in 120 * 6 = 720
7! ends in 720 * 7 = 5040 = 040
8! ends in 040 * 8 = 320
9! ends in 320 * 9 = 2880 = 880
10! ends in 880 * 10 = 8800 = 800
11! ends in 800 * 11 = 8800 = 800
12! ends in 800 * 12 = 9600 = 600
Why does it get lower the number?
i do not understand your question.
Why is 13 factorial lower than 12 factorial?
Why is 13 factorial a smaller number than 12 factorial?
it isn't.
But I was using your method?
5! ends in 120
6! ends in 120 * 6 = 720
7! ends in 720 * 7 = 5040 = 040
8! ends in 040 * 8 = 320
9! ends in 320 * 9 = 2880 = 880
10! ends in 880 * 10 = 8800 = 800
11! ends in 800 * 11 = 8800 = 800
12! ends in 800 * 12 = 9600 = 600
and in doing so, you got that 13! ends in what?
Wait
5! ends in 120
6! ends in 120 * 6 = 720
7! ends in 720 * 7 = 5040 = 040
8! ends in 040 * 8 = 320
9! ends in 320 * 9 = 2880 = 880
10! ends in 880 * 10 = 8800 = 800
11! ends in 800 * 11 = 8800 = 800
12! ends in 800 * 12 = 9600 = 600
It ends with 800
13! ends in 800.
so then why do you ask me such 2+2=5 type questions
you use it to find the last three digits of 14!, and then of 15!, and then of 16!, and so on...
you asked me "Why is 13! < 12!" knowing full well the statement 13! < 12! is false
That was a accident
you said it twice
Ok so find 15 factorial then
so it cannot be an accident
Show me how you would find 15 factorial
i would not find 15! itself
for the purposes of this problem i would only find its last 3 digits
13! ends in 800
Show me how you would find the last 3 digits of 15 factorial
14! ends in 800 * 14 = 11200 = 200,
and 15! ends in 200 * 15 = 3000 = 000.
yes, exactly.
I was adding
this answers neither of my questions.
4! is not 218 nor is it 12
why are you putting 5! in your sum in the first place?
again how did you get this
I added the even factorials
show, don't tell!
this isn't what i wanted you to do.
show me exactly what numbers you added and exactly what your result is.
i think you've forgotten a term.
which word in my message needs explaining? "i", "think", "you", "have", "forgotten", or "term"?
I didn't forget any term?
yes you did.
Which one?
2!
do you think i am going to remember this after several pages between you posting the pic and now
you're gonna have to repost it -- ideally in proper orientation this time
,rccw
I have my pencil ready for all the math
k
if you didn't care whether or not the rectangles were touching, could you count the ways to shade exactly two of them in
I'm confused what do you mean?
the way you worded it was a bit confusing i’m pretty sure
Who are you? please, someone is already helping, a genious
okay mb nvm
@hasty idol if your problem read
In how many ways can exactly two of these nine rectangles be shaded?
would you be able to do it? yes or no
how did you count that?
I can send a picture of the way I did it
you realize that, for the purposes of finding PAIRS OF RECTANGLES, you've grossly undercounted them, right?
How?
you counted only these pairs:
1, 2
2, 3
4, 5
5, 9
6, 7
7, 8
and other pairs such as 1, 3 or 1, 4 or 1, 5 are nowhere to be seen
How does 1 three work?
Ok so since there is a number there is 8 rectangles
6 pairs
8 + 6
it works by shading in rectangle #1 and rectangle #3 ...
no, you're confusing yourself!
How?
you're overthinking it and confusing yourself!
i deliberately simplified the problem to NOT care about how the rectangles are placed!
Oh ok
there are just 9 rectangles suspended in midair! how many ways are there to select 2 of them?
6
incorrect!
How?
1 2
1 3
1 4
1 5
1 6
1 7
1 8
1 9
2 3
2 4
2 5
...
Do you mean 1, 2, and 3 shaded
I feel like he doesn't know how enumeration works at it's core
I didn't ready the question properly
god fuck i need a moment
C CZSdMZD BMZGD' STGF
JTDGFNS
GTFmDGFHN
STH
HOW THE FUNSFKDHG
Want me to pick it up from here ?
I understand
how the hell do you expect to get help if you don't read things people say to you properly
yeah go ahead imma go try to cool down
Sorry!!!!
Okay so
No ?
Are you even good at math?
I suppose so, I'm pursuing my masters degree
I guess yeah
There's no triangle
Rectangle sorry
Yes
And then try and enumerate all the other rectangles you can select to form a pair
What does enumerate mean?
Basically just write down or jot down
This is going to be long at first
But I need you to do this to understand how it works
Ok
So
Go ahead
So
1, 3
1, 5
1, 9
1, 6
1,7
1,8
That's already 6
Now
2, 6
2, 7
2,8
3
Now
3,1
3,4
3,5
3,8
3,7
3,6
That's 6
Now
?
1,3 and 3,1 is the same pair of two rectangles
5,6
5,1
Sounds like you've accidentally stumbled upon the formula
That's 2
Didn't you notice anything ?
6,1
6,2
6,3
6,5
6,9
6,7
Look at what @vast shale said
Why AFTER ?
Because so I know which ones to remove after
Look at each single rectangle and figure out which ones are touching
Ok now remove the pairs that occur more than once
Jesus christ all mighty
What?
What you just wrote right now
Copy and paste all of it in a text editor or whatever
And remove the pairs that occur more than once
I know I need to write them on paper
Why would you need to that ?
This is not even how you're supposed to solve this question
I need to write them on paper so I can cross out the same ones
Okay if you need to..
Take your time
Do you have the answer?
Yeah I do man
Mb I thought 7 and 9 touched each other on the picture
It wasn't clear enough to see
Its 22 right?
Now if you notice there a pattern. You didn't cross out anything on the first rectangle, nor the second
But the further you went the more you crossed out
It's 21
how?
You were correct here
Yep
Ok I have another question
Yes
,rccw
23 was supposed to be 22. 23 is a mistake
Do you know what a prime number is ?
Yes
Good can you give me ONE example where a + b + c = 22 and with a, b and c are prime
a, b and c are all smaller than 22
Yes
So I'm going to need you to list me all the primes that are smaller
Out of all of these do you think you can find ANOTHER combination that will sum up to 22 ?
2 + 3 + 17
You figured it out.
There are only three numbers that amount to 22 when added
2, 3 and 17
Now here's the thing
What?
We are asked a number of triples (a, b, c) which means that (2, 3, 17) isn't (3, 2, 17)
Do you agree ?
Yes
Give me all the combinations
Ok
Your ones and
(2,3,17)
3,2,17
17,3,2
17,2,3
2,17,3
3,17,2
Those are all of them
6
There you go
Where did you get 12 from ?
.
YOU forgot about them
Why
Yeah so 6 from each
Yep yep, my bad, I'm sorry I got really frustrated
,rccw
Seconds
Are you sure ?
How did you reach that
By using my brain
Nice
100*4 = 400
How do I calculate distance?
It is given to you
4 metres per second
By reading
It's written there
It's written there
Your train of thought needs to be clear
Ok
Of course it's written but you need to show how you got there besides "my brain"
So 100 seconds for one lap
for person A
And 80 seconds for one lap for person B
Yeah so you need to figure out how much distance they're going to cover in 10 minutes
So in 10 minutes how many laps is person A going to run
And B ?
We're not gonna count halfs here
Ok
So we'll just say 7
Ok
Which is the floor value
They will pass each other 13 times
Is that correct?
Yes
WOOW
And we can prove this by using another method
THANK YOU
I AM SO HAPPY I FINALLY UNDERSTAND THIS TYPE OF QUESTION
Ok let's prove it now
Yes
Yes
And since they move in opposite direction (which is a key element)
They have 400m between
5400/400 =
13.5
It is 13
There you go
Ok I have another question
I have class in like 2 min hahhah
: (
Wait
What class do you have
Graph Theory 2
Why?

?
Because he's my teacher
Okay so you want me to ask my teacher to help someone on a discord server
I have to go now, good luck
No wait
Pls bring someone else
Aww
@paper depot
I need help
I will now listen to you sincerely
<@&286206848099549185>
Hello
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tried this derivative problem 2 diff methods got stuck on both idk how to do it
am i doing it right to begin with tho?
Hi again @still anvil ! While the top part looks fine, you have an error on the bottom part. Remember, g is the denominator, not the numerator.
the stuff on the left is wrong because the derivative doesnt just "expand" into the denominator like that
the stuff on the right would work, but you somehow mixed up ur f and g in there?
where do i mix it up i thought i did the one with 7 as the denominator as g?
o
nvm
first off (and sorry if you did bu I can't read well the paper), separate the constant and put outside the derivative
ok so forg g^2 do i use the power rule on (7 3sqrtx^2)^2 to make it 2*7sqrtx^2
this derivative you do in 3 steps if you do the steps correctly
do what i said
which is the constant 5 or 7
who is acting over x?
7
5 isn't?
aha
5/7 * x??
so that out
eliminate the numerator
wait what why?
do it
you can see things clearly
when you don't have fractions
what is 1/a
isn't it a^(-1)?
yeah
so what's 1/x^(2/3)
so x^-2/3
derivate that
huh how
derivative of x^a?
nope
a
?
a*x^a-1
-2/3 * x^1/3
no
-1/3?
what's -2/3 - 1?
2/3
do you know how to substract fractions?
how much is -2/3 -1
No -5/3
you can say numbers until u want, no problem
the idea is you think before answering
it clicked at -5/3
so you have -2/3 * x^(-5/3)
don't forget about the constant you left out
can you add all the things together now?
you finished
rewrite back in fraction form
-10/(21x^(-5/3))
check your signs
so convert the -5/3 ?
what do you think?
yeah
use the definitions you have
you don't have to ask if 2+2 is 4
if you know something, use it
np
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Hey, what graphing calculator do you recommend for high school to university?
I could get the Ti-Nspire or Ti-84 plus CE, and they both are extremely appealing options
But im wondering what yall would go with and what you think its important to note
I will not be using it in any high school tests, but i do plan on using it in college physics and for personal studying
Im not particularly interested in programming it or gaming on it
I realize the Nspire is far better performance wise, but I've heard some say its overly complex to use, or that the buttons are too small
Nspire is a little heavy for most uses, but i guess if youre programming on it, it makes sense
In my opinion 84 is the best choice, it can do everything you need
I realize its a bit much, but I'd like something that lasts me through college, as paying 100+$ is something i like avoiding when possible
The 84 lasted me all the way through college, depending on the course, some would allow up to the ti 36pro
@vast shale Has your question been resolved?
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what have you tried?
it might help to solve to smaller right triangle, if you need a place to start
by smaller i mean the one formed by the shack, the ground, and the hypotenuse connecting the two
yea
sorry i was drivin
so lemme see here
im with you up to the 21.09
then i think we would say something like
$\tan \qty(21.09) = \frac{16+x}{100}$
janniku
then $x = 100\tan(21.09)-16$
janniku
yea, looks good
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Determine the values of a and b so that 𝑦 = (𝑎 + 𝑏)𝑒
𝑎+𝑏𝑥 note that 𝑥1 =
𝑎 − 1𝑏 a local minimum and at 𝑥2 = 𝑎 − 13𝑏 a point of Inflection (P.I.)
do you mean $$ y = (a+b)e^{a + bx} $$
even order group => solvable
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Can someone check if these two are correct
they are correct
what's amplitude defined as? crest to trough or crest to average?
thank u
crest to trough
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if it’s crest to trough, then it should be 10, so none are correct
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wait idk it just asks that
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Can someone check if this is correct
one way to check if it’s right is to substitute x = 0 into the function and see if it gives the right y
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Is this correct?
i am assuming 8
So is it correct?
I honestly haven’t learned this yet so I’m trying to find a yt video about how to know if a sequence is divergent or not
I haven’t found anything yet
If r > 1 or r < -1, it's divergent
yes this is correct
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✅
Is this also correct?
no 
I see, so iii is false then?
interesting that you went to iii
yes, iii is false, but why? can you think of a counterexample?
Iv seems false to me and i too
It'll approach the limit, but it'll never touch it. That's what I was thinking in mind
i mean "most" of the time that will be true, that a limit will equal the function's value
eg $\lim_{x\to 3} \frac{x-1}{x-2}$
Hayley
So was that thing saying that the value of f(a) is the same as L?
yeah (iii) is asking whether it's true that f(a) = L means the limit will go to the same place
which is true for continuous functions, that's the definition of continuity actually
So then it's false because f(a) isn't always = L?
we are in this case trying to find a function that has a point where it's approaching one spot but the actual value isn't at that spot
A side unrelated question, does (1/2)^infinite = 1?
Hayley
One last question, is this correct?
Alr, thx for the help!
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can someone check this
questions 8 and 9
@stuck harness Has your question been resolved?
these two entries don't look right
also the ink is kinda wearing off the paper, which makes it hard to read some things
im not sure how to fix it because i dont get the relationships
this stuff should be easy but out of everything in geometry im struggling with it
can you tell me how to find each measure for 45 45 90 given each value
you don't understand the relationship between x and y here?
$y = x\sqrt{2}$
Saccharine
yeah?
suppose you give me a number, and I multiply it by 2
so if you give me 3, I give you 6
then suppose you forgot what number you just gave me, but I give you 100
how might you figure out what number you gave me?
im not sure
i wanna say put 100 in radical for 10 but idk how that would help
yeah im stumped
bruh what
im not sure
let's simulate the doubling machine
stop jumbling the symbols in your head and just pay attention to the doubling machine
if you put a number in, it prints out a piece of paper with double that number on it
so if you write 23 on a piece of paper and put it into the machine, what do you get out?
46
okay now suppose I write something on a paper in secret and put it in the machine
and the machine spits out a paper with 66 on it
what number did I write?
maybe 33?
how did you figure out 33
66/2
so you just did something to reverse the multiplication by 2, right?
yea
so conceivably, we could invent a machine called the "halfway doubling" machine, which just multiplies the input by sqrt 2
how might you figure out what got put in, just given what the halfway doubling machine spit out?
[in a completely irrelevant detail, it's called the "halfway doubling" machine because if you put the output of the machine back in, you get double the original number]
saccharine i dont think this method of learning helps me because english isnt my first language 😭
can you teach the way a textbook would like this equals that and blah blah blah
i get that that method of teaching doesnt let u fully understand how things work and you just mindlessly do them but thats how im used to it
I mean if you want that, you should probably consult a textbook
I'm trying to explain something that you didn't get out of a textbook
thats my mistake cuz i dont think its very easy for u to teach if my english isnt the best
thank you anyway
I think you're missing something fundamental here
otherwise I'm not sure what's hanging you up
like if you know that y is sqrt(2) times x
then how is it that you can't figure out how to find x from y?
the doubling machine example I gave was when y is 2 times x
i found out how to do it, like if y = 24 i just divide it by rad 2 then simplify which turns into 12(rad2)
i think im just burnt out because summer school is not fun
an entire course of geometry squeezed into 6 weeks which means we skim the material and you dont get a true undersanding of how geometry works
wait until you get to college lol
then it's an entire intro to linear algebra or something squished into 10 weeks, and you have like 5 of them
is the course originally a years worth of stuff?
cuz for our summer school its incredibly scuffed like a regular school year condensed for college credit
well, they teach it in a year in high school, but in college, things get much faster paced
something might just be a semester or a quarter
thats gonna suck because im going into computer science
i think i need high math req
computer science doesn't require super high math
CS majors got a free math minor at my school
just your introductory undergraduate stuff I think
damn sign me up
the math minor was such a pain that I gave up
and a math major was like +6 classes or so
anyways i think my problem is resolved
make sure you fix the other table too
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i need help again on this
ive been doing hw for almost 8 hours because this is our most packed chapter yet
try finding the length of this side
6
then use the fact that the triangle is a 30-60-90 triangle
i tried x(rad3) = 6
it didntwork
nvm
i got a simple answer of 2(rad3)
y = 2 and x = 2(rad3)?? @rapid swift
i am actually so confused
so after i found out the side u told me to find was 2(rad3)
that means the shortest side or Y needs to be 2
right"??
which side is 2 rad 3
longer side
didn't you just say that side was 6
why is that the case
because when i set xrad3 to 6 u get 2 rad 3 = x
why should x rad 3 = 6?
sacharrine i dont think any of my solutions are making sense i am not lying when i said ive been nonstop working for 8 hours
well, it's not particularly productive to work on something 8 hours if you're making it nowhere
try taking breaks and do something else
also try not to stare at something for so long and perhaps review other things
same thing happens with university math homework
you read the problems, then you ponder something else
staring at them doesn't help