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#help-26
turbid sable
dry rune
that doesnt make sense
turbid sable
what
turbid sable
not entirely sure sorry
thorny flameBOT
turbid sable
let me know if this makes sense
turbid sable
its cool fr
dry rune
ty for the help
topaz sinewBOT
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muted steeple
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dawn shell
hey quick question about a power series
topaz sinewBOT
hey quick question about a power series
dawn shell
$\sum^{\infty}_{n=1} \frac{(2n-1)^n(x-1)^n}{2^{n-1}n^n}$
thorny flameBOT
Mephisto
dawn shell
I have to find the convergence radius and interval
I tried doing the root test, but it doesn't seem to work
drowsy sonnet
Ratio
dawn shell
doing the root test gave me $\sqrt[n]{2}\frac{(2n-1)(x-1)}{2n}$
thorny flameBOT
Mephisto
drowsy sonnet
That simplifies a lot just break up the first term
dawn shell
and taking the limit of n going to infinity, just gives 0?
dawn shell
for n going to infinity
we can devide 2n by 2n
because the -1 is neglegable I think
dawn shell
which would give $\sqrt[n]{2}(x-1)}$
thorny flameBOT
Mephisto
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dawn shell
Break up the 2n-1
oh as in distribute?
drowsy sonnet
**Mephisto**
Compile Error! Click the <:errors:504266556054962196> reaction for ...
Also this limit is not 0
dawn shell
ohhh
so the limit has to be smaller then one
meaning we can take both side to the n'th power
which gives the inequality
2*(x-1)^n < 1^n
drowsy sonnet
what?
dawn shell
0.81
dawn shell
hold on
dawn shell
0.729 yeah
dawn shell
it converges for everything smaller then one
drowsy sonnet
yup
dawn shell
I know, but how do I solve for this x?
drowsy sonnet
**Mephisto**
Compile Error! Click the <:errors:504266556054962196> reaction for ...
so in this if you have 2^(1/n)
in the limit as that goes to inf
can it be 0?
dawn shell
omg I just realised
I feel so dumb
I completely forgot about how roots are just fractions...
yeah it should converge to 1
drowsy sonnet
which means your radius is?
drowsy sonnet
yeah
dawn shell
and the interval of convergence then?
dawn shell
ah alright
meaning
x + 1 < 1
so one bound is 0
or atleast x < 0
and the second one should be x < 2 right?
I feel like I've got smtg wrong here
yeah it's been a while since I've used absolute values...
|x -1 | < 1
meaning for x < 0
we get -x + 1 < 1
- x < 2 meaning x > -2
for x > 0
... yeah I'm lost here...
I though |x -1 | < 1
was -x + 1 < 1 for x smaller then 0
and x-1 < 1 for x larger then 0
but this gives me x > 0 and x < 2
while it should be x < 0 and x > -2
nvm I typed the question wrong... it should've been x+1 < 1 ....
.close
topaz sinewBOT
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cursive cloak
topaz sinewBOT
cursive cloak
i tried solving it with a(1-r^n)/(1-r) but i kept getting a different number then what was in the solution
also i made a typo by saying a sub n instead of s sub n
oops in the bottom line i said 9/5 instead of 9/4, the actual answer should be -324, but thats just the negative of what the series at n=0 specifically so its still wrong
topaz sinewBOT
@cursive cloak Has your question been resolved?
topaz sinewBOT
cursive cloak
.reopen
wooden osprey
Given $M(2,3,1)$ and the sphere $(x-1)^2+(y-1)^2+z^2=4$. From M, infinitely many tangent lines of the sphere can be drawn. All the points of tangency creates a circle, calculate its radius
topaz sinewBOT
Given $M(2,3,1)$ and the sphere $(x-1)^2+(y-1)^2+z^2=4$. From M, infinitely many...
thorny flameBOT
wooden osprey
not sure what to do here
topaz sinewBOT
topaz sinewBOT
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tidal frigate
- A US company wants to pay its import bill to its supplier in New Zealand and requires NZ$ 15,575000 in 90 days. Assume it can buy NZ$ at the spot rate for NZ$1 = $0.72. However, the US company enters a currency forward contract with HSBC and accordingly the US company will purchase requires NZ$ 15,575000 in 90 days at the exchange rate NZ$1 = $0.81. Imagine that on the day of maturity of the currency forward contract the spot rate is NZ$1 = $0.88, calculate total gain/loss of the US company.
idk my brain is lagging on this one can some one help me
topaz sinewBOT
3. A US company wants to pay its import bill to its supplier in New Zealand and ...
thorny flameBOT
Hermoso
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lucid crypt
show work
tidal frigate
Forward contract:
NZ$15,575,000 * $0.81 = $12,614,250
the day of maturity:
Cost = NZ$15,575,000 * $0.88 = $13,682,000
= $12,614,250 - $13,682,000
= -$1,067,750
thorny flameBOT
Hermoso
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tidal frigate
iam not sure is this correct xD i asked coopilot for help if its correct it gave me an different answer
and i asked chat gpt too and that too gave different asnwer
lucid crypt
ok, so the company couldve bought at .72, but instead they buy at .81
on the day they buy at .81, the price is .88
am i reading this right?
topaz sinewBOT
@tidal frigate Has your question been resolved?
manic flax
topaz sinewBOT
topaz sinewBOT
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lunar oriole
can someone help me with this question i have no idea how to do it
topaz sinewBOT
can someone help me with this question i have no idea how to do it
@lunar oriole Has your question been resolved?
lunar oriole
<@&286206848099549185>
topaz sinewBOT
@lunar oriole Has your question been resolved?
topaz sinewBOT
@lunar oriole Has your question been resolved?
prime storm
How do I do 6iii?
topaz sinewBOT
How do I do 6iii?
prime storm
I am currently disecting it
And I got
Couple (5C1)(2)
Others (4C1)(3)
But I am thinking how to in-coperate the couple into the general group
Since if thinking as a pure math question, i.e. only 2 individuals and 3 individuals wo the hotel part that should be 2!(4!)
topaz sinewBOT
@prime storm Has your question been resolved?
smoky sparrow
Couple (5C1)(2)
Others (4C1)(3)
The others part should just be 4 * 4 * 4 = 4^3
There's no requirement that each of the 3 other people has to stay in a different hotel
So just 5 (for the couple) * 4^3
it's not 5 * 2
topaz sinewBOT
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tawdry root
hii, can someone please help me solve this integral? I know what the answer should be but i think i'm on the wrong track doing the working out
topaz sinewBOT
hii, can someone please help me solve this integral? I know what the answer shou...
gritty birch
you can't do this using integration by parts, because you can't integrate (x^2+1)^-1
tawdry root
im confused because the question says to do it with integration by parts
supple oxide
it is arctan x
gritty birch
with your v' = (x^2+1)^-1, you need to integrate v' but you differentiated v' instead
tawdry root
but before i tried it with partial fractions and i managed to get the -3 atan (x) part
supple oxide
if u integrate 1/(xx+1) with trig sub u get arctan
thorny flameBOT
half edge
u can do simple integration using this ^
tawdry root
okay thanks guys ill give it another go
supple oxide
oh nice
topaz sinewBOT
neon iron
am i supposed to know this
yeah u can quickly prove it if u wanna
topaz sinewBOT
Available help channel!
thin adder
I need help understanding this:
smoky sparrow
I need help understanding this:
They are using the fact that Var(X + Y) = Var(X) + Var(Y) if X, Y are indepedent
X, Y are indeed independent as the covariance of x_1i and x_2i is 0
Here's a derivation
safe jacinth
Can I have some help?
topaz sinewBOT
Can I have some help?
safe jacinth
actually I figured that one out
Can I have help with this?
I did 8 x 8.5/2
Whig was 34 ft
ok
what about this one
Ok
I did
8.7
Ok so now what
8.7
Oh ok
I don’t know
AC
Ok
Im confused tbh
BC
Opposite?
Yes
Using 30 degrees
?
Wym
5.3 x 5.3?
I got 51.15
As my answer
For the whole question
Was I correct?
ok
how do I do it correctly.
Ok
Sin
Sin 30?
Ok
So it’s bc = sin30/10
What do I do I thought I multiplies both sides
Ok
topaz sinewBOT
@safe jacinth Has your question been resolved?
waxen vine
How to prove by induction that 7n^41 + 6n^31 + 34n^3 + 8n is divisible by 5 for all n in N
topaz sinewBOT
How to prove by induction that 7n^41 + 6n^31 + 34n^3 + 8n is divisible by 5 for ...
waxen vine
Tried proving it for n+1, but got to the fact that 7(n+1)^41 + 6(n+1)^31 + 34(n+1)^3 + 8(n+1) = 7n^41 + 6n^31 + 34n^3 + 8n + (. . .), with the first terms being a multiple of 5 by inductive hypothesis, but then got stuck
topaz sinewBOT
@waxen vine Has your question been resolved?
waxen vine
<@&286206848099549185>
waxen vine
<@&286206848099549185>
waxen vine
can you help me
waxen vine
How to prove by induction that 7n^41 + 6n^31 + 34n^3 + 8n is divisible by 5 for ...
to solve this
waxen vine
ok
hot shard
To show that ( 7n^{41} + 6n^{31} + 34n^3 + 8n ) is divisible by 5 for all positive integers ( n ), we can use mathematical induction.
Base Case:
First, we check the base case ( n = 1 ).
[
7(1)^{41} + 6(1)^{31} + 34(1)^3 + 8(1) = 7 + 6 + 34 + 8 = 55
]
Since ( 55 ) is divisible by 5, the base case holds.
Inductive Step:
Assume that the statement is true for some integer ( n ), i.e.,
[
7n^{41} + 6n^{31} + 34n^3 + 8n \equiv 0 \pmod{5}
]
We need to show that it also holds for ( n + 1 ). Consider the expression for ( n + 1 ):
[
7(n+1)^{41} + 6(n+1)^{31} + 34(n+1)^3 + 8(n+1)
]
Using the binomial theorem, expand each term:
- ( (n+1)^{41} = n^{41} + 41n^{40} + \cdots + 41n + 1 )
- ( (n+1)^{31} = n^{31} + 31n^{30} + \cdots + 31n + 1 )
- ( (n+1)^3 = n^3 + 3n^2 + 3n + 1 )
Substitute these expansions into the expression:
[
7(n+1)^{41} = 7 \left( n^{41} + 41n^{40} + \cdots + 41n + 1 \right)
]
[
6(n+1)^{31} = 6 \left( n^{31} + 31n^{30} + \cdots + 31n + 1 \right)
]
[
34(n+1)^3 = 34 \left( n^3 + 3n^2 + 3n + 1 \right)
]
[
8(n+1) = 8n + 8
]
Combining these, we get:
[
7(n+1)^{41} + 6(n+1)^{31} + 34(n+1)^3 + 8(n+1) = 7 \left( n^{41} + 41n^{40} + \cdots + 41n + 1 \right) + 6 \left( n^{31} + 31n^{30} + \cdots + 31n + 1 \right) + 34 \left( n^3 + 3n^2 + 3n + 1 \right) + 8n + 8
]
Separate the terms involving ( n ) from those involving constants:
[
= 7n^{41} + 7 \cdot 41n^{40} + \cdots + 7 \cdot 41n + 7 + 6n^{31} + 6 \cdot 31n^{30} + \cdots + 6 \cdot 31n + 6 + 34n^3 + 34 \cdot 3n^2 + 34 \cdot 3n + 34 + 8n + 8
]
Group and simplify the terms:
[
= \left( 7n^{41} + 6n^{31} + 34n^3 + 8n \right) + \left( 7 \cdot 41n^{40} + \cdots + 7 \cdot 41n + 6 \cdot 31n^{30} + \cdots + 6 \cdot 31n + 34 \cdot 3n^2 + 34 \cdot 3n \right) + \left( 7 + 6 + 34 + 8 \right)
]
We know from the inductive hypothesis that ( 7n^{41} + 6n^{31} + 34n^3 + 8n ) is divisible by 5. So we focus on the remaining terms.
thorny flameBOT
Daksh_GamerYT
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waxen vine
yes
waxen vine
Tried proving it for n+1, but got to the fact that 7(n+1)^41 + 6(n+1)^31 + 34(n+...
i had actually sort of got to this point
but idk how to prove that 5 | the other terms
crimson prairie
To prove (7n^{41} + 6n^{31} + 34n^3 + 8n) is divisible by 5 for all (n \in \mathbb{N}):
- Base Case: For (n = 1), the expression equals 55 which is divisible by 5
- Inductive Step: Assume true for (n = k), then prove for (n = k+1).
- Show that (7(k+1)^{41} + 6(k+1)^{31} + 34(k+1)^3 + 8(k+1)) is divisible by 5.
By expanding and factoring, each term is divisible by 5, so the expression is divisible by 5 for all (n \in \mathbb{N})
thorny flameBOT
ika
crimson prairie
I did something similar like that in the past. I hope this helps
waxen vine
mh i need to find a way to factor the remaining terms
crimson prairie
Ok.. I'll work on that
Each term has a common factor of k, so the expression is divisible by k.
crimson prairie
To prove by induction that ( 7n^{41} + 6n^{31} + 34n^3 + 8n ) is divisible by 5 for all ( n ) in the set of natural numbers, you typically follow these steps:
- Base case: Show that the statement holds true for the initial value of ( n ).
- Inductive hypothesis: Assume that the statement is true for some arbitrary positive integer ( k ).
- Inductive step: Prove that if the statement holds for ( k ), it also holds for ( k+1 ).
Let's proceed:
Base case:
For ( n = 1 ):
[ 7(1)^{41} + 6(1)^{31} + 34(1)^3 + 8(1) = 7 + 6 + 34 + 8 = 55 ]
which is divisible by 5.
Inductive hypothesis:
Assume that ( 7k^{41} + 6k^{31} + 34k^3 + 8k ) is divisible by 5 for some positive integer ( k ).
Inductive step:
Now, let's consider ( n = k + 1 ):
[ 7(k + 1)^{41} + 6(k + 1)^{31} + 34(k + 1)^3 + 8(k + 1) ]
[ = 7(k^{41} + 41k^{40} + ...) + 6(k^{31} + 31k^{30} + ...) + 34(k^3 + 3k^2 + 3k + 1) + 8k + 8 ]
Now, we need to show that this expression is divisible by 5.
The first two terms are multiples of ( k^{41} ) and ( k^{31} ), respectively, which we assume are divisible by 5 by the inductive hypothesis.
The third term is ( 34(k^3 + 3k^2 + 3k + 1) ), which simplifies to ( 34k^3 + 102k^2 + 102k + 34 ). Notice that ( 34k^3 + 102k^2 + 102k ) is clearly divisible by 5 (since every term has a factor of 5), and adding 34 doesn't change that.
Finally, ( 8k + 8 ) is clearly divisible by 5.
Therefore, the expression ( 7(k + 1)^{41} + 6(k + 1)^{31} + 34(k + 1)^3 + 8(k + 1) ) is divisible by 5.
By mathematical induction, we conclude that ( 7n^{41} + 6n^{31} + 34n^3 + 8n ) is divisible by 5 for all ( n ) in the set. So it factored is:
[ 7n^{41} + 6n^{31} + 34n^3 + 8n ]
[ = n(7n^{40}) + n(6n^{30}) + 34n(n^2) + 8n ]
[ = n(7n^{40} + 6n^{30} + 34n^2 + 8) ]
So, the expression is divisible by 5 because it can be expressed as ( n ) multiplied by an integer value.
I think this will help
thorny flameBOT
ika
pastel oracle
To prove by induction that \( 7n^{41} + 6n^{31} + 34n^3 + 8n \) is divisible by ...
how is 8k + 8 divisible by 5? take k=1, then 8k + 8 = 16
crimson prairie
Wha.
pastel oracle
you said "(8k + 8) is clearly divisible by 5". that is generally not true
crimson prairie
Wait I'm so sorry I think my brain is dying
Listen I'm so sorry. I think I'll just have to go to sleep and think about this in the morning.
waxen vine
Okok no problem
crimson prairie
Any tips on sleeping tough? /J
pastel oracle
for n+1, you know the first and last terms of the expansion are divisible by 5 (the first terms 7n^41 + 6n^31 + 34n3 + 8n by the induction hypothesis; the last terms are just 7 + 6 + 34 + 8 = 55)
so you're left with $7(41n^{40} + \cdots + 41n) + 6(31n^{30} + \cdots + 31n) + 34(3n^2 + 3n)$. this is the tough part. for each of those terms, you can still factor out $n$, but I'm not sure if that will get you anywhere
thorny flameBOT
cwatson
pastel oracle
you could try asking in #proofs-and-logic , maybe someone there can give some hints
are you supposed to use properties of binomial coefficients/the binomial theorem?
waxen vine
I dont know because this was in a modular arithmetic book, which has an easy proof via modulo 5, but in the end It says that this can also be proved via induction
candid dove
not much of a math question, but could anyone help me to solve the circuit by kirchoff's law?
topaz sinewBOT
not much of a math question, but could anyone help me to solve the circuit by ki...
topaz sinewBOT
not much of a math question, but could anyone help me to solve the circuit by ki...
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
am actually having trouble tryna apply it, so i would say pretty much the first step
if i were to consider the 30V cell as the source, I'd write:
30V -6V -3i1 -5i2 -6(i1+i2)=0
which gives me 24V= 9i1 +11i2
well what do i do after that? Just consider the upper loop and take only i1?
static scarab
ill give it a try
static scarab
will do
static scarab
kindly ping me while replying
do you have the answer key?
candid dove
ohh yea the answer is 2 amps
static scarab
ok then im pretty sure i got this right, the first thing is that you cant consider only the 30V one as your source, both the 6V and the 30V are sources
have you set up kirchoffs laws?
candid dove
ohh
what exactly do y mean by that tho
like what does the mathematical formulation of that look like?
candid dove
if i were to consider the 30V cell as the source, I'd write:
30V -6V -3i1 -5i2 -...
not like this i suppose?
static scarab
ok so you have to think of the circut as 2 loops
candid dove
alrightt
static scarab
i will get discord on my ipad so i can put in some drawings
candid dove
thanks a lottt :D
static scarab
np, i need the refresher i have an exam in electromagnetism next week
coarse flint
candid dove
haha i see, all the best of ur exam!
coarse flint
can someone help pls
coarse flint
when i catch u prince gumball
candid dove
huh
candid dove
yeahh i always get confused regarding the second equation
static scarab
for each loop you have to set up $V_s - V_r - V_r = 0$
thorny flameBOT
caspar
candid dove
so for the second eqn, we can write smthn like 30V+6V-5(i1+i2)-3i1-6i2=0?
static scarab
almost, but remember the 3 ohm resistor is not in the loop containing the 30V source
so'
candid dove
yesyes, that's why i multiplied it only by i1 and not by i2
thorny flameBOT
caspar
candid dove
ohhhhhhhhhh right that makes more sense
thorny flameBOT
caspar
static scarab
so when you use kirchoffs 2nd law, remember only 1 loop
candid dove
yeah i included the second battery and resistor in my loop as well, got it
static scarab
and then its as easy as solving for $I_1 and I_2$
candid dove
sometimes these just seem so ambiguous
thorny flameBOT
caspar
static scarab
yeah it is tricky, but it gets easier just going loop by loop and with practice
candid dove
okii, thanks again and best of luck for ur exam!
static scarab
thank you!
candid dove
.close
topaz sinewBOT
topaz sinewBOT
Available help channel!
topaz sinewBOT
uh didnt work or
neon iron
there
i barely studied this
but isnt basically
i draw triangle
between the 3 dots\like join them together
Then find mid point of each line
like AB BC and CA
then draw the perpendicular line in mid point of each line
then draw straight line until they line up
or am i missing smth
prisma mesa
then draw the perpendicular line in mid point of each line
sounds correct
the point is, that the intersection will be the center of circumscribed circle
prisma mesa
since A, B and C lie on the circle, they are same distance away from the center
topaz sinewBOT
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Available help channel!
junior rain
do I send questions here?
topaz sinewBOT
do I send questions here?
fiery badge
Yes
neon iron
do I send questions here?
ye ofc/
fiery badge
neon iron
yea
junior rain
I have got the answer correct
But I don't understand the algebra behind it
I did it by letting x = 1 as a length
but I tried this with x =2 and got a different answer
would appreciate if I could get the logic behind it
neon iron
lol
fiery badge
What's the formula for a cube's volume?
fiery badge
Right so V = x^3
junior rain
yes
fiery badge
If V increases to V1
Then let's say that x increases to x1
1 being index
for example
junior rain
yeah I get it
fiery badge
so the new cube V_1 has volume x_1^3 right?
fiery badge
Yeah, we'll get there
neon iron
The volume
𝑉
V of a cube is
𝑠
3
s
3
, where
𝑠
s is the length of one side.
If the volume increases by 25%, the new volume is
1.25
×
𝑉
1.25×V.
The original side length is
𝑠
s, and the increased length is
𝑠
new
(
1.25
)
1
/
3
×
𝑠
s
new
=(1.25)
1/3
×s.
So, the increase in side length is approximately
0.077
×
𝑠
0.077×s.
In summary, if the volume of a cube increases by 25%, the side length increases by approximately 7.7% of its original length.
neon iron
fuk it cropped wrong 😦
junior rain
yes
junior rain
yes I see
fiery badge
So now we can cube root this
and we'll get the percentual change in length
more or less
You'll get how much longer the new sides are compared to the old ones
but you can reformulate that
junior rain
a1/a = sqr (1.25)
so a1/a = 1.077....
so what do I do nexxt
a1/a = 1.077/1
so a = 1?
fiery badge
uhh no
junior rain
then 1.077 -1
junior rain
oh yeah I see
fiery badge
a/a + L/a = 1.077
and then you get
L/a = 0.077
which means the side a got increased by 0.077 it's original length
or rather, 7,7...%
junior rain
a/a + L/a = 1.077
how do you do that
a+l/a = 1.077
l = 1.077a-a
l = a(0.077)
l/a = 0.077 so now what?
junior rain
I see
Thank you
Thank's a lot
You eplained it really well
I just realised this works?
topaz sinewBOT
@junior rain Has your question been resolved?
viral mirage
topaz sinewBOT
viral mirage
need help understanding this Q
there’s no given z here, it’s it just a flat surface at z=0?
the answer key says no CV, but i found (0,0) as one and haven’t checked the edges and corners. but i assumed they were also 0 because no z component
knotty ledge
What's f?
noble laurel
viral mirage
?
noble laurel
f(x,y)= ?
knotty ledge
This is part of a larger question
noble laurel
z=f(x,y)=...?
knotty ledge
f should be defined somewhere
knotty ledge
Yeah so you're now considering f only defined in that region
viral mirage
yea i got it now thanks
topaz sinewBOT
Available help channel!
quiet sorrel
the answer is D but idk y its D. I thot it was C because i seperated the x and y compontents, so X wud be 100cos 30 and Y wud be 100 cos 60. 100cos 30 > 100 cos 60 so i concluded that X goes farther horizontally and j assumed it was the opposite for Y
topaz sinewBOT
the answer is D but idk y its D. I thot it was C because i seperated the x and y...
rigid nimbus
You cannot forget about gravity here
Separating the x and y components is a good first step, but it does not tell you everything you need to solve the problem
quiet sorrel
well sure but i was focusing on the x compontents which is independent of gravity
icy sky
when it lands is based on when the y component is 0
correction: when its y coordinate is 0
rigid nimbus
well sure but i was focusing on the x compontents which is independent of gravit...
The range is also dependent on how long the projectiles are in the air, which is dependent on the y position, which is dependent on gravity
rigid nimbus
You would need to determine the time that each projectile is in the air
quiet sorrel
ok wait i just realized i apporched this question wrong
😭
i think
i know
what to do
now
thanks
.close
topaz sinewBOT
topaz sinewBOT
Available help channel!
neon iron
topaz sinewBOT
iron marsh
create equations
neon iron
how
iron marsh
map out the relationships
you should set each color of marble to a variable or just name each so you know what u are referring to
neon iron
ok
topaz sinewBOT
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neon iron
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neon iron
how to solve this problem?
lucid junco
What hav u tried?
teal gust
split the integral
neon iron
i dont know what to do with the numbers in the denominator
teal gust
take them out of the individual integrals
lucid junco
split the integral
Why bother with 3 integrals when u can just get one common denominator
teal gust
since they are constants
teal gust
Why bother with 3 integrals when u can just get one common denominator
i usually see final answers for these types split up
lucid junco
Eh
teal gust
so i assumed common denominators would be wasteful
plus by splitting it you can individually evaluate each term
helps for beginners ig
lucid junco
It’s not wasteful if it’s more time efficient and easier 
teal gust
but its nto a definite integral you dont need one chunky combined ans
teal gust
Doesn’t matter tho
yes
teal gust
yes
neon iron
what is the next step?
teal gust
you can take out the constants if youd like
as in for the first one it'd be -1/2∫x^4 dx
so your final answer would be -1/2 * (integral of x^4)
for the first bit
topaz sinewBOT
@neon iron Has your question been resolved?
neon iron
<@&286206848099549185>
yes
but remove the integral sign
second line
because you already integrated
ohh okey, thanks for response
how about this?
dm me
okeyy
topaz sinewBOT
@neon iron Has your question been resolved?
tawny gyro
it says the answer is sin(theta/2) but I have no idea why!! could someone explain it to me?
topaz sinewBOT
it says the answer is sin(theta/2) but I have no idea why!! could someone explai...
verbal hound
it says the answer is sin(theta/2) but I have no idea why!! could someone explai...
so what have you tried till now?
tawny gyro
well, the theorum the book gave me is that any angle subtended on the edge of a circle by chord AB will be equal, no matter where it is
but that doesn't help
and, there's no right angles so I can't really use sin, cos, or tan
i know that the length OA and OB are equal
and 1/2
verbal hound
i know that the length OA and OB are equal
right
whenever there are trigo ratios think about right angles
verbal hound
what can you tell about a bisector of the chord AB drawn from the centre of the circle O
yes
tawny gyro
ok lemme try that real quick
verbal hound
what property does that have?
tawny gyro
ok so that creates two right triangles. Then looking at the top triangle, the hypotenuse will be 1/2, and the angle opposite now theta/2 will be 1/2AB
verbal hound
right sorry
tawny gyro
ok so sin theta/2 should be 1/2AB / 1/2
tawny gyro
niceeeeee
ok thank you sm!!
makes sense now
didnt even think about splitting them
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noble bane
Looking to have some work checked here. Just making sure this is what I’m supposed to be doing here.
topaz sinewBOT
Looking to have some work checked here. Just making sure this is what I’m suppos...
noble bane
Prof requested precision to the first decimal, so no worry about sigfigs
verbal hound
Prof requested precision to the first decimal, so no worry about sigfigs
i didnt check the calculation but the method is correct
noble bane
I’m pretty confident about the calculations, but I wanted to confirm the logic and method. Thanks.
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vale burrow
Can someone please explain how it was broken down?
topaz sinewBOT
Can someone please explain how it was broken down?
vale burrow
after the 25/ 5 root 5
verbal hound
ok so 25 = 5*5
so 25/ 5 root 5 = 5*5/5 root 5
=5/ root 5 (just cancelling the 5 in the num and den)
5= root 5 * root 5
vale burrow
I am not understanding this
verbal hound
so 5/ root 5 = root 5 * root 5/ root 5= root 5/1 (just cancelling root 5 in the num and den)
verbal hound
happy to help!
rancid arrow
So my problem is:
Show that the implicit function xy(x+y) = C is the general solution to the differential equation (2xy+y^2)dx + (2xy+x^2)dy = 0
Could someone walk me through step by step if possible exactly what I need to do?*
topaz sinewBOT
So my problem is:
Show that the implicit function xy(x+y) = C is the general sol...
rancid arrow
someone also told me that:
Its homogenous
Use y=vx and simplify
but I'm not sure how to use this information
topaz sinewBOT
@rancid arrow Has your question been resolved?
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rancid arrow
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coarse tusk
So my problem is:
Show that the implicit function xy(x+y) = C is the general sol...
that's not how this server works
we can give you tips on how to do the question, but not the entire question
!status
topaz sinewBOT
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
coarse tusk
do you know what an exact differential equation is and how to solve one?
rancid arrow
do you know what an exact differential equation is and how to solve one?
is it something like we have a differrential equation and then we have a condition what would be the function to solve it?
coarse tusk
yeah something like that
we try to convert $f(x,y)dx+g(x,y)dy=0$ to $\frac{d}{dx}F(x,y)=0$
thorny flameBOT
kheerii
coarse tusk
there are a few conditions for this
which you can learn
i think blackpenredpen has good videos on this topic
quiet bobcat
Yoo, I'm trying to isolate µ, but Geogebra doesn't accept e as Euler's number, so please help, I didn't have log and that stuff yet.
topaz sinewBOT
Yoo, I'm trying to isolate µ, but Geogebra doesn't accept e as Euler's number, s...
quiet bobcat
I want to get d(1/2), but I know how to do the rest
quiet bobcat
Yoo, I'm trying to isolate µ, but Geogebra doesn't accept e as Euler's number, s...
<@&286206848099549185> someone knows how to solve this with me?
I really just need to get µ on the other side alone, <@&286206848099549185>
if I use variables for it, the tool says that
but if I then insert some values it gives this when solving in x
which is not what I want to write on my physics protocol
uhh, please explain it, I don't understand
are you in the wrong channel or does this have something to do with what I wrote, I am not aware of that
no problem
still does someone know?
.close
topaz sinewBOT
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still goblet
what frequency tell us?
topaz sinewBOT
what frequency tell us?
still goblet
in the section of fourier / laplace transform
topaz sinewBOT
@still goblet Has your question been resolved?
topaz sinewBOT
@still goblet Has your question been resolved?
topaz sinewBOT
@still goblet Has your question been resolved?
topaz sinewBOT
@still goblet Has your question been resolved?
topaz sinewBOT
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sweet roost
hi guys can somebody please help me with this question
topaz sinewBOT
hi guys can somebody please help me with this question
sweet roost
They're equal
okay where does the 13C7 come from though?
elfin sparrow
Intuitively, I'm not sure, I would have done it your way
But you can prove that yours is equal to the given answer
by just expanding the nCr expressions
sweet roost
by just expanding the nCr expressions
did you expand them? because either i messed up my calculations or the solutions are not equal
elfin sparrow
Can you show what you did?
sweet roost
but the answer is 13C7 x (10!)^2
13C7 = 1716 = 11x12x13
therefore 10C3 x 3! x 10C4 x 4! must be equal to 10!
10C3 x 3! = 720 = 6!
so 10C4 x 4! = 7x8x9x10
5040 = 5040
nvm im dumb
thats right
elfin sparrow
If you don't have a calculator, you can still show they're the same, like
$$_{10} C_3 \cdot 3! \cdot {10} C_4 \cdot 4! \cdot 13!$$
$$\frac{10!}{3!\cdot 7!}\cdot 3! \cdot \frac{10!}{4!\cdot 6!}\cdot 4! \cdot 13!$$
$$\frac{13!}{7!\cdot 6!} \cdot (10!)^2$$
$${13} C_7 \cdot (10!)^2$$
thorny flameBOT
tatpoj
elfin sparrow
remember the definition of $_n C_r$ is $\frac{n!}{r!(n-r)!}$
thorny flameBOT
tatpoj
elfin sparrow
sorry that took me a sec, my latex is terrible lol
sweet roost
**tatpoj**
yea yea right
i just didnt want to get super big numbers and try to solve it by logic
sweet roost
okay where does the 13C7 come from though?
anyway this is still weird
elfin sparrow
**tatpoj**
If you do it like this you don't have to deal with any big numbers tho
I mean 13! is a big number but you don't need to actually calculate it
sweet roost
**tatpoj**
damn thats smart i just realized
elfin sparrow
I wouldn't have thought to do that though, I only found that because I was trying to turn your answer into the given answer
Like your answer is good too
sweet roost
I wouldn't have thought to do that though, I only found that because I was tryin...
alright thanks very much for the help
i really appreciate it 🙂
elfin sparrow
No problem 👍
sweet roost
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polar stump
topaz sinewBOT
polar stump
The Task is in German so ill translate:
Let K be compact and g differentiable in x*. Show that for all EPSILON > 0 exists a LAMBDA > 0 such that |...| for all x
Idk if LAMBDA is allowed to be inf.
But if so its like super easy, i fear im missing something
topaz sinewBOT
@polar stump Has your question been resolved?
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@polar stump Has your question been resolved?
topaz sinewBOT
@polar stump Has your question been resolved?
topaz sinewBOT
@polar stump Has your question been resolved?
polar stump
Send help pls
polar stump
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lucid sail
Hello, I am trying to graph a function using a handful of limits, I do not believe that my work is correct. I am not sure i am conceptually understanding what f(-5)=2 means for a limit, and also what limit are able to both be true of a specific line(ie; under what circumstances must a new line be made due to two conflicting limits) thank you. (Calc 1)
topaz sinewBOT
Hello, I am trying to graph a function using a handful of limits, I do not belie...
@lucid sail Has your question been resolved?
lucid sail
<@&286206848099549185>
topaz sinewBOT
@lucid sail Has your question been resolved?
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lucid junco
A diffraction grating is used to investigate an atomic spectrum which is known to contain a wavelength of $430nm$ and one other longer wavelength. In the interference pattern produced by the grating, the fourth-order maxima of the $430nm$ line is observed to fall at the same place as the third maxima of the longer wavelength line.
\vspace{1em}
a) Find the wavelength of the second line emitted from the atomic spectrum.
topaz sinewBOT
A diffraction grating is used to investigate an atomic spectrum which is known t...
thorny flameBOT
water beam
lucid junco
how do i start
loud oasis
use the formula for the position of an intensity maximum & set them equal
lucid junco
use the formula for the position of an intensity maximum & set them equal
Set what equal?
wary tulip
loud oasis
you have that the location of each maximum is givn by $d \sin \theta = m\lambda$ where $d$ is the distance between slits, $\theta$ indicates the angular position of the maximum, $m$ is the order of the maximum and $\lambda$ is the wavelength
thorny flameBOT
cloud
loud oasis
think about which of those things are the same for both
lucid junco
<:dorime:1175657706149912677>
slayyyyyyyyyyyyyyyyyyyyyyyyyyyylaaaaaaaaaaaaaaaaaaaaaaaaaa
lucid junco
think about which of those things are the same for both
the wavelength?
wary tulip
loud oasis
we are looking at two different wavelengthss which have a coincident position
lucid junco
we are looking at two different wavelengthss which have a coincident position
is it d then?
loud oasis
that's common to anything involving the same grating, yes
lucid junco
so is that what im looking for?
loud oasis
there is another thing that is common between both situations, based on the setup of the problem
topaz sinewBOT
@lucid junco Has your question been resolved?
lucid junco
there is another thing that is common between both situations, based on the setu...
Fringe spacing?
loud oasis
the spacing between fringes is not the same for different wavelengths
but we are considering a situation where two fringes from different wavelengths are in the same position
lucid junco
but if the wavelengths are in the same position how can they be different?
loud oasis
lucid junco
so what value is this>
loud oasis
think about which variable out of m, d, lambda, theta relates to the physical position of the fringe
lucid junco
so the angle is the same then?
loud oasis
yes
loud oasis
so we have the equation [ d \sin \theta = m_1 \lambda_1] and also the equation [ d \sin \theta = m_2 \lambda_2 ] where $d$ and $\theta$ are common, $m$ and $\lambda$ are distinct
thorny flameBOT
cloud
lucid junco
right
loud oasis
so then given $m_1$, $m_2$, and $\lambda_1$, you can solve for $\lambda_2$
thorny flameBOT
cloud
lucid junco
okay
lucid junco
so then given $m_1$, $m_2$, and $\lambda_1$, you can solve for $\lambda_2$
i got lambda 2 = 573 NM?
is that rite
ok i checked
its rite
thanks so much
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topaz sinewBOT
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trail dragon
I need to map this out
topaz sinewBOT
I need to map this out
trail dragon
its meant to be like parabolas
how do i know
when to
seperate each lines out
i was told to not double up in the x axis
but im not sure what that means
topaz sinewBOT
Available help channel!
rose bough
The domain of this function is x∈[0,inf),x ≠ √3, right?
topaz sinewBOT
The domain of this function is x∈[0,inf),x ≠ √3, right?
gritty birch
The domain of this function is x∈[0,inf),x ≠ √3, right?
no, if x is for example 1, what happens?
rose bough
no, if x is for example 1, what happens?
ohh right the number can't be negative in the square root
x∈(√3,inf) right?
knotty ledge
what about x = -2?
rose bough
what about x = -2?
that'd be -2/√1, I don't see a problem with that
gritty birch
that'd be -2/√1, I don't see a problem with that
exactly, if there's no problem at x=-2, that suggests the function is also defined there right?
rose bough
exactly, if there's no problem at x=-2, that suggests the function is also defin...
right
that still falls into x∈(√3,inf) though
√3 is smaller than 2
gritty birch
-2 is in (sqrt(3), inf)?
gritty birch
so
x∈(-∞,-√3)U(√3,∞)
yup
candid cape
hi
candid cape
i dont know where i went wrong on finding the bearing
someone told me last night how to find it, and from what i can remember, they said to add both angles that are 90
which equal 180
so do i add 180+40?
nvm i figured it out
lol
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versed notch
topaz sinewBOT
molten harness
!status
topaz sinewBOT
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
@versed notch Has your question been resolved?
versed notch
3
molten harness
so whats your answer
versed notch
for part a i got 950 yeah
molten harness
and for b)?
versed notch
2567.6
molten harness
ah i see
you have to calculate the other way around
dont divide with the number but multiply
versed notch
ah alright, ill give it a try thanks
molten harness
answer should be smth in the 300s
molten harness
you're welcome!
versed notch
.close
azure grove
Hello, I got this question and am stuck. I couldn't think of what to do so i just wrote all the angles in the shape and wrote which sides are equal. The shape on the left is a clean slate if the one on the right is a big messy. Any hints would be amazing. Pls ping me initially when u wanna say smth.
topaz sinewBOT
Hello, I got this question and am stuck. I couldn't think of what to do so i jus...
azure grove
Oh yeah, i forgot smth, im not allowed to use a calculator since its from a non calculator past paper
azure grove
yes
fiery badge
so we can basically disregard the other half of this image and forcus on the left part alone
azure grove
alright, so let me make drawing of that to visualise, should only take min
alr, we got this right now
that 15 is for ABE, i put it in there for now in case we need it
fiery badge
Right, so now we've simplified the problem down to one paralelgram with point P
and we know the angles of the paralelogram
we also know AB = x
azure grove
yep
azure grove
ye, we dont know its length
fiery badge
But there's an easy trick to get around that
If I tell you, it'll basically solve the whole problem
try to find it yourself first
azure grove
alright, ill think on it for a bit
fiery badge
focus on what you have and what you wish you had
azure grove
thanks for the help so far
fiery badge
remember AP is known too
azure grove
wait, how do we know AP?
fiery badge
Oh sorry, I meant BP
azure grove
ah, alr, np
fiery badge
ah, alr, np
Did you manage to find it?
azure grove
not yet, rn im just alternating through possible ways
fiery badge
The inconvenient part here is the distance PF, it would be convenient to get rid of it
azure grove
the method im thinking of rn involves some trig, it would find the length of AF but i cant do it without a calculator,
azure grove
oh yeah, i was thinking of right angle triangles, srry
fiery badge
So here's the hint of all hints
well, it's the answer really
draw a line parallel to FE through P which intersects BE at point G
Observe parallelogram ABPG
Here's another hint, the solution has x as a squared term
azure grove
wait, r we trying to find AP here?
fiery badge
No
We're trying to find a way to relate angle ABF to x
You can alternatively observe triangle ABP
azure grove
Alright
fiery badge
Also note that BP is a diagonal to ABPG
and angles in a parallelogram are halved by the diagonals
azure grove
oh, that completely changes things
fiery badge
Wait no
azure grove
ABP = 7.5 degrees i think?
azure grove
alr, that fine
fiery badge
But there is some other property to be looked at here
It doesn't halve it, but it does split it in such a way that angles APB and ABP added up make the angle ABG
azure grove
we know tht BAP is 165 degrees bc of co interior angles, since the other 2 angles must be equal to 180-165 which is 15. Thats the only relation i see
oh, and APG is 15 for the same reasons as ABG
wait, ima search up parallelogram rules, i think im forgetting something easy here
fiery badge
Ignore the paralelogram, we're just looking at the triangle now
the approach here is to simplify the problem to the most convenient shape
We started with a hexagon, got to a parallelogram and now we have a triangle, a fully defined triangle at that
Now just play around with laws of sines and laws of cosines
and some trig identities probably
and you'll get your answer
azure grove
one thing, we haven't learnt anything beyond sine law and cosine law, trig identities are at college level.
azure grove
yeah, im sorry, i dont think thats in my curriculum, some trig identities are in year 12 - 13, which are 2 - 3 yrs beyond where i am right now. A friend of mine has the mark sceme so ill just ask him to show me.
i can send my curriculum if u want to skim through it
fiery badge
Well then it might be solvable with sin and cos laws
but I'm pretty sure you should have done like
sin(x+y) = sin(x)*cos(y) + sin(y)cos(x) by now
azure grove
im sorry, we haven't yet, ill show a pic of all the trig stuff we have learnt so u have a better idea of what im at.
this is the only stuff i know
on trig
fiery badge
huh, so you need to solve this at your grade?
azure grove
yep
wait, i have the answer, ill send it here so u can see
this is the mark scheme
fiery badge
I see
Well I took a different approach because I assumed we're dealing with all of trig available here
azure grove
yeah, thats fair
ill close this help channel since i have the answer now, thanks for the help regardless
.close
topaz sinewBOT
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wispy mountain
Hi I am stuck on this question (part b). I am a bit stuck on the method to achieving the answer, more specifically steps 3 and 4 indicated. The markscheme is not particularly clear and I would like a bit of help understanding what to do. Thanks.
topaz sinewBOT
Hi I am stuck on this question (part b). I am a bit stuck on the method to achie...
wispy mountain
fiery badge
Which part do you not understand?
vestal totem
What do you not understand?
wispy mountain
I do not understand steps 3 and 4 of the markscheme
fiery badge
So what do you think happened at step 3?
wispy mountain
Well I know that the (x-1) is multiplyed up, but from there I am lost
vestal totem
Honestly they just ate a lot of steps.
wispy mountain
Yeah, just got it thanks. The fact they just decided to skips showing steps is really unhelpful. Anyways, thanks.
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topaz sinewBOT
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neon iron
I need help with parabolas
neon iron
I don’t know how to find the equation of a parabola by looking at a parabola
I believe there is different forms of parabolas
glossy arrow
Well you could find it’s min and max and use the complete the square form
neon iron
y=ax^2 +bx + c standard form
y=(x-p)(x-q) Factorised form
y=a(x-h)^2 + k Vertex (TP) form
What do you mean it’s min and max
wraith iron
I don’t know how to find the equation of a parabola by looking at a parabola
what kind of parabolas
do you mean polynomials?
fiery badge
What parabola are you looking at?