i prefer equation
#help-13
crimson sedge
golden raven
the height is the range, the domain is the width i believe
the terms width and height are usually not used to describe equations however, they are usually reserved for rectangular objects
um i would do width=max domain-min domain maybe?
and height=max value of function across domain-min value of function across domain
in this case the domain is [-14,-8]U[8,14]
you alraedy have it written that way
width would just maybe be 14-(-14)=28?
gilded wyvern
yo
gilded wyvern
any help for finding the limit of this?
not allowed to use the sum of k² = n(n+1)(2n+1)/6
cedar kilnBOT
@gilded wyvern Has your question been resolved?
gilded wyvern
<@&286206848099549185>
cedar kilnBOT
@gilded wyvern Has your question been resolved?
fathom condor
cedar kilnBOT
haughty wraith
yes
fathom condor
Ok
He world problem was that Daniel was twice the age of bob 4 years ago
Bob is y
So bob was 0 and Daniel was 2
?
velvet mortar
This is a worldwide problem of course
fathom condor
show the full question
Idk it was on like a test
I did
It was like
4 years ago daniel was twice the age of bob.
6 years from now bob will be the same age Daniel is now
How old are they
@haughty wraith
So Daniel was 2 and bob was 0
4 years ago
fathom condor
Daniel current age = x, bobs current age = y
y-4 = 2(x-4)
y+6 = x
Welp
I screwed that up
haughty wraith
😦
haughty wraith
you can .close
fathom condor
.close
strong jacinth
hi
haughty wraith
what have u tried
wary kelp
set one of the legs as x and get all the other legs in terms of x and add them add and equate to 153 to find x and all the other legs
cedar kilnBOT
cedar kilnBOT
Available help channel!
crystal mason
Let (G, *),(H, ·) be groups. Show or refute the following:
i) If G is isomorphic to H, then G is abelian if and only if H is abelian.
The answer just seems so obvious idk what im supposed to do
long swan
why not try showing a contradiction
crystal mason
oh i thought it was true
crystal raptor
It is true
If you want a direct approach, G being abelian means g*h = h*g for all h and g, see what happens if you apply the isomorphism to both sides
cedar kilnBOT
@crystal mason Has your question been resolved?
crystal mason
okay so from what i know
the mapping is isomorphic if it fulfills:\
- $f (x * y) = f (x) \cdot f (y)$\
- its bijective
wraith daggerBOT
Levens
crystal mason
..with x and y being in G
so im gonna assume u mean applying the first one so:\
$g_1*g_2 = g_2 *g_1 \longrightarrow f(g_1) \cdot f(g_2) = f(g_2) \cdot f(g_1)$ with $g_1,g_2 \in G$ and $f(g_1),f(g_2) \in H$?
crystal raptor
Right and the things in the second equation are elements of H
Because f is bijective (in particular surjective) every element in H looks like that
(h was a bad letter to pick for an element of g lol sorry forgot we had H)
crystal mason
we can do g_1 g_2
crystal raptor
Sure, same logic applies
wraith daggerBOT
Levens
crystal mason
Because f is bijective (in particular surjective) every element in H looks like ...
okay... so how does it show that if one is abelian, so is the other one?
crystal raptor
Well we've shown any two elements in H commute
By assuming any two in G commute
The other direction of the if and only if is practically identical
crystal mason
**Levens**
but like, isnt this just jumping straight to showing that H is commutative?
crystal mason
<@&286206848099549185>
cedar kilnBOT
@crystal mason Has your question been resolved?
crystal mason
.close
cedar plume
I need help
calm sierra
.close
sudden vortex
considering ∃x∀yP(x, y) to be false, how can I find the logical value of ∀x∀yP(x, y) and ∀x∃yP(x, y)?
craggy oriole
∀x∀yP(x, y) is also false because you can instantiate the x with the one that makes it false
∀x∃yP(x, y) could be true, is that all the details you're given?
sudden vortex
yep
sudden vortex
∀x∀yP(x, y) is also false because you can instantiate the x with the one that ma...
how so?
craggy oriole
Wait, I misread lol
sudden vortex
I was thinking about going with negations ¬, but cant find equivalence between the propositions
craggy oriole
You can move a negation in by flipping the quantifier so ¬(∃x∀yP(x, y)) becomes ∀x¬(∀yP(x, y))
sudden vortex
∀x¬(∀yP(x, y)) became true here right?
craggy oriole
Yes, we start by the negation of the thing we know false
sudden vortex
but I can't see how to shape the other propositions to fit this
the first one if negate will become ∃x∃y¬P(x,y) | ∃x¬(∀yP(x, y))
sudden vortex
the second will be true then?
if yes, I can see how
could you explain how it will contradict the first?
sudden vortex
Repeating the process you know that ∀x∃y¬P(x, y)
it's just because the ∃y part in this?
craggy oriole
Yes, it says for any x there's a y such that P(x,y) is false
So it's not true that for any x and any y P(x,y)
The second could still be true 🤔
sudden vortex
thinking that the second could be an invalid proposition, but that makes no sense i guess
to put in a exercise
cedar kilnBOT
@sudden vortex Has your question been resolved?
sudden vortex
using the negation of the second (alias to 2) exists an x for any y ¬P(x,y), so using the false given proposition (alias to 0) that says exists an x for any y P(x,y), therefore if there's at least an x that fulfil the prop 0 there's at least one that don't?
making the prop 2 true? (∀x∃yP(x, y))
cedar kilnBOT
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Available help channel!
wary nebula
Probabilities
cedar kilnBOT
**Probabilities**
wary nebula
1.Probability of the addition of the number of 6 dice rolls equals 7
bright agate
1.Probability of the addition of the number of 6 dice rolls equals 7
,tex \left(\frac {1}{6^{5}}\right
wary nebula
bruh
wraith daggerBOT
ayushch80
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
dire geode
@bright agate stop and go to #latex-testing
bright agate
<@913540108442042419> stop and go to <#844681108473118750>
😅
wary nebula
can you help me @dire geode
bright agate
1.Probability of the addition of the number of 6 dice rolls equals 7
possible outcomes that equals to 7
{1,1,1,1,1,2}
{1,1,1,1,2,1}
{1,1,1,2,1,1}
{1,1,2,1,1,1}
{1,2,1,1,1,1}
{2,1,1,1,1,1}
total no. of possible outcomes = 6^6
wary nebula
.close
cedar kilnBOT
cedar kilnBOT
Available help channel!
oak moat
cedar kilnBOT
oak moat
If I have a function like this
Do the difference quotient, and from both sides it gives me the same number
Does that mean that the derivative exists? Bc in x=0 the derivative will be 0
oak moat
Just use definiton of derivative?
So if that gives me 2 from both sides, it doesnt matter if when x=1 is not equal to the latterals value of the derivative?
elfin hemlock
f’(0)=lim h->0 of (f(h)-f(0))/h
This is definition
Nothing other than definition should matter
cedar kilnBOT
@oak moat Has your question been resolved?
bronze token
cedar kilnBOT
bronze token
What are these numbers man
I’m about to eat the monitor
So I took the derivative of r
Then found its length
Then placed it in a definite integral from 0 to 1
That’s what I’m supposed to do to find the arc length, right?
But the integral itself is making me lose my mind
I made an online calculator do the work to see if I had at least done it right to this point, and it didn’t work
granite wyvern
But the integral itself is making me lose my mind
The expression under the square root is a perfect....
bronze token
A perfect way to lose my mind?
granite wyvern
A perfect way to lose my mind?
perfect square
lean shell
Im trying to prove that the union of 2Z and 3Z are not a subgroup of Z, however when i plug it into the criteria for when a subset is a subgroup, i only get that it would be a subgroup. Whats the argument to say 2Z u 3Z isnt a subgroup?
crimson delta
Well is 2+3 in the union?
lean shell
Its not, since 2Zu3Z is 2,3,4,6,8,9,10,12 ... etc but that shouldnt affect this no?
I mean these should be the criteria right?
- the neutral element is in the subgroup
- the inverse of every element of the subgroup is also in the subgroup
- if a and b are in the subgroup then ab is in the subgroup
Oh since i could plug in + or * both products need to be inside the subgroup? Is that what you mean
random shale
groups are closed by the operation they are defined on
so if our operation is addition then the union is closed by addition if its a subgroup
if our operation is multiplication it needs to be closed under multiplication
lean shell
My task didnt define the operation, so is that for me to decide?
Its kind of vague:
Zeigen Sie, dass die Vereinigung 2Z ∪ 3Z keine Untergruppe von Z ist.
"Show that union 2Z u 3Z isnt a subgroup of Z"
sudden hawk
lean shell
Kind of, that should be my last question
crystal raptor
well they are subgroups of Z, the group operation on Z is addition
leaden snow
Prove $$\abs{x} + \abs{y} \leq \abs{x+y} + \abs{x-y}$$
So far, I have shown this by splitting up into 4 possible cases but I was told this is provable in a quicker way using the triangle inequality. Currently, I'm not seeing the trick, though. Any hints appreciated.
wraith daggerBOT
nvx
hollow minnow
I feel like you could look at 2|x| and. 2|y| separately then add them
By noting that 2x=(x+y)+(x-y) and find similar expression for 2y then apply the triangle inequality
@leaden snow
cedar kilnBOT
@leaden snow Has your question been resolved?
leaden snow
\begin{align*}
2 \abs{x} &\leq \abs{x+y} + \abs{x-y} \
2 \abs{y} &\leq \abs{x+y} + \abs{y-x}\
2 \qty( \abs{x} + \abs{y} ) &\leq 2\qty( \abs{x+y} + \abs{x-y} ) \
\abs{x} + \abs{y} &\leq \abs{x+y} + \abs{x-y}
\end{align*}
wraith daggerBOT
nvx
hollow minnow
novel ginkgo
hilp
novel ginkgo
how to do this
south tundra
Draw a line perpendicular to OD that passes through S and another line perpendicular to DV passing through S
south tundra
You should get right triangles where 2 of the sides are known
novel ginkgo
is lo and ve congruent @south tundra ?
south tundra
Yes
novel ginkgo
can i ask is LO=VE=2 square root of 239?
cedar kilnBOT
@novel ginkgo Has your question been resolved?
static hound
cedar kilnBOT
leaden snow
implicitly differentiate and then plug in 4/3 for dy/dx (or y')
crimson sedge
To converte cosine angle into sine angle i need to use sin x = cos(pi/2 - x) right?
crimson sedge
so if cosine is pi/3, is sin x value gonna be pi/6?
wraith daggerBOT
wraith daggerBOT
toxic moat
,w graph sin
crimson sedge
ty
wraith daggerBOT
toxic moat
,w graph cos
wraith daggerBOT
toxic moat
u can prob observe from here what translation to get from 1 to the other
there are a few i guess
a few common ones
theres infinitely many LOL
crimson sedge
but ther eis also one formula ```ctg x = ctg(pi/2 + x - pi/2) = -tg(x-pi/2)
like whats the deifference between this and that one?
toxic moat
yea um
tangent is kinda different right
but normally u can observe the graph as before
crimson sedge
so this is particular for tangent and cotangent?
toxic moat
and try to see what translations would get from 1 to the other
toxic moat
so this is particular for tangent and cotangent?
yes
crimson sedge
alr
toxic moat
idk some translation
u can prob figure it
some flip about y axis n translation im guessing
so that gets the -x i guess
crimson sedge
,w ctg(5pi/7)
wraith daggerBOT
crimson sedge
,w -tg(3pi/14)
wraith daggerBOT
toxic moat
also if u just doing it urself im pretty sure u can just google it
its faster
but use
crimson sedge
ok
toxic moat
tan and cot
crimson sedge
but ther eis also one formula ```ctg x = ctg(pi/2 + x - pi/2) = -tg(x-pi/2)
```
so tan and cot use this
this shit is confusing me
everything is the same
toxic moat
ok google doesnt recognise cot but idk wolfram or similar online cld prob get u answers faster
i kinda did it here so u can see what i entered as well i guess
crimson sedge
,w tg(pi/3)
wraith daggerBOT
crimson sedge
,w cot(pi/2 - pi/3)
wraith daggerBOT
crimson sedge
but ther eis also one formula ```ctg x = ctg(pi/2 + x - pi/2) = -tg(x-pi/2)
```
aw man if this is same with this then why do we need this formula?
toxic moat
but ther eis also one formula ```ctg x = ctg(pi/2 + x - pi/2) = -tg(x-pi/2)
```
like this and
toxic moat
To converte cosine angle into sine angle i need to use ```sin x = cos(pi/2 - x)`...
this?
crimson sedge
ye
toxic moat
i mean its quite different right
i mean i guess u can say the 2nd one is the first 1 applied
but is a bit much
crimson sedge
what im tryna do is to find something like arccos(sin(x))
so i can use both right?
toxic moat
err
not rly?
u wouldnt use cot here
i think what u would do is change that first sin to a cos
using that first rule
then uve
arccos (cos (stuff)) = stuff
under some restricted domain
wraith daggerBOT
wraith daggerBOT
crimson sedge
bruh i confused myself
marsh pond
cedar kilnBOT
marsh pond
This is feeling really wrong.. is the second step legal?
Also am I allowed to remove 6x from the denominator like that?
I couldn’t factor the quadratic so I must have done something wrong
The second step is an experiment I did for cross multiplication and keeping proportions
But I don’t think it’s allowed with two terms are added or subtracted?
tropic oxide
$\frac{1}{3} \neq \frac{6}{3x}$ and $\frac{x}{2} \neq \frac{6x}{2x}$
wraith daggerBOT
Ann
tropic oxide
"so the step you pointed out is the issue?"
it's less "you're not allowed to do this" and more "what you did is a good idea but you screwed up in its execution"
marsh pond
Oh let me check again
marsh pond
it's less "you're not allowed to do this" and more "what you did is a good idea ...
Should I be cross multiplying two terms on the left side like this instead?
Oops sorry that should be 3 in middle denominator
Just trying to understand the the idea of cross multiplying to preserve proportions and if it’s even possible if more than 2 fractions. It can be single rational expressions on both sides I get that part but not sure about more than 2 fractions
crimson sedge
The rule is generally multiply each fraction by something that equals out to one, bc a number multiplied by 1 is still the same number
Ie) x/x, 2/2, 6/6, etc.
marsh pond
Right yeah I could have done that and it would have been much easier to solve
This is more out of curiosity if cross multiplication can even work here
Or if it’s strictly for 2 fractions only
lethal jackal
to answer that question, you need to see if you can construct a justification for what you're trying to do
not trying random symbol manipulations and then asking some Oracle of Delphi whether it's right
marsh pond
not trying random symbol manipulations and then asking some Oracle of Delphi whe...
Errr, Oracle of Delphi? Sorry, I’m not following.. just asking a question about “can cross multiplication be used for more than two unique fractions? If so, is my method correct?”
lethal jackal
it's a mythological reference
my point is that you shouldn't be randomly pushing symbols around the paper and having to ask someone whether that's "correct"
marsh pond
@lethal jackal shouldn’t we recommend to everyone to stay curious about math and to ask questions?
lethal jackal
yeah, but not random, unguided questions
marsh pond
I’m not randomly there is method to it
lethal jackal
if I ask whether f(x) > 1 and then whether f(x) > 2 and then whether f(x) > 3, ... then I'll have wasted a ton of time
I'm saying that you should be able to figure this out yourself by exploring the logical justifications for why "cross-multiplication" is a valid thing to do
and the hint is that it usually involves common-sense steps like "multiply both sides of the equation by the same thing" or "divide both sides of the equation by the same thing"
and in a more general sense, you can do whatever you want mathematically as long as you have some sort of reasoning to back it up
marsh pond
Alright… I understand for single fractions. But don’t quite understand it for multiple fractions. If it’s possible to do.
lethal jackal
well, I'll enlighten you a little about what this "cross-multiplication" means
Let's say you have a/b = c/d
why is it that you can "cross-multiply" to get ad = bc?
well, first we can conclude that a = bc/d because multiplying both sides by b doesn't change the truthfulness of the equation
and then we can conclude that ad = bc, because multiplying both sides by d doesn't change anything
but the point here is that I'm not just randomly pushing the letters around and then having to ask "is this okay? / should I memorize I can do this?"
what you're asking is basically like trying to solve something like x/3 = 6/7, and then asking "can I add 11 to both denominators? is that right?" and changing the equation to x/311 = 6/711
there's absolutely no logical justification for doing that
lethal jackal
Should I be cross multiplying two terms on the left side like this instead?
and likewise with these types of questions -- there's no SHOULD you do anything. You're free to do whatever you want, as long as each step logically follows from the previous. There are multiple ways to do many things.
and likewise, if you don't have a solid logical reason for doing something, you cannot do it, even if you think it will yield a correct answer
marsh pond
Thank you for the time to write it out. I really do appreciate your time and I will go through your notes in detail.
Can I ask you a simple question, maybe thinking back to when you were first learning math: do you feel that sometimes math teachers can be condescending to new learners?
lethal jackal
I think that the math teachers that I had when I was first learning math were almost all universally terrible
and that's why so many people think that they can't do math or despise it or try to memorize their way out of it
math is not about memorizing some procedure like is taught in lots of elementary classes
math is about constructing logical arguments from things that you know to prove other things
it's not about arithmetic, or numbers, or algebra, or remembering whether you do a + or - there or remembering to cross-multiply or divide here
marsh pond
I think that the math teachers that I had when I was first learning math were al...
That’s unfortunate to hear. It seems so common place these days for new students. Math is a topic that is difficult to learn, and if not more so it is very difficult to teach, and to teach effectively. It has one of the highest fail rates of any subject. But once you put the time into it and understand it, also the highest passing grades too. I will keep plugging away at it, I apologize if my question comes across as logical and easy. “Just use your brain cells and stop being lazy” I found that answer so bizarre when a student asks my professor a question.
lethal jackal
it's not about being lazy, and I apologise if you got that message
it's about not trying to memorize every single truth and taking it as true just because someone told you to do so
or someone told you that it's right
part of building mathematical intuition and understanding is asking yourself why things work
being able to put into your own words the underlying logical argument
for example, with cross-multiplication, it's a convenient mnemonic for remembering things, but I would never say "I can cross-multiply, just because." I ultimately understand that ross-multiplying a/b = c/d is just multiplying both sides by bd
marsh pond
https://youtu.be/59r_yhp9e_g
it is only thanks to Professor Leonard that I am sticking with it.. his videos are very well articulated and he is one of the most effective math teachers I have come across
cedar kilnBOT
@marsh pond Has your question been resolved?
marsh pond
Just FYI in case anyone is curious: “No, you cannot cross multiply when adding fractions. Cross multiply only when you need to determine if one fraction is greater than another, or if you are trying to find a missing numerator or denominator in equivalent fractions.”
.close
crimson sedge
i have a question
cedar kilnBOT
i have a question
crimson sedge
i have 0 and 1 on y axis so if i have to find where 0.25 is how should i plot that onto it
like i understood for 0.5 and like 0.8 i am not getting 0.25 and 0.35 thing
is there an easier way to recognise it
0 and 1 on y axis
lethal jackal
for 0.25
find 0.2 and find 0.3
it's right in the middle
if it's like 0.22, put it closer to 0.2 than 0.3
there's guesswork involved
crimson sedge
i am mostly stuck when i have to deal with numbers such as 0.25 and 0.35
crimson sedge
find 0.2 and find 0.3
oh
0.22 will be closer to 0.2
did not understand the logic of this one
gleaming cloud
0.2 = 0.20
lethal jackal
0.22 is closer to 0.2 than it is to 0.3, yeah?
gleaming cloud
What's 0.22 - 0.2?
crimson sedge
0.22 is closer to 0.2 than it is to 0.3, yeah?
it looks like the logic
crimson sedge
What's 0.22 - 0.2?
0.02
gleaming cloud
What's 0.3 - 0.22?
crimson sedge
What's 0.3 - 0.22?
0.03
lethal jackal
0.08
crimson sedge
What's 0.3 - 0.22?
0.08
crimson sedge
0.22 is closer to 0.2 than it is to 0.3, yeah?
so if they say 0.45
crimson sedge
i understood it
i will try this in my next questions
thank you for the help
.close
cedar plume
Hi I need help
cedar kilnBOT
Hi I need help
lethal jackal
likewise I need help with my head
cedar plume
cedar kilnBOT
@cedar plume Has your question been resolved?
lethal jackal
probably should state what you're thinking and what you've tried
because right now, looks like you're just asking what to put in the blanks
cedar plume
I don't understand how to do this proof
lethal jackal
that's not helpful information
do you know what everything already written there means?
do you understand what "complete the proof that -snip-" means?
cedar plume
do you know what everything already written there means?
Yes
lethal jackal
then what statement are you asked to prove, and why is it true?
cedar plume
I am asked to prove STW congruent to UVW
lethal jackal
and why is it true
sonic thistle
@cedar plume you have to type .close
ivory finch
why is 1 to power of infinity indeterminate?
cedar kilnBOT
why is 1 to power of infinity indeterminate?
south tundra
Take a look at limits 1^x as x approaches infinity and (1 + 1/x)^x as x approaches infinity for example
In both cases you have forms 1^infinity and different answers
south tundra
wont it always be 1 anyways?
No
In the latter case the limit is equal to e
Which is not 1
mighty drift
If you adapt it to be (1+1/x)^kx for varying k, you can actually get any number and still 1^inf
ivory finch
im saying that power is like multiplying the same thing
for example 1^2 is 1*1
so for any number itd always be 1
is 0^infinity an indeterminate form?
south tundra
is 0^infinity an indeterminate form?
I think not
ivory finch
1^99999999999999999=1
ivory finch
0^1000=0
south tundra
It's approaching 1
ivory finch
0^9999999999999=0
mighty drift
It's like 1.001^1000
,w 1.001^1000
wraith daggerBOT
mighty drift
(1+1/x)^x doesn't go to 1
ivory finch
how did it change to 1.001?
static crag
(1+1/x)^x doesn't go to 1
it goes to e right?
south tundra
how did it change to 1.001?
Do you know what a limit is?
south tundra
Pretty much
mighty drift
It's provable. It's far from easy at your level though
ivory finch
like basically i believe 1 times any positive number is still 1
and idk why 0^infinity isnt indeterminate either then
sonic thistle
because
mighty drift
like basically i believe 1 times any positive number is still 1
It's not exactly 1
sonic thistle
infinity isnt a number
mighty drift
That's what you don't understand
ivory finch
yeah, i am baffled as to how the problem changed
mighty drift
Any minor difference can be amplified infinitely hard by putting it to a very high power
mighty drift
yeah, i am baffled as to how the problem changed
Formulate it rigorously then ?
The usual way would be to consider all numbers of the form x^y and look at what happens when x goes to 1 and y goes to infinity
cedar kilnBOT
@ivory finch Has your question been resolved?
cedar kilnBOT
cedar kilnBOT
Available help channel!
digital bloom
cedar kilnBOT
digital bloom
Why is my answer to c wrong?
balmy geode
Why is my answer to c wrong?
The first line is itself incorrect: $E^c \cup F$ is not equivalent to $E \cap F^c$. Instead, the "rule" would say that [(E^c \cup F)^c = E \cap F^c]So you're missing a complement there...
wraith daggerBOT
Sid
cedar kilnBOT
@digital bloom Has your question been resolved?
cedar kilnBOT
cedar kilnBOT
Available help channel!
cedar plume
cedar kilnBOT
cedar plume
is this correct i am not sure
cedar kilnBOT
@cedar plume Has your question been resolved?
jovial anchor
is this correct i am not sure
line 8 is pointless
cedar plume
.close
sudden escarp
why did it become (x-1)^2 not (x+1)^2?
cedar kilnBOT
why did it become (x-1)^2 not (x+1)^2?
sudden escarp
the positive one comes from (-2/2)^2 , which is the rule for completing the square
nvm i did the rule wrong
sorry for the inconvenience\
.close
dense wyvern
Okey so
dense wyvern
Theres a hotel
In the hotel theres 2 kind of bedrooms
one with 3 beds
one with 5 beds
in total in the hotel theres 59 beds
and we also know that there is 1 3-bed-bedroom more then a 5-bed-bedroom
now you need to figure out how many 3-bed-bedrooms and how many 5-bed-bedrooms there is
could it be x+1+y=59
the starting thing ?
steep meadow
Let's say the amount of 5-bed-bedrooms is equal to x. Then obviously the amount of 3-bed-bedrooms is equal to x+1. The total of the 5-bed-bedrooms and 3-bed-bedrooms is then x+x+1
Does that help?
steep meadow
Well yes they're different types of rooms
But that's not really what we're counting right?
steep meadow
Oh my bad
Oh no not my bad
Got confused rhere6
There**
Once you solve x
You know how many 5-bed-bedrooms there are right?
Since we assumed x is equal to the amount of 5-bed-bedrooms
And given is that there's exactly 1 more 3-bed-bedroom
steep meadow
So the amount would be x+1
tropic oxide
But shouldn't it be x and y?
are you under explicit instructions to define two variables? if not then you don't need to do that.
though of course you still could.
steep meadow
I suppose you could also define it like x+x+1=y
Which gives y=2x+1
Then we want to calculate for which x the value y is equal to 59
So we get 59=2x+1
dense wyvern
58:2
steep meadow
Right which is?
steep meadow
Well no y would still be 59 which is the total beds
dense wyvern
oh
dense wyvern
yeha that's what I meant
steep meadow
Well then great :)
steep meadow
Right
dense wyvern
How can the hotel have only 59 beds?
steep meadow
Now that's a good cuz I didn't think about the total beds but the total rooms lol
In that case you have 5x and 3(x+1)
Since x is the total 5 bed bedrooms then the total beds is 5 times x which is 5x
dense wyvern
mhm mhm
steep meadow
And x+1 is the total 3 bed bedrooms
Then the total beds in that case is 3 times x+1 which is 3(x+1)
dense wyvern
u mean 3*x+1?
steep meadow
Nope
dense wyvern
yyp
steep meadow
The parentheses are important here
dense wyvern
Yup
steep meadow
And from there solve for x again given y = 59
steep meadow
In this case that should be correct now
dense wyvern
mhm
Then the answer in my book is incorrect
cuz it says 8 3 bed bedrooms
and 7 5 bed bedrooms
steep meadow
Or I'm making a mistake lol
dense wyvern
so 7 5 bed bedrooms
steep meadow
Exactly
steep meadow
Exactly
dense wyvern
Yeha so I was correct
cedar kilnBOT
cedar kilnBOT
Available help channel!
misty rain
How would I find the domain of this function?
cedar kilnBOT
How would I find the domain of this function?
dire geode
differentiate the equation
misty rain
differentiate the equation
How would I differentiate the equation?
Like what would be the setup
dire geode
differentiate both sides of the equation
misty rain
how would I differentiate f(x)?
dire geode
and evaluate the initial condition f(2)
cosmic steppe
Fundamental theorem of calculus
dire geode
How would I find the domain of this function?
then y=f(x)
dire geode
how do I render the equation on discord
read #latex-testing for examples
wraith daggerBOT
Nathaan
Compile Error! Click the reaction for more information.
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misty rain
oh it worked
misty rain
read <#844681108473118750> for examples
what would go in the integral though?
like would it be y(cbrt{t})
or something else?
dire geode
Fundamental theorem of calculus
.
cosmic steppe
Le ping
misty rain
Oh is it like so F(x^3)-F(8)
wait thats not right
wait is it?
oh wait is it F(x)-F(2)?
misty rain
ok so i know that integral a^b f(x) is F(b)-F(a)
but I'm not sure how it would apply when its liek integral a^b f(cbrt{x})
would it be F(cbrt{b})-F(cbrt{a})?
dire geode
use u-sub
misty rain
Ok so $3+3sqrt[3]{t^2}(F(x^3)-F(8))$
wraith daggerBOT
Nathaan
dire geode
show your steps and justify each one
.
misty rain
so I u sub the f(cbrt{t})
dire geode
$f(x) = 3 + \int_8^{x^3} f(\sqrt[3]{t})dt$
misty rain
so u=cbrt{t}
wraith daggerBOT
riemann
misty rain
$f(x) = 3 + \int_8^{x^3} f(u)dt$
wraith daggerBOT
Nathaan
misty rain
and then
cedar kilnBOT
@misty rain Has your question been resolved?
ember imp
Hi, sorry, I asked about this question here earlier but I'm still not able to get the correct answer. Can someone try calculating this so I can compare what it should be?
bold vine
If you can, show your work
ember imp
Sure, will do. Let me actually close this room since I've got to attend a meeting right away, will reopen with my work shown after the meeting is over
.close
sharp vale
cedar kilnBOT
sharp vale
anyone able to help with this?
cedar kilnBOT
@sharp vale Has your question been resolved?
sharp vale
<@&286206848099549185>
sweet sluice
well for the first step a and b are right angles so they must equal _
sharp vale
.close
crimson sedge
cedar kilnBOT
crimson sedge
Forgot to ask about the question below
what does it mean by m<ABD
don’t see an m anywhere in the problem
livid hound
m< is notation for measure of the angle
crimson sedge
So it’s just asking what ABD is equal to?
which would just be the same answer
as the question above
crimson sedge
what do I have to find in order to solve this problem
livid hound
m<ABD = x-8
digital gazelle
cedar kilnBOT
digital gazelle
dont understand how they suddenly made this an equation
or why r^4 is equal to that
cedar kilnBOT
@digital gazelle Has your question been resolved?
cedar kilnBOT
cedar kilnBOT
Available help channel!
oblique canyon
cedar kilnBOT
gaunt hamlet
!15m
cedar kilnBOT
Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.
crimson sedge
Mate stop it it’s once every fifteen, respect server rules
oblique canyon
can you help?
crimson sedge
Also sorry I suck at math in English
Got no clue what sec is altho I’ve seen it on the server
But have you just looked online for relations between sec and cos ?
oblique canyon
secant
oblique canyon
sec is 1/cos
crimson sedge
Ohh
We call that sec as well I think actually 💀
And + and - there mean that it tends toward orrr…
oblique canyon
wym
crimson sedge
Like $sec (θ+1) or sec(θ) +1$
wraith daggerBOT
ಠ_ಠ
crimson sedge
That kinda changes everything
oblique canyon
second one
crimson sedge
Ok
oblique canyon
sec(x) +1
crimson sedge
So
I guess the first step
Is saying that
And then
But then again idk
You introduce cos
Unless there’s a specific property for sec that helps you solve immediately from that ?
oblique canyon
uuhh
crimson sedge
Lmao idk it’s 4 am I am malfunctioning
I told you it’s not a good idea 😭😂
I mean basically you have
But that doesn’t help ah all
Idk factorizing and sayin 2>0 so it has to be sec part =0 seems more fun to me 😭😂
Anyways good luck lol 😂🤝
warm cedar
hi
warm cedar
tìm m để B tồn tại ma trận nghịch đảo >>>>>>>>>>>>>>>>>>>>>>>>>>>> Finding M for B inverse.
I need someone help me this one pls
warm cedar
take determinant
so
round dew
warm cedar
I find det B= 4m-3
warm cedar
take determinant
I think it is adjon?
dull oxide
just solve det B != 0 and that's your answer.
As long as det B != 0, then the matrix is invertible
warm cedar
As long as det B != 0, then the matrix is invertible
ok. Thanks
cedar kilnBOT
@warm cedar Has your question been resolved?
cedar kilnBOT
@warm cedar Has your question been resolved?
cedar kilnBOT
cedar kilnBOT
Available help channel!
round dew
hi
round dew
/roles
oblique pier
hi
oblique pier
May you help me
im sorry i dont know the language that the question is in
round dew
im sorry i dont know the language that the question is in
Finding M for Det A= -9. A=(m,2,3)
round dew
idk how to do that question
thanks. It is Linear Algebra
oblique pier
yeah i dont know linear algebra, i can recomend another server with a linear algebra section
round dew
yeah i dont know linear algebra, i can recomend another server with a linear alg...
can you recommend it
round dew
yeah i dont know linear algebra, i can recomend another server with a linear alg...
pls. Thanks
🥺🥺🥺
cedar kilnBOT
@round dew Has your question been resolved?
bold compass
The answer is C. F''(x) will always be positive. how come B wouldnt work aswell. F(x) is concave up when f" > 0.
haughty wraith
The answer is C. F''(x) will always be positive. how come B wouldnt work aswell....
how did you get f''(x) will always be positive?
bold compass
f double prime is 12ax^2 +2b and its ab > 0
haughty wraith
yes
bold compass
so f double prime will positive for all values of x
haughty wraith
what is -1 * -1
bold compass
1?
bold compass
.close
cedar kilnBOT
cedar kilnBOT
Available help channel!
slate creek
cedar kilnBOT
slate creek
i don't understand how I got this wrong
gusty forum
i don't understand how I got this wrong
you're asked the answer in terms of [mean]± error
you gave your answer as bounds of your confidence interval
buoyant latch
you're asked the answer in terms of [**mean**]± **error**
you gave your answer ...
What’s the difference
gusty forum
there is no difference, they just gave the wrong form
if your instructors expected you to answer {2} ± {1} and you didn't pay attention and answer {1} ± {3} intending the interval [1,3] you got the right idea but are getting a 0
buoyant latch
But then it should be in the form [1,3]?
gusty forum
yes, but they were asked in the form of {mean} ± {error}
buoyant latch
Ah I see
If you found the confidence interval would you say like the average of the confidence interval range +- value to reach the boundaries
For the form {mean} +- {error}
cedar kilnBOT
@slate creek Has your question been resolved?
warm vector
How should I do this?
cedar kilnBOT
How should I do this?
warm vector
should I use ar^n-1 ?
ember geode
mhm
crimson sedge
how to find slope of secant line in terms of x and h for this function: f(x)=9x^2+6x
cedar kilnBOT
how to find slope of secant line in terms of x and h for this function: f(x)=9x^...
@crimson sedge Has your question been resolved?
shy girder
Can someone help please show that if u.v=0 then |u+v|=|u-v|
cedar kilnBOT
Can someone help please show that if u.v=0 then |u+v|=|u-v|
gusty forum
evaluate the right equality directly
shy girder
like bring it all to one side?
crimson delta
remember that |x|^2 = x.x
crystal raptor
^2?
shy girder
so what should I do?
dull oxide
Alternative method: If uv=0, what could u or v be?
frosty ocean
I thought of u and v to be vectors, and dot products
shy girder
yep they are
crimson delta
|u+v|^2 = (u+v).(u+v)
gusty forum
remember that |x|^2 = x.x
use this and evalute the right equality
shy girder
when u mean other side do u mean for |u-v|?
crimson delta
yes
frosty ocean
Yes
frosty ocean
.close if you have no more doubts
shy girder
.close
dull oxide
At @shy girder here's a nice visual proof too
cedar kilnBOT
At <@561106885739020297> here's a nice visual proof too
dull oxide
.close
cedar kilnBOT
cedar kilnBOT
Available help channel!
pale mulch
if we have f,g 2 endomorphisms... is there something quick to deduce from this?
pale mulch
what should i understand
the question is if g is diagonisable show that f is nilpotent
vocal sparrow
you can show that f maps eigenvectors of g to eigenvectors of g
cedar kilnBOT
@pale mulch Has your question been resolved?
pale mulch
you can show that f maps eigenvectors of g to eigenvectors of g
okay but what does the equality tell us in general
fg-gf=af
what should i deduce as soon as i see this
vocal sparrow
that f maps eigenvectors of g to eigenvectors of g
pale mulch
that f maps eigenvectors of g to eigenvectors of g
How did you deduce this
vocal sparrow
I wrote "suppose g(x) = lambda x" and then I wrote the identity for such an x
pale mulch
I wrote "suppose g(x) = lambda x" and then I wrote the identity for such an x
why did you think of doing this
vocal sparrow
because I was trained to
pale mulch
I wrote "suppose g(x) = lambda x" and then I wrote the identity for such an x
the identity ?
wdym identity for such an x
vocal sparrow
f°g - g°f = alpha f
pseudo merlin
hi!
neon moon
Hint: Consider the curve of the target function above x-axis.
pseudo merlin
Tyy
cedar kilnBOT
@pseudo merlin Has your question been resolved?
pseudo merlin
Hi i have a question!
cedar kilnBOT
Hi i have a question!
pseudo merlin
How come when converting cm cubed to litres its the same as metres cubed?
Like 1000cm^3 is equal to 1000L and 1000m^3 is equal to 1000L
It's really confusing me
And how do i do this?
bright agate
Like 1000cm^3 is equal to 1000L and 1000m^3 is equal to 1000L
,tex 1m^{3} = 1 L
wraith daggerBOT
ayushch80
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(You may edit your message to recompile.)
gusty forum
Like 1000cm^3 is equal to 1000L and 1000m^3 is equal to 1000L
1000cm^3 (10cm cube)= 1L
bright agate
,tex 1000cm^{3} = 1 L = 1m^{3}
wraith daggerBOT
ayushch80
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
gusty forum
what no
1m^3 is 1000L or a ton
pseudo merlin
Ohhhhh
pseudo merlin
And how do i do this?
And this question please!
gusty forum
complete the square
pseudo merlin
Like how?
gusty forum
intuition:
suppose you have a quadratic (x-a)^2
expanding gives you x^2-2ax+a^2
going backwards:
halving the coefficient of the x term and shove that number into a square gives you (x-a)^2 and then bring the excess outside
gusty forum
halving -4 gives you -2
squaring -2 gives you 4
so your square term is x^2-4x+4 = (x-2)^2
but you added 4 inside the brackets so subtract that 4 outside the brackets to keep equality
and you have what's shown in the working out