#help-23
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Yes
@naive osprey Has your question been resolved?
No
@naive osprey do vu have to find the area of the whole shape
Yes
7
okk
√6^2 - 2^2
= 5,66
5,66•4•1/2
11,32•7
= 79,24
But ChatGPT says the answer is 84
Because of the Pythagorean
11,32 ig?
pythagoras works on right angled triangle right
yes
is the bigger triangle a right angle triangle
it isn’t the height
the method chatgpt tried
was
0.5* perimeter * apothem
apothem is center to the side
this is only applicable for regular polygons
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"crypto casino"
Very sad
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My professor sent me this slide and the middle definition is confusing me, I'm not sure what it meant by bit-level and pattern-level, I would like to see some examples to clarify this.
- Does "NOT (x AND y)" has the pattern level of 2?
- Does "NOT (x OR y OR z)" has the pattern level of 2 or 3?
- Does "x NAND y" has the pattern level of 1 or 2 (since NAND could be said to be made up from NOT and AND)
-
- But since NAND is the universal logic operation, does that mean the opposite that "x AND y" is actually "NOT (x NAND y)" and therefore has pattern-level of 2?
- Does "1011 AND 1001" has the bit-level of 2, or 4?
- Does "(1011 OR 1110) AND 1011" has the bit level of 12, or 8, or 2 if the original statement is "(x OR y) AND x"?
the slide doesn't say there is a number associated with bit level or pattern level
it just means a logical operation at the bit level is an operation on a single bit, e.g. 0 AND 1 = 0, and on the pattern level is a bunch of bit level operations on corresponsing bits, e.g. 011 AND 101 = 001
a pattern-level operation here simply means an operation applied to one or two bitstrings, whereas a bit-level operation is an operstion on individual bits.
Mind me asking what kind of math that is
digital logic, mostly.
Ahhh okay thanks
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@tidal haven Has your question been resolved?
Differentiate the Pythagorean theorem with respect to time
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@tidal haven Has your question been resolved?
however this will only give for 1 time where the car is moving towards the intersection
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My teacher got 81k
you went from 3^-3 in line 2 to 3^-2 in line 3
ye that is fine
Js take negative sign out from numerator
yea
.close
Always
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i love your handwriting
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i need to use the iroc formula
@meager igloo
im sorry its 3 in the morning i have an exam tomorrow i really need this
huh what's the bot doing
no, geography
this is gr12 so it might be easy for u
damn mb
js pls
help
this is the formula i need to use @mighty mango
you surely are supposed to use differnetiation rules... right?
I can help if you know the diuffernetiation rules
otherwise, I have quite hte trauma with that specific formula
whats differnetiation rules
☠️
okay... yeah I'm sorry, I cant differnetiate with that specific formula
damn
,tex .diff trig
Xavier 🌺
i dont think this is what its about
This combined with chain rule should get you there
they have to differnetiate using the limit defn
You're showing us the answers to 4
the textbook has the question answers mixed up
question 4 is the answer to question 5 its confusing
but u just gota figure it out by looking at the answer
oh wait
nvm that is the right answer
The unit of the equation isn't years anywhere
Yes
so how is the answer "answers may vary"
cus theres different methods correct?
i need to use this formula, and h=0.0001
I mean in that case it looks like you've got your work cut out for you
is what youd think
but i got 1800 as my answer
which is pretty different than 4.19
Did you remember to set the calculator to degrees
(or to convert the stuff to radians)
wel its an exam, and a bunch of units need it to be in radians
so my calc is in radians
Well then did you remember to convert 120° to radians
oh
You either change your calculator to work with degrees or change the formula to be in radians
(don't do both)
yeah so how would i edit the formula to be in radians
Do you know how to convert from degrees to radians
we ac we were never told that
pie minus 60 degrees?
wiat ik pie is 180 degrees
bro im confused as shit
When a question has degrees in it, it needs degrees
but we always used radians for the degree questions like
theta and beta
quadrants
all that
this is our lesson if that helps
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Help me
,rccq
,rccw
,rc
thanks mate
np
set up the integral then you use integration by parts i believe
i dont want to brute force it , i was thinking of some clever way around it
i haven't written it out but i don't think it would be that ugly
i got the first result but the second one is something i cant get hold of
how far have you gone?
i think i got it, i may have made a mistake
im just running loops at this point
ok and .....?]
you start from the equation you get from the first part
yeah a(n) <1/n-1
$A_n < \frac{1}{2n-2}$ is equivalent to showing $A_n<\frac{1}{2}\left(A_n+A_{n-2}\right)$
Green
after simplifying the inequality, consider the interval over which we are integrating
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would the majority imply "in relation to the other content" or no?
in my mind 1ist statement is correct because it has the largest proportions relative to each other category
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for the question just want to make sure i'm doing this correctly.
to find out how far he is horizontally you do tan(35) = 125/a
and then for the direct distance you're trying to find the hyp so you take what you got for the horizontal^2 + 130^2 then take the root of it?
it should be 120 bcz the sailor is seen at the light
lighthouse is 50m and the cliff is 80m but the light from the lighthouse is 5m from the top
so 50+80-5 right?
Would be helpful to everyone to draw a diagram
right
?
125
does what i did look right scoob
yh
That's wrong no?
The lighthouse is 50m and the light is 5m below the peak
right so that's why i used 125 in the first part
I have it drawn out on my paper but I can't send a pic lolol
125 is correct, and scoob's diagram, while a bit awkward and off-scale (why is the 50 so much longer than the 80) also looks correct from the way he placed the 5
lmfaoooo ur awesome scoob
ah it labelled wrong but you get the feel
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Consider set S={1,2,3....50}, where m,n belong to S, if P is number of ways such that 6 to power m + 9 to power n is divisible by 5 and Q is the number of ways in which (m+n) is the square of prime number, find(P+Q)
i couldnt find the question online and wanted to verify my answer as i have a different answer from where i got the question
First of all, are for example (1,3) and (3,1) counted as 1?
It does not seem like the first part and the second part of your solution agree on this
no i thought theyd be different
counting as 2
which part did i count them same as? first or second?
The second part
for m+n to be 4 its either 1,3 or 3,1 and i did write 2 ways
9=
1+8
2+7
3+6
4+5
5+4
6+3
7+2
8+1
n can only be odd right as we calculated in the first part
Or does it?
whats the answer with your way?
It's the wording fault
i think it was 1333 where i got the answer id have to recheck
Idk I haven't done it, I have just read your
4 will have 3, 9 will have 8, 25 will have 24, and 49 will have 48
Q=83
so yeah 1333
thank you!
np
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Let two circles of equal radii with centres A and B intersect at C and D. Let EDF || AB as shown. IDH is another line not parallel to AB.
I lies on the minor arc ED between E and D and H on the minor arc FJ between F and J.
Q.1
ID < ED and DH > DF.
Is this true?
Q.2 If yes, how can one prove this?
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Does anyone know at which step I'm going wrong here?
You didn't multiply 3T by 5
Ohhh
When you multiply both sides, you do it for the whole sum
So wqhen you need to multiply something on one side, you multiply everything in the expression?
you multiply the ENTIRE left side and the ENTIRE right side by that
so really the fully 100% formally correct step would be $$5 \paren{3T + \frac{8T-8}{5}} = 5 \cdot 7$$ which then simplifies to $$5 \cdot 3T + 5 \cdot \frac{8T-8}{5} = 5 \cdot 7$$
Ann
This makes so much more sense
indeed! refer to the lovely distribution property you probably learned.
b-
okay 🤩
should be 8t - 8
good that you noticed
- -8
indeed
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i'm having trouble with this
prove that for any x,y in Z
if x in Z+ and y not in Z+ then xy not in Z+
if xy = 1 then either x=y=1 or x=y=-1
using the equivalence class definition and properties of Z
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
I know what I need to do but I dont know how to make it fully work without assuming things that rent explicity proven
by def, x = (a,b) (with a stick on top) with a>b, and y = (c,d) (with a stick on top) with d≥c
so i can multiply those and get (ac+bd, ad+bc) with a stick on top
but idk how to prove that it's not in Z+ by using a>b and d≥c and N inequality properties
2
ok I found a way for the first one even if it's too long its the only one that satisfies em
@formal grail Has your question been resolved?
You may find page 2 here to be helpful: https://crawford.elmhurst.edu/classes/m301/2019Sp/wk4.4_Divisibility.pdf
thank you! it might be honestly, i didnt consider that approach since we didnt really delve into divisors
they are defined in the material there just wasnt really any exercises around them
and i can use this, which isnt in the material, because its the logical conclusion of the first thign proposed
for the first one i did this, it's correct right
If i imagine all sets of parentheses have bars on top
x is in Z+ so x is (a,b) such that a>b
y is not in Z+ so y is (c,d) such that d≥c -> either d=c or d>c
its a known that for everything in Z always exactly 1 of in Z+ in Z- or =0 is true (the material calls this trichotomy)
And product in Z is such as (a,b)(c,d) = (ac+bd, ad+bc)
case 1: c=d
if c=d then (c,d) ~ (1,1) so it's 0 in Z
since for all Z, x*0 = 0, xy = 0 and 0 not in Z+ (the material says x*(1,1) = (1,1) which is the same thing but more convoluted)
case 2: d>c
the > in N is defined such that
if a>b, then there exists n in N such that a = b+n
same thing -> d = c+m
so ac + bd = (b+n)c + b(c+m)
= 2bc + nc + bm
and ad + bc = (b+n)(c+m) + bc
= 2bc + nc + bm + nm (only difference)
so using those two it can be seen that ac+bd = ad+bc +nm, and nm is in N (duh), so by def of > it implies that
ad+bc > ac+bd
so since the right side of the equivalence class that is the product > the right side, the product is in Z- and not in Z+ in case 2
so no matter what the product of a Z+ and a not in Z+ will not be in Z+
idk if im allowed to use that (-m)(-k) = mk
so best to not mess w signs imo
9)b)
let x = (a,b) and y = (c,d), with a,b,c,d all in N ofc
1 = (2,1), so
(ac+bd, ad+bc) must be = (2,1), and by equivalence it means
ac + bd +2 = ad+bc + 1
ac + bd +1 = ad+bc (since 2=1+1) !!!!
if a=b, then bc+bd = bd+bc+1 not possible in N
same for if c=d, not possible
Case 1: a>b
there exists n in N such that a=b+n. substituting
(b+n)c + bd = (b+n)d + bc + 1
develop and cancel ->
nc = nd +1
so either c<d or c>d
case 1.a.: c<d so there would be k in N such that d=c+m so
nc = n(c+m) +1
nc = nc + nm + 1 not possible, so necessarily in case 1 c>d
so there exists k in N duch that c = d+k.
n(d+k) = nd +1
nd + nk = nd +1
nk = 1
since in N 1*a = a always, it means n=k=1
so a = b+1 and c = d+1
so a=suc(b) and c=suc(d)
so x = (suc(b),b) = 1 and y = (suc(d),d) = 1
Case 2 is the same by analogy but flipping a and b and c and d respectively results in
b = a+1 and d = c+1
so b=suc(a) and d=suc(c)
so x = (a,suc(a)) = -1 and y = (c,suc(c)) = -1
so if xy=1, either x=y=1 or x=y=-1
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GUYS TELL me what is this jacobimatrix good for?
do you know derivatives?
yes i know how to build the gradient as well
do you know what slope is in 1 dimension?
gradient is just the generalization to n dimensions
In vector calculus, the Jacobian matrix (, ) of a vector-valued function of several variables is the matrix of all its first-order partial derivatives. If this matrix is square, that is, if the number of variables equals the number of components of function values, then its determinant is called the Jacobian determinant. Both the matrix and (if ...
slope is steigung i guess yeah i will need to know what this matrix is good for
yes then there are 3 applications here
yeah i guess dunno the dude in our exam preparation just showed us how to do it but not how it is used lol
guess the use is secondary in our case
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is the first one true or false? I know how to prove that it's true but also how to prove that it's false
give a counterexample (one is enough)
nvm ugh
I know that the first one is true but some things I see keep attempting to give a counterexample even if there isnt because they wrongly interpret N as including 0
which would make the first impossible by saying they aren't the same kind of structure if (N,+) has neuter and (Z-,+) doesnt
But obviously (N,+) doesnt include 0 so this thing doesnt matter and it's easy to find a function that proves they are homomorphic
for the latter though it can be easily disproven by saying that (Z-, *) isnt closed so it cant be possible, right? @glass carbon
what is this operator ? same set? same cardinality?
if you define the set N as {1,2,3, ...} (without zero) then yes
wdym by this
everything in the class is based on this definition yes
no it means that they are algebraic isomorphisms
or that they of the same size?
that the operation + in the set Z- is algebraically isomorphic to the operation + in the set N
and the same for * (although there I need to disprove since it's false)
yeah I already knew how to prove it
.
great-o
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This is from the answer key for an exam review package
If you plug x/2 + pi/6 into g(x) you'll obtain -cos(x) + 3
In other words, it is the undo of these transformations, the same with -y+3
All you need to do is plug the values from the first table for x and calculate accordingly, e.g. for x = 0 we have x/2 + pi/6 = 0/2 + pi/6 = pi/6, y was 1, so now it will be -1+3=2 etc.
but how do we get that x/2 + pi/6 from the original equation
"undoing the transformation" 2(x - pi/6)
where did (pi/6) / 2 come from
if i look at 2(x - pi/6) i think (2x - (2pi)/6) OH
OHHHH
so if you divide it by 2 it's pi/6
i'm still struggling to pull it out of the original equatioon
@mortal sundial Has your question been resolved?
I'll just hope
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Is this correct? I believe it is I just want someone else to double check me.
It does not have to be fully simplified or even simplified at all, I just simplified it enough that people wont have an aneurysm reading it
(a+b)^2 isn't a^2 + b^2
Can you show me what you're refering to
g²
Oh fair enough
Wdym it doesn't get affected
these are not integrals
(x^2+1)^2 is not x^4 + 1^2
Yes
tyty
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Idk if this question is too short to be asked here but is there a difference between the two sine laws or can I just always use one and I’ll get the same answers
one is enough (though it depends), you use it depending on your needs
There is no difference, none of the denominators will ever be zero
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What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
You just dumped like 20 problems there. pick one and try it
(also please don't spam your question across multiple channels, you have this one for yourself, so there's no need to disrupt other people with their questions
)
You have to be trolling
this is not appropriate and you know that.
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can someone tell me where to start number 7?
and also if possible pls teach me how to find direction of force exerted on a hinge because they didnt teach it in the book yet many of the past paper questions uses it
First, what are the forces that act on the rod?
hey sorry, so i found out how to do it
i just gotta resolve the forces vertically and horizontally, find moment on a, get y and x and find the magnitude, and use tantheta=y/x
Ah ok gr8 :)
thanks man
Np
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bruh......
you're trolling us right
you meant to use your alt
but you ended up using the same account
to reply to yourself
that solution is clearly wrong
the y axis is at x = 0
so when you're crossing the y axis youd just be at x = 0
,w graph y=3x^2+6x, y=0

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Could u help me
Factorize??
yes
also denominator is under square root
yh now you can simplify
,,\sqrt{\frac{(a-1)^2}{a}} = \frac{a-1}{\sqrt{a}}
scoob
now try to match it with RHS
i would suggest split fraction
,, \frac{a+b}{c} = \frac{a}{c} + \frac{b}{c}
scoob
what do you mean by separate?
@fickle mantle Has your question been resolved?
,rccw
what have you tried
,, x = \sqrt{x}\cdot \sqrt{x}
scoob
or you could just cancel x in numerator and denominator here
yh you cannot add numerator and denominator to add fractions
,, \frac{a}{b} + \frac{c}{d} \neq \frac{a+c}{b+d}
scoob
go with this cancel x in numerator and denominator
👍
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Hlw i can solve integral dx/((sqrt(6x+1) +2)
try rationalizing the denominator
by multiplying by the conjugate
or perhaps directly substituting could be quicker
i would think u := sqrt(6x+1) would be a better path tbh
!status
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5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
Where is my mistake?
you might wanna write the characters u, 4 and 6 in a way that doesn't make them all look similar
I dont like write but is same i got 6x -5
But i got a mistake
and is it sqrt(6x+1) or sqrt(6x-1)?
Where?
6x-1
I am so dumb i cant see my mistake i dont know why syntethic division cant work but wether large division
hmm
thats not really common with a square root
and ur result is sqrt(u) - 4/(u-4)?
$u^{\frac{1}{2}}-\frac{4}{u-4}$
MathIsAlwaysRight
Sure, I'd probably just start with this
either u := sqrt(6x - 1) or u := sqrt(6x - 1) + 2
I can continue wirh u message i need 15 minutes me back
<@&268886789983436800>
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How I can solve thiw
at very start of the question, Put
$u = \sqrt{6 x - 1}$
Shikhar
But i can multiply for conjugat3
@vague rain Has your question been resolved?
dt = 6/sqrt(6x-1) dx
Wdym?
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I stumbled across this question on a set of integral questions: A toilet roll has an inner radius r = 2 cm and an outer radius R = 7 cm. The thickness of the paper is 0.5 mm. What is the length of the paper when it is completely unrolled?
i mean technically
the paper is in a spiral
so the outer radius of 7cm fluctuates depending on where you take the point
but let’s say we assume that the end of the spiral is 7 cm
yeah ive come so far, and so for every layer the diameter increases by 2*whatever variable
is that right?
im quite literally takin a shi scrolling math rn
so a bit ironic 😅
and i was about to solve it until i realised it was a spiral
hahahaha
well I solved it without using integrals, however I really want to solve it using it as well
not sure what this means tbh
i was thinking of a calculus solution too
because we want to take the arc length of the spiral
I think he means cuz the paper laps around twice so its effectively adding 1mm each time it goes around
the circumference is 2pir^2
to be completely honest, i dont really understand
why does it lap twice?
that’s the diameter, isn’t it?
Yes
we seem to only be concerned with the radius
for the radius it increases by one thickness, however when its the diameter, it gets added to the other side as well
right
The integral is int(r>R) (2pix)/thickness I believe
and so i used a sequence for the circumference Cn = pi*(d+2nt)
so can we agree that assuming this
we have 5/0.5 = 10 laps of toilet paper
ok so we’ll stick with this
yup
not sure about that haha, its just 2nd semester economics maths
well we’re being exact here, aren’t we? 😅
I think you use this
For this level
But yea
is that the surface area integral
Yes
Then you have the thickness so area is just thickness x length
So you just divide the area by the thickness
I think
i think thats where i fell off a bit
are we taking the integral in a way that sums up all the circumferences
at each "layer"
Yk you can just do this with basic level math?
yeah I have done that
Oh ok
but thats the idea of it?
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Call a nonnegative integer simple if its only nonzero digits are ones. Suppose S is simple, and R is the number obtained by reversing the order of its digits. If 901·S is simple, 901·R must also be simple. Why?
i solved and the only simple integers works are repunit integers with more than 2 digits such as 111, 1111, ... also can end with 0 as 1110,11110,....
but i couldn't express in words so can someone help
what do you mean when you say "simple" integer
do you mean a number that's divisble by 1, itself and nothing else?
Call a nonnegative integer simple if its only nonzero digits are ones. such as 0, 1, 10 , 11, 100, 101
oh nevermind
ok i thought you might've mistranslated from another language and meant prime
my bad
no worries
901S = (1000 - 100 + 1)S. this seems like it would be relevant.
iam confused can u explain it
try to write out some simple integers and their products with 901
see if anything pops out
yes i did
.
are you sure that something like 111110000011111000 with the runs of 1's far enough apart wouldn't also work?
what exactly breaks down if you try it this way
i just tried it and it didd work
yes i said it does
no worries thanks btw
@tropic umbra Has your question been resolved?
just react to the bot. the bot can't read text replies to that question
oh my bad😭
so simple integers are just binary strings that don't start with 0?....
yes
ig
like 0,1,1011,100,101
in base 10
@tropic umbra Has your question been resolved?
so what is left
ah yes
,w 111000111000*901
why is it base 2 by default bruh
any simple number S with no single 1,0 or no consecutive pair 11 or 00 satisfy that 901*S is a simple integer
I think
S*901=900*S+S
so if S have a string abc111def we can well do addition there would be
...... 9 9 9 ....
........ 1 1 1 .....
there has to be at least 3 consecutive 1 or else when do addition
.....0 9 9 0 ....
.......0 1 1 0 .....
a 9 would appear
similarly for string ...1001...
...... 9 0 0 9 ....
........ 1 0 0 1.....
when do addition a 9 could appear, or
...... 9 0 0 9 9 9 ....
........ 1 0 0 1 1 9......
.............. 2 0 0 0....
a 2 would appear
ofc when we reverse the order there aren't anything changed with the number of consecutive 1's and 0's
So obviously its reversed in order multiply by 900 would also be simple integer
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Could someone please help me with this question?
are you familiar with radians?
Do you know the formula of angular speed?
And ik that one revolution is 2pi
!noai
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if one revolution is 2 pi radians, then how many radians is 20 revolutions? and how many seconds pass in one minute? then you can find the number of radians per amount of seconds passed
10 pie /3 m/s
are you familiar with the concept that the distance traversed on a circle is the product of the angle covered and radius?
I am not
when you go around a circle, you are going across a fraction of its perimeter (2 pi r)
when you travel a complete revolution of the circle you are travelling 2 pi radians multiplied by r resulting in 2 pi r
so when you travel any "T" radians across a circle, you are actually travelling T*r length
Okkk
speed = distance /time = 40 . 5 pie /60 =10/3pie metre per second
in your case you can find its rate by dividing with units of time to get linear speed = angular speed * radius
So my radius is 5 so the perimeter would be 10pi
you found the angular speed in step 1 of the problem in radians per second, this represents how fast the merry go round spins about its centre (how many radians it covers in one second)
Ok...
for any point on the edge of the merry go round, you would multiply that speed with how far you are from the centre (the radius) to get its linear speed
So liner speed is angular velocity x 2?
this is correct but not the answer you are looking for for part 2
not quite
Linear speed is angular velocity x r
yes...
yes good job!
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so u solved it ?
Not really, it's not a complete solution but you get the idea
@tropic umbra Has your question been resolved?
I feel stupid still trying to
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I have a question. When confirming local extremes, is it correct to say that ‘function(point) is a local extremum’ or that ‘the point is a local extremum’? Or what would be the correct way to say it in an exam?
function(point) is a local extremum
point is a critical point
You can say extremum point for the x value and extremum value or just extremum for the y value
Thank you! I have another related question. For this function, I had concluded that (0; 0) is a saddle point, but in an online solution, they stated that (0; 0; f(0; 0)) is the saddle point. Which one is correct, or are both wrong?
The exercise asks you to analyse whether f(0; 0) is a local extreme.
Details really
(0,0) is the point in the 2D xy-plane which gives a saddle point in the 3D graph of the function
(0,0,f(0,0)) being that saddle point
Calling (0,0) a saddle point is perfectly fine
That makes sense, thanks
In this case, would it be incorrect to say that f(0; 0) is a saddle point?
f(0,0) is the value at that point, so I'd say it's technically incorrect
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Hello, could someone check if this proof looks good please?
kind of confused by your argument here
the expression youre saying equals 6l doesn't seem to match the 5^2*k - 1 expression that youre substituting 6l for below
Typo, fixed 👌
but...in the top expression you have a k in the exponent
then below you have multiplication by k
ok
also need to fix the middle one
and i dunno just some thoughts for clarification, you are kind of changing n with k a lot
Fixed 👌
\begin{Definition}[Natural numbers]
The set of the \emph{natural numbers} is the set $\bN = \{1,2,3,4,5,\dots\}$.
\end{Definition}
\begin{Theorem}
Suppose $n \in \bN$, then $6 \mid (5^{2n} - 1)$.
\end{Theorem}
\begin{proof}
Let $n \in \bN$. We proceed by induction.\\
\underline{Base Case.} The base case is when $n = 1$, and $6 \mid (5^2 - 1)$ as desired.\\
\underline{Inductive Hypothesis.} Let $k \in \bN$, and assume $6 \mid (5^{2k} - 1)$.\\
\underline{Induction step.} We aim to prove that the result holds for $k + 1$.
That is, we wish to show that $6 \mid (5^{2(k + 1)} - 1)$.
To do this, we begin with the expression $5^{2(k + 1)} - 1$:
\begin{align*}
5^{2(k + 1)} - 1 &= 5^2(5^{2k}) - 1 \\
&= 24(5^{2k}) + 5^2k - 1 \\
\end{align*}
By the inductive hypothesis, $6 \mid (5^{2k} - 1)$, and so, by definition of divisibility,
$5^{2k} - 1 = 6l$ for some integer $l$.
Then,
\begin{align*}
24(5^{2k}) + 5^{2k} - 1 &= 24(5^{2k}) + 6l \\
&= 6(4(5^{2k}) + l) \\
\end{align*}
Since $k$ and $l$ are integers, so is $4(5^{2k}) + l$.
Then, $5^{2(n+1)} - 1 = 6m$ where $m = 4(5^{2k}) + l$.
Thus, by the definition of divisibility, $6 \mid (5^{2(k+1)} - 1)$.
Therefore, by induction, $6 \mid (5^{2n} - 1)$ for all $n \in \bN$.
\end{proof}
in the hypothesis line, assume the statement is only true for n <= k because the way you phrased the induction hypothesis currently, it sounds like youre assuming the whole thing youre trying to prove
changing n to k there made it worse
Mor Bras
my suggestion for hypothesis: assume $6 \mid (5^{2n} - 1)$ for $n\in\mathbb{N}, n\leq k$
Soosh
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Why wouldn't
$$
t = 5(log25/log0.5)
$$
be correct?
Sep
Compile Error! Click the
reaction for more information.
(You may edit your message to recompile.)
i suspect they want the "percent" to be given as a decimal between 0 and 1
yes
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I finally tried u-sub but I'm stuck trying to figure out the integral of 3^u
idk
learn derivatives before integrals
I got a c in calc 1
I'm thinking of withdrawling and learning calc 1 all over again
first
its before the deadline?
you think it's doable ?
I'm good at math it's just last semster was hell
.
3^x / ln(x) ?
oh right
basically undoing that factor of ln b
so that it cancels
you're welcome
btw you could've done 3^(ln x) = (e^ln 3)^(ln x) = (e^ln x)^(ln 3) = x^ln(3)
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do you just want confirmation?
^ (sorry, forgot to reply)
yep
you always want to make it into an "if xxx, then yyy"
(well you don't always have to, but this always works)
lemme see
there are more ways to express this
do you know how to turn an implication into an OR statement and vice versa?

ah pls don't delete modpings even if they've been handled, it gets super confusing when mods show up
as evidenced
(you can edit them to say resolved or sth)
i just did not think of that — but off-topic
@drowsy vale sorry for everything that has happened in this channel lmao, you still here?
A mod telling you off is never off topic 
lol i wasnt clear, but im meaning that my set of excuses that i were about to dump out is off-topic
i got a ghost message lmao
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confirmation ||your answer is wrong, one of A / B / C / D is correct||
do you know how to express implication as OR
wait are you sure
||A?|| fuck.
clearly i'm not in the mental state of doing this rn
your answer is right marcel
yeah. it turns out that p -> q is exactly equal to (not p) or q
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Can you guys check this please
you will benefit from making your z and 2 look less similar, this much i can tell you immediately
your final answer of (-1, 3, 5) satisfies the system, so you have made an even number of mistakes
Gotcha, will do
You can write z with a stroke "ƶ", I personally use it all the time when doing math problems
one really nice thing about systems of eqns is that you can easily check your own work
by simply plugging the solution you get into your starting equations
another option is to write your 2 with a bit of a loop to it
which is what i did and reported the result :P
@twin fulcrum Has your question been resolved?
New
<@&286206848099549185>
tried diff format still got same result, not sure what im doing wrong
Why do you believe your result is incorrect
,w solve 2x - y + z = 0, 3x - 2y + 4z = 11, 5x + y - 6z = -32
oh, sorry someone said there was a mistake
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i need help with this one
the goal is to find DE?
yup
do you have progress so far?
,rccw
hmm might be difficult this way
here's a better idea: drop a perpendicular from B into AC. call it BH for example. then consider triangles ADE and ABH

look at triangles ADE and ABH
there is something special about them
seeing as DE and BH are parallel
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I need help starting with this.
do you know how spherical (R^3) coords work
uh i think so
ok can you state the conversion from cartesian to polar in R^3
r = sqrt( x^2 + y^2 + z^2)
theta = arccos(z/r)
phi = arctan(y/x)
misquoted and opposite to what i actually asked
what about now
did you mean cylindrical coordinates
no, i meant spherical
i expected:
x = r cos(θ)
y = r sin(θ) cos(φ)
z = r sin(θ) sin(φ)
or some permutation of this
ok i guess actually i am the fool here because i asked the question in the opposite way to my own intentions
okay yeah i was going to say this but you asked going from cartesian to polar and i was really confused lol
so what do we do with this
can you come up with a generalization to 4 coordinates in the same way i showed
use x_1, x_2, x_3, x_4 for the rectangular coords and r, theta_1, theta_2, theta_3 for the spherical coords
(x1)^2 + (x2)^2 + (x3)^3 + (x4)^4 = r^2
x1 = rcos(theta1)
x2 = rsin(theta1)cos(theta2)
and
this one is unnecessary
but didnt you say to use x1,x2,...
i did, but the equation i specifically replied to is not what im looking for.
ok great
thats good now
can you write down the generalization of this to n coordinates
you'll have one radial and n-1 angular coordinates on the polar side of things
also you will want to write down the ranges that these thetas go over
for integration purposes
x_n = rsin(theta1)sin(theta2)...sin(theta_n-2)sin(theta_n-1)
x_n-1= rsin(theta1)sin(theta2)...sin(theta_n-2)cos(theta_n-1)
theta 1 and theta 2 from 0 to pi maybe
@split fulcrum Has your question been resolved?
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What have you tried
why do you need integration?
See it’s a little confusing what you’ve written
On the right side you have an integral with respect to t, and the integrand is a function of t, that’s perfect
But on the left side you have that v is a function of t
But there’s no dependency on t on the right side
Once you integrate the t will disappear and v(t) will always be a constant
It’s more like $v(t) = \int_0^t a(\mathcolor{green}{t}),d\mathcolor{green}{t}$
frosst
i dont understand why you are using calculus at all when acceleration is constant
@quaint marlin Has your question been resolved?
No
I forgot that v(6)=45
$\int_0^2 a(t) \dd{t} = v(2) - v(0)$, and you don't know $v(0)$ (yet).
Ann
Did you understand anything I said?
Yep
So you understand this?
I mean why i can do this directly
Where did v(6) = 45 come from
I forgot i'm sorry
Why can you or why can’t you
Where in the question does it say that





