-close
#help-10
proven flame
native inlet
.
proven flame
.close
golden current
heyy can someone help me with this?
obtuse pebbleBOT
heyy can someone help me with this?
golden current
anyone know anything about weighted averages? 😭
all i know is that:
the weights of points A and B with respect to X is 0.5
the weights of points C and D with respect to Y is 0.5
the weights of points X and Y with respect to Z is 0.25 and 0.75, respectively
i drew a pic too
obtuse pebbleBOT
@golden current Has your question been resolved?
frigid prism
nocturne minnow
,rotate
warm shaleBOT
obtuse pebbleBOT
@timid silo Has your question been resolved?
obtuse pebbleBOT
@craggy canopy Has your question been resolved?
tardy epoch
It's already one term
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
steel lion
how do we solve this
obtuse pebbleBOT
how do we solve this
glossy basalt
very interesting question
i have been staring at it for 10 mins lol
i was trying something like
$4^{\sqrt{\frac{\log 5}{\log 4}}}\left(\frac{5^{\sqrt{\frac{\log 4}{\log 5}}}}{4^{\sqrt{\frac{\log 5}{\log 4}}}}-1\right)$
warm shaleBOT
OldBiscuit
glossy basalt
where i use log as natural log
glossy basalt
I'm still thinking 🤔
worn yoke
$5^{\sqrt{\log_5 4}} = (e^{\ln 5})^{\sqrt{\log_5 4}} = e^{\ln 5 \cdot \sqrt{\frac{\ln 4}{\ln5}}}$
warm shaleBOT
cloud
worn yoke
bring the ln(5) inside the square root and do a similar thing for the other term
glossy basalt
👍
steel lion
bring the ln(5) inside the square root and do a similar thing for the other term
okay let me try this
wait
they both are equal now
so the answer is 0
steel lion
okay i didnt expect it to be that easy
steel lion
i feel so dumb
same
glossy basalt
steel lion
this question was on my test day before
and i had no clue
@worn yoke tysm for the solution
worn yoke
yw
steel lion
.close
drifting nova
number 8 please
obtuse pebbleBOT
number 8 please
icy tartan
number 8 please
what do we have to find exactly?
drifting nova
the cross section perpendicular to y -axis using semicircles
drifting nova
what do we have to find exactly?
and that is the graph of the 'base' when changed to x =
icy tartan
were you able to solve the 7th one?
could you give me the solution to that? it would help me better understand the question
drifting nova
could you give me the solution to that? it would help me better understand the q...
radical3 /4 integral -3 to 0 (-x^2-3x)^2
icy tartan
radical3 /4 integral -3 to 0 (-x^2-3x)^2
so we plug in the area bounded by the 2-x^2 and 3x + 2 into the formula for the area of the triangle? how does that work?
drifting nova
because the s is the area bounded by those two equations
icy tartan
if thats what the question is, can we plug it in to Pi * r^2 /2 as well?
drifting nova
if thats what the question is, can we plug it in to Pi * r^2 /2 as well?
that is the equation, but from what interval?
icy tartan
-3 to 0 right cos that gives you the area
drifting nova
-3 to 0 right cos that gives you the area
this is in terms of x
cause its about the y axis
icy tartan
im sorry but im having a hard time understanding this question. Are you sure theyre asking for the cross section?
icy tartan
radical3 /4 integral -3 to 0 (-x^2-3x)^2
if they are this shouldnt be the answer. the units dont match
drifting nova
if they are this shouldnt be the answer. the units dont match
ur confused, that answer was with y= and the one i am doing now is in x=
icy tartan
no Im just saying subbing in the area in root3/4*s^2 gives the answer in quartic units
but either way im just wasting your time with all this. Good luck, hope you find the answer
royal frigate
I have no idea where and how to get PQ
obtuse pebbleBOT
I have no idea where and how to get PQ
balmy cosmos
Id say use tales. You should be able to find BA with pytragoras
obtuse pebbleBOT
@royal frigate Has your question been resolved?
royal frigate
Id say use tales. You should be able to find BA with pytragoras
I'm sorry, but what is tales?
royal frigate
Is it this thing?
Do you mean this? @balmy cosmos
I can try and use this even though we haven't learnt this yet. It's just an MC question, so
?
royal frigate
it's okay
Should I try mentioning the helpers, or would that be a bother?
I can't seem to properly find an answer-
obtuse pebbleBOT
@royal frigate Has your question been resolved?
grizzled shore
@royal frigate hello :3
royal frigate
oh hi, it's you
Help, do you know how to do this
I'm honestly still trying to do it
I just chose C
I haven't got a proper calculation yet
grizzled shore
lemme have a look
yucky geometry >:(
hmm i have a few very yucky ways to do this
trying to see if there's an easy way
grizzled shore
you cna probably use I here
idk it's kinda disgusting i odnt have a good way to solve this
you can show that I is 1/4 of the way of CF
then similarly you can find PI
not very elegant though
if he told u a good way show me XD
royal frigate
if he told u a good way show me XD
Sure, give me a second
we set up points B, C, E and F in the Cartesian coordinate system. I've chosen C to be the origin so that the whole figure would lie in the first quadrant, and we won't have to deal with negative numbers.
you can work out the coordinates of points A, D, then P and Q using the internal section formula in coordinate geometry. the rest should be straightforward exercise of Pythagoras Theorem
grizzled shore
lmao
that's the even worse way
HAHA
you can nearly always solve these geometric problems by overlaying them on top of cartesian coordinates
but that's like the SUPER scuffed way to do geometry
royal frigate
True
grizzled shore
i know you can do this but it's like last resort you know?
royal frigate
Understandable
grizzled shore
i try to keep geometry problems done using geometry as was intended by the question
royal frigate
That's what I've been trying to do too
grizzled shore
you cna probably use I here
does this help?
you can prove that since G is the midpoint of BA and P is the midpoint of BG
that if u drew a line from P perpendicular to BC (intersects at point X) that triangle BPX (new point is X, the intersect) is similar to BAD
and is 1 quarter in size
they're similar and the hypotenuse has shrunk by 1/4 so all the sides have as well
royal frigate
I've tried doing that before, but I'm not sure how to proceed
royal frigate
royal frigate
11 I think
royal frigate
Yes
grizzled shore
now consider the middle part
let X be the midpoint of PQ
we know X is on DA
that's just some simple triangle similarity thing
consider triangle PYX
since VB = 11, BD = 44-11 = 33
PY = 33
then we know YX = 21 from the bottom
you can pythag for PX
then just add it to CP and double it
grizzled shore
you dont even need midpoint
you could consider this triangle as well
PXQ in this case
royal frigate
we know X is on DA
It's just when you said this, that's when I realised what you meant-
grizzled shore
ah
grizzled shore
sometimes i like to imagine in my head what happens when i move the points
like i know they've given BC = 88 and CF = 84
but they could've just said BC = x, CF = y
find the red line in terms of x and y
so given the constraints (without the measurement of BC and CF)
if i move the points around i can make it taller or wider
so the path stays geometric on some triangles
and they are always similar at the ratio given
hollow epoch
obtuse pebbleBOT
hollow epoch
How do I do this
obtuse pebbleBOT
@hollow epoch Has your question been resolved?
hollow epoch
<@&286206848099549185>
blissful finch
i means x, j means y, k means z
hollow epoch
yep
blissful finch
-4 means a vertical thing happens
obtuse pebbleBOT
can anyone help me with this?
clear shadow
.close
sacred kite
Is this seriously the right answer?
obtuse pebbleBOT
Is this seriously the right answer?
sacred kite
cause I think its just too "simple" you know?
sage geode
Yeah that's it
sacred kite
oh I just had a litte writing error anyway thanks!
waiiit since Ive already asked how about this one I did earlier?
the process was too long so I became unsure
sage geode
Looks good as well
sacred kite
hmmm buttt don't you think the square root sin(25x) should have been at the top?
or nah?
sage geode
There are no square roots though
sacred kite
I guess I did right then? niceee
last question, how do you suppose I can start this problem?
sorry it took a while, slow internet and all
'cause my answer was -(1/4)cot(x) +C
but im having second doubts that there shoud'nt have been a negative there
oh wait I think I remember that there was rule that I should show my solution so I guess I broke that haha sorryyy ^^;
.close
vivid lodge
im so confused on the table part c?
obtuse pebbleBOT
im so confused on the table part c?
vivid lodge
it says simulate choosing 10 participatns but theres only 4 simulations..
idk if im dumb or
WIAT LOL IM DUMB
nvm
.close
chilly knoll
i need help with this question
obtuse pebbleBOT
i need help with this question
oak anchor
I have the answer to this, but why do I have to prove f(x)+f(1-x) = 1
obtuse pebbleBOT
I have the answer to this, but why do I have to prove f(x)+f(1-x) = 1
oak anchor
Given f(x) = 4^x/ 4^x +2, calculate the sum: S= f(1/2017)+ f(2/2017)+ … + f(2016/2017)
Can’t I just immediately do summations
high lily
do you plan on tediously evaluating the function at 2016 values by hand?
and combine/simplify on top of that?
odd ice
can anyone help with this, laplace equation -> Y(s) = X(s)(1-s^2)/1+as+X(s). I want to get Y(s)/X(s) independent of X(s) on RHS
feral acorn
- If A is an n × n matrix prove that λ is an eigenvalue of A if and
only if λ is an eigenvalue of AT
. (Hint: Use det(BT
) = det(B).)
obtuse pebbleBOT
3. If A is an n × n matrix prove that λ is an eigenvalue of A if and
only if λ i...
feral acorn
A^T btw
odd ice
can anyone help with this, laplace equation -> Y(s) = X(s)(1-s^2)/1+as+X(s). I want to get Y(s)/X(s) independent of X(s) on RHS
obtuse pebbleBOT
@feral acorn Has your question been resolved?
hidden tinsel
Can anyone explain to me how they expanded this triple sum with dependent indices? Here, a is a real number for which |a|<1, if that helps in any way.
obtuse pebbleBOT
Can anyone explain to me how they expanded this triple sum with dependent indice...
@hidden tinsel Has your question been resolved?
hidden tinsel
<@&286206848099549185>
obtuse pebbleBOT
@hidden tinsel Has your question been resolved?
hidden tinsel
sorry but i gotta @ruby fulcrum as i still havent got a reply
obtuse pebbleBOT
@hidden tinsel Has your question been resolved?
hidden tinsel
Nope bot
brazen viper
@hidden tinsel if it helps any, it does seem to hold, it does not seem to be dependent on the value of a, and I have verified through multiplication of the first few terms and addition of the first few terms (up to i<=j<k<100 with a CAS system), that the coefficients of the various powers of a are the same from both the triple sum and the products.
hidden tinsel
Well, thanks! It does help, as I now only need to worry about the logic behind it, not whether its even correct(expected it to be somewhat wrong at first)!
brazen viper
I've been kinda playing around with it for the past hour or so, but yeah, there some trick that I'm not seeing, but it definitely seems to be a true equality
obtuse pebbleBOT
@hidden tinsel Has your question been resolved?
obtuse pebbleBOT
@hidden tinsel Has your question been resolved?
hidden tinsel
I mean it should be, others have verified it
Yeah but its a sum, shouldnt it be (a^1a^2a^3) + the rest
dense solar
hello
obtuse pebbleBOT
@hidden tinsel Has your question been resolved?
hidden tinsel
Can anyone explain to me how they expanded this triple sum with dependent indice...
Heres my question <@&286206848099549185>
brazen viper
@hidden tinsel I figured it out
typesetting
\begin{align*}
\sum_{1 \le i < j < k} a^i a^j a^k &= \sum_{1 \le i} a^i \qty(\sum_{i < j} a^j \qty(\sum_{j < k} a^k)) \
&= \sum_{1 \le i} a^i \qty(\sum_{i < j} a^i a^{j-i} \qty(\sum_{j < k} a^i a^{j-i} a^{k-j})) \
&= \sum_{1 \le i} a^{3i} \qty(\sum_{i < j} a^{2(j-i)} \qty(\sum_{j < k} a^{k-j})) \
&= \sum_{1 \le i} a^{3i} \qty(\sum_{1 \le (j - i)} a^{2(j-i)} \qty(\sum_{1 \le (k - j)} a^{k-j})) \
&= \sum_{1 \le i} a^{3i} \sum_{1 \le j')} a^{2j'} \sum_{1 \le k'} a^{k'} \
&= (a^3 + a^6 + a^9 + \ldots) (a^2 + a^4 + a^6 + \ldots) (a + a^2 + a^3 + \ldots)
\end{align*}
hidden tinsel
Genius
warm shaleBOT
OmnipotentEntity
brazen viper
You're welcome, glad I could help!
hidden tinsel
.close
foggy veldt
so i think i'm doing someting wrong here, but i can't put my finger on what
obtuse pebbleBOT
so i think i'm doing someting wrong here, but i can't put my finger on what
@foggy veldt Has your question been resolved?
foggy veldt
<@&286206848099549185>
foggy veldt
.close
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
rough bough
obtuse pebbleBOT
rough bough
Could someone please explain to me why (b)
So like I got the answer but just taking the positive differentiation of f(x) for x
but then I thought it would be the negative of the absolute value
because as x approaches to 10, the absoluter value becomes negative
I'm lost
gilded needle
notice that if x is less than 10 but close to 10, the expression inside the absolute value is negative, so for the purpose of part b, you could just replace it with -(x^2 - 17x + 70)
,w plot x^2 - 17x + 70
warm shaleBOT
rough bough
,w plot x^2 - 17x + 70
why did you replace with the positive version?
and not -(x^2-17+70)?
gilded needle
and not -(x^2-17+70)?
no, you do replace with the negative version
gilded needle
notice that if x is less than 10 but close to 10, the expression inside the abso...
^
i just plotted the original version to show that it goes negative for x<10
gilded needle
sure
i was just giving a simplification that allows you to get rid of the absolute value
that makes it easier to see what is going on, and straightforward to differentiate
rough bough
i was just giving a simplification that allows you to get rid of the absolute va...
But when we take the absolute value don't we want to take the side where it is negative
and when it is negative you want to negative it so that it is no longer negative
gilded needle
i'm not sure i understand your question
do you agree that if we have some number y<0, then |y| = -y?
rough bough
do you agree that if we have some number y<0, then |y| = -y?
yesr
rough bough
no, you *do* replace with the negative version
Like isn't (x^2-17+70) the version where as x approaches to 10, it is going to negative
and we do not want that
We want this instead?
gilded needle
right
rough bough
I'm confused why did you just get rid of the absolute value, its makes no sense to me]
gilded needle
and we can go from the first graph to the second in one of two equivalent ways:
either take the abs value as in the problem statement
or multiply by -1 which is equivalent for a negative number
gilded needle
i didn't just get rid of the abs value, i compensated by also multiplying by -1
rough bough
The issue here is that I don't get the right answer when I replace the absolute value with -1?
rough bough
i didn't just get rid of the abs value, i compensated by also multiplying by -1
hmm okay, but how can you solve this with multiply by -1?
@gilded needle Could you please explain what to do next?
obtuse pebbleBOT
@rough bough Has your question been resolved?
obtuse pebbleBOT
@rough bough Has your question been resolved?
static beacon
find angle BAC
remember angles in a tirangle add up to 180
seems right
Now are you familiar with sine rule?
obtuse pebbleBOT
@timid silo Has your question been resolved?
thorny forge
why is the answer no?
obtuse pebbleBOT
why is the answer no?
brisk matrix
what is the definition of an inverse
brisk matrix
let's be more rigorous
thorny forge
is it the domian and range thing
brisk matrix
a function f:X->Y has an inverse f^-1:Y -> X if f o f^-1 is the identity on Y, and f^-1 o f is the identity on X
and yes
brisk matrix
the main issue here is that the domain of f(x) doesn't match the codomain of arcsin(x)
exotic venture
why is the answer no?
what’s arcsin(0) for example?
thorny forge
pi/2
exotic venture
bro?
exotic venture
that’s arcos(0)
exotic venture
Yeah, but what’s sin(2*pi)
exotic venture
Yep,
thorny forge
???
exotic venture
Functions can only map any given value to 1 value,
thorny forge
so
exotic venture
it’s the inverse of sin(x) (-pi/2 <= x <= pi/2) for
thorny forge
its only the invrse from -pi/2 to pi/2
exotic venture
yep
exotic venture
mhmm
sin(x) is not injective over the reals and hence doesn’t have an inverse basically
brisk matrix
go back to what we said before
exotic venture
Ah, then don’t worry about it
brisk matrix
the domain of sin is different from the range of arcsin
fathom flicker
sin(x) has more than one inputs that produce the same output
brisk matrix
so they can't be inverses of each other
fathom flicker
so if you take any given output, and you want to find its input
exotic venture
the domain of sin is different from the range of arcsin
^^
fathom flicker
you can't be sure
brisk matrix
the injective thing
being injective means that the function would pass the horizontal line test, if you're familiar with that term
brisk matrix
bijective is injective and surjective
brisk matrix
the inverse is a function
injective functions are invertible if their codomain is restricted to their range
brisk matrix
whats surjective
it maps to every element in the codomain
for example
f: R -> R, f(x) = 0 is not surjective
exotic venture
it just means for every x \in A, for an injective function f: A -> B, there exists a unique y \in B where f(x) = y
brisk matrix
because we don't map to every value in R
but f: R -> R, f(x) = x is surjective
because we can map to every real number
fathom flicker
I don't mean to interrupt you guys here, but you might be using some more advanced vocab/terminology than ranga is used to. Maybe try to simplify it down to someone who might just be taking a trig class XD
thorny forge
nws
exotic venture
Idk what kids in school learn, but basically inverse means you should be able to get the x you put in to get the y
but putting it in the inverse function
- by
exotic venture
nah, just say range of arcsin doesn’t cover all the real numbers
or something like that
Make your life easier with words
instead of notation soup
high vortex
Greetings, I am writing an independent math research paper on various topics. I have a bit of a formatting nightmare. I need help making the document flow and presentable. Is there anyone that can help me, or make suggestions on what I can do to tidy it up? Thank you for your time
obtuse pebbleBOT
Greetings, I am writing an independent math research paper on various topics. I ...
brisk matrix
what software are you using to write it?
high vortex
Overleaf
brisk matrix
then you're asking for quite generic help, maybe pinpoint what exactly you need help with, with respect to formatting, and post about it in #latex-help
high vortex
I just wasn't very careful in keeping the notation consistent, etc. I pulled from a variety of sources. I cited, but still should be a bit more consitent there as well.
No, I don't need help with the Latex portion.
I need help just making it presentable, It isn't for a graduate paper, but I want to learn how to write research documents at that level
and what I need to get to that point.
brisk matrix
"making it presentable" is again very generic
if you could narrow down what you're trying to achieve maybe someone can help
high vortex
Well, what I mean is if I put it in a portfolio or somthing. I don't what them to immediately say, oh "this shouldn't be done like this"... I don' know exactly what that would mean, as I don't understand academic writing that well yet.
thorny forge
just write formally and make sure it looks polished
brisk matrix
it's still very difficult to say what would make it "presentable", especially with no sense of what the paper is. as long as your grammar, notation, styling are all consistent and accurate, there isn't really an issue
the rest is basically stylistic choice by you
high vortex
It started off with a lot of linear algebra and the Fourier Series, then it gets into Control theory.
brisk matrix
what kind of advice are you trying to get
because it really is all stylistic choice
as long as the underlying concepts and notations are correct
high vortex
I just want help to make sure, I'm not forming incorrect mathematical definitions
not using consistent mathematical language
things like that
high vortex
Right, yah. Would there be somewhere I could go to get that type of help?
brisk matrix
that is not an invitation for you to post every single definition and bit of notation you present in the paper, because i doubt anyone will want to parse through all of that
brisk matrix
Right, yah. Would there be somewhere I could go to get that type of help?
an advisor
and also just double checking yourself with your resources
high vortex
okay, thankx
obtuse pebbleBOT
Available help channel!
obsidian isle
Let's say we have the matrix equation:
AX^2 + BX + C = 0
Where A,B,C,X are all square matrices with the same dimensions.
If A,B,C are known, can we solve for X?
obtuse pebbleBOT
Let's say we have the matrix equation:
AX^2 + BX + C = 0
Where A,B,C,X are all s...
brisk matrix
you should be able to get some restraints on the solutions
just compute X^2, AX^2, and BX, and you’ll get a system of equations
obsidian isle
I'm looking to solve formally if possible
brisk matrix
well if you let A B and C be made up of some arbitrary constants, you can probably find a solution in terms of A B and C
obsidian isle
maybe there's a version of the quadratic formula that works over noncommutative rings
brisk matrix
well the quadratic formula is for polynomials of 1 variable
by letting X be a square matrix you’re making this much more complicated
is there a specific reason you want to solve this
obsidian isle
It's called algebra
brisk matrix
what
obsidian isle
Algebra is its own reward
tardy epoch
AX=B alone is hard
https://www.maths.manchester.ac.uk/~higham/narep/narep347.pdf
But here you go
obsidian isle
Huh this is really neat
There have to be some results on noncommutative algebras in general
obtuse pebbleBOT
@obsidian isle Has your question been resolved?
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
tawdry vortex
Hey what am i doing wrong in this problem i find my self stuck and unaware of what to do next
obtuse pebbleBOT
Hey what am i doing wrong in this problem i find my self stuck and unaware of wh...
jolly ginkgo
for number 13
Solve x²-x-20=0
worn yoke
you should be able to factor that
jolly ginkgo
Doesn't matter
tawdry vortex
Ok👍
obtuse pebbleBOT
@tawdry vortex Has your question been resolved?
sweet lake
Help me find the angles
obtuse pebbleBOT
Help me find the angles
sweet lake
tall wind
here's a big hint to start you off: when you have two lines crossing, their opposing angles are equal. What I recommend you do is pick two lines that cross, and look at which angles are opposing like the ones in the image given
sweet lake
i think i got all of them but it just looks a lil complicated when its all drawn out like that
thats prob why i was confused
tall wind
yeah, so to make it easier I want you to only pick two lines and focus on them
ignore the others
if you're having trouble choosing, I've highlighted two lines for you
obtuse pebbleBOT
@sweet lake Has your question been resolved?
elder snow
how is it possible???how can we determine the distance of the ferries wheel from the ground with the give info
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
full heart
Anyone know why this is wrong?
obtuse pebbleBOT
Anyone know why this is wrong?
full heart
the height is just 4cos(x^2) so the area for the shell is just that squared times pi
is what I thought
but apparently that's incorrect??
elfin burrow
the area is $2\pi(\text{radius})(\text{height})$
warm shaleBOT
tushar
languid stream
I’m trying to integrate but I can’t figure out the inside function, is it just x^2 or is is the entire top half?
obtuse pebbleBOT
I’m trying to integrate but I can’t figure out the **inside function**, is it ju...
ruby path
$\sqrt[3]{x^2} = x^{\frac{2}{3}}$
warm shaleBOT
NEONPerseus
obtuse pebbleBOT
Available help channel!
digital geode
f(x) = x + 10*sqrt(x) + 3
Find the domain of the inverse function.
obtuse pebbleBOT
f(x) = x + 10*sqrt(x) + 3
Find the domain of the inverse function.
digital geode
.close
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
wide spade
obtuse pebbleBOT
wide spade
I don't get it
I believe the answer is B
The correct answer is D. But the mark scheme doesnt give any working out or info since its a MCQ
Its either B or D since 8/0--20 = 0.4
But the displacement
If s is 0 then that must be the fixed point
Therefore at time -20 is the fixed point
west sierra
Fixed point is O?
wide spade
fixed point i assume is when displacement is zero
wide spade
What I understand is that basically, the person just started the timer 20 seconds after the moved
So 20 seconds before the vehicle is at the fixed point
in that case how is it that 20 seconds later the displacement is 16
-20+20 = 0
therefore is msut be 8
west sierra
What I understand is that basically, the person just started the timer 20 second...
Well from the graph the timer still is there for negative displacement but nvm
wide spade
am i misunderstanding it. or am i missing smth
There is no negative displacement
There is negative time
Which is impossible
Time is a scalar
west sierra
There is no negative displacement
Y axis is the displacement so below origin there is negative displacement
wide spade
Can't have direction or any form of negative value, unless you take the perspective of that much time needed
yes. My B. How does that affect the question though>
?
west sierra
Time is a scalar
Yeah time can not be negative but from graph you can see that time is there to quantify the displacement
west sierra
Can't have direction or any form of negative value, unless you take the perspect...
Yeah
wide spade
Ok, so then why is it D?
west sierra
Wait, before that the fixed point should be x intercept if it's the point where displacement=0
wide spade
Correct, therefore 20 mins before the time = 0 the displacement is 0 therefore the vehicle is at the fixed point
west sierra
Slope would be the velocity and D's velocity is correct in this case
wide spade
Therefore its logical to conclude that the timer starts 20 seconds after the vehicle left the fixed point
so 8m
Yes
D or B
Both result in 0.4
west sierra
Therefore its logical to conclude that the timer starts 20 seconds after the veh...
Timer was always there, from perspective of what's asked, timer does start 20 seconds after fixed point
wide spade
Which I know is 0.4 since.... gradient. 8/20
west sierra
Which I know is 0.4 since.... gradient. 8/20
Yeah gradient/slope
wide spade
Yes
west sierra
Answer should be B
west sierra
If the fixed point is where displacement=0
wide spade
That is what is very confusing. The only logical way for the answer to be D would be if the fixed point is at where the time = 0
Which is 8 m away from the fixed point
so it contradicts itself
How can a point be 0 m and 8 m away from itself.
Phyics......
And maths....
Because then the displacement would be 16, since 20 seconds away its coords are (20.16)
But in the question it state 20s from the fixed point
not 20s from when the timer began
So its very messed up
west sierra
That is what is very confusing. The only logical way for the answer to be D woul...
There is no logic in it being D
The gradient is constant and so for constant time period of 20 seconds, the displacement would always be the same=8m
wide spade
Precisely my point
west sierra
Yeah so you can consult it to whoever made the answer selected to D
It's definitely a human error
wide spade
Lets get other people involved. Make this into a big discussion since it cant be an error. This is a past paper. from an official exam board
<@&286206848099549185>
Anyone who comes Please read above. Our convo helps you to understand the situation
west sierra
The answer would be D if the question had a slight edit.."40 second after it passes the fixed point"
wide spade
yes or 20 seconds since the timer started
west sierra
yes or 20 seconds since the timer started
I don't understand, what timer you're talking about
wide spade
The time value is zero. No such thing as -20s. Therefore the timer used to obtain the value for the time. in this encounter. starts at the y intercept
since time = 0
Therefore 20 seconds after time = 0 then the displacement would be 8
west sierra
The time value is zero. No such thing as -20s. Therefore the timer used to obtai...
I see. when time=0 that's when the timer started
wide spade
Yes
west sierra
Ok
wide spade
That is my logical assumption
west sierra
So fixed point is either at t=0 or s=0 and both gives different fixed point but answer doesn't change because gradient is constant
wide spade
yes
west sierra
The time value is zero. No such thing as -20s. Therefore the timer used to obtai...
Actually in the graph such thing as -20 is written on x axis..welp it's the graph that's saying it lol
wide spade
BUT what is -time?
west sierra
BUT what is -time?
You said time can not be negative but here it's negative...it's the past time I guess?
wide spade
i guess
west sierra
Future time is shown as positive, present time=0 and past time= negative(guessing)
<@&286206848099549185>
west sierra
Anyone who comes Please read above. Our convo helps you to understand the situat...
Please refer to this
vernal root
I think the 'fixed point ' is not clear in this question
If you assume it to be -20 then B
But if fixed point is zero then answer would be D
west sierra
But if fixed point is zero then answer would be D
Why must the displacement be the one on y axis, and not the one which is covered in those 20 seconds that is the displacement=final point - initial point in which case it would still be B
obtuse pebbleBOT
@wide spade Has your question been resolved?
wide spade
<@&286206848099549185>
obtuse pebbleBOT
@wide spade Has your question been resolved?
native shard
i just need help finishing this problem
obtuse pebbleBOT
i just need help finishing this problem
tired nacelle
i just need help finishing this problem
reduce (x-30) 🤔?
native shard
yeah I wasn't sure if I had combined my factors right
i feel like I got most of the second brackets problem done its just the factoring at the very end which I'm confused about
obtuse pebbleBOT
@native shard Has your question been resolved?
obtuse pebbleBOT
@native shard Has your question been resolved?
native shard
.close
fallow cove
Idk how to start
obtuse pebbleBOT
Idk how to start
fallow cove
Help
glass nova
hi
deft crescent
I'm dying reading that handwriting
fallow cove
sry lol
Given 2 non-zero vectors, (~x) and (~y) are not parallel. Find the values of m and n if (m+2n+1)(~x)+(2m+3n-3)(~y)=0
obtuse pebbleBOT
@fallow cove Has your question been resolved?
fallow cove
<@&286206848099549185>
vital frost
obtuse pebbleBOT
vital frost
cousbt this be l/ft = w/d
viral blade
the top thing could be changed into that
The bottom thing can't be, it actually has less information than the top one
vital frost
The bottom thing can't be, it actually has less information than the top one
not sure what u mean
viral blade
You can't turn the bottom thing into that
The bottom equation tells you less
It's a looser restriction on those variables
vital frost
ok so pic is correct
viral blade
It is yeah
vital frost
thanks
viral blade
np
obtuse pebbleBOT
@vital frost Has your question been resolved?
solemn flax
obtuse pebbleBOT
solemn flax
Can someone help me integrate this?
obtuse pebbleBOT
@solemn flax Has your question been resolved?
obtuse pebbleBOT
@solemn flax Has your question been resolved?
high fiber
i have no idea where to start
timid silo
Can you write f(x)?
use the geometric interpretation of the integral
ie area under the graph
high fiber
you mean the area under the x axis?
timid silo
The area between the curve and the x axis
high fiber
The area between the curve and the x axis
so that is under y=-3
would "a" be 4? @timid silo
because there are 4 shaded regions
timid silo
would "a" be 4? <@456226577798135808>
yes
timid silo
because there are 4 shaded regions
That is not the reason
Do you know what the integral means?
high fiber
Do you know what the integral means?
no
timid silo
Ok that's a problem lol
timid silo
Well explain what you think the integral means
high fiber
b looks like it would be -4?
timid silo
Or more specifically a definite integral
timid silo
b looks like it would be -4?
This is wrong
high fiber
Or more specifically a definite integral
that is the total sum of a function over a specific interval
timid silo
Ok so for b) what is the interval
high fiber
(-1,2)
timid silo
Ok, what is the area in the region bounded by (-1,1)
high fiber
Ok, what is the area in the region bounded by (-1,1)
8
timid silo
According to the problem you posted it's 4 lol
high fiber
According to the problem you posted it's 4 lol
ohhh, i was thinking that it was 2 regions due to the y axis lol
so b would be 0
timid silo
Well if you read the question it says there are 3 regions
But yeah
There you go
You can do the whole problem
high fiber
3 regions
"e" i dont understand
there is no literal average value visibly
it cant be 4
timid silo
Not visibly, but since you know d, you can find the average value of any function over an interval by calculating the integral over that interval and then multiplying by 1/(length of the interval)
You may have missed that in class or something
high fiber
yeah
timid silo
I can't write out the answer but the formula is (total area)*(1/(length of the interval))
Oh my bad, d isn't what you need, c is
timid silo
Close
high fiber
8?
timid silo
Yes
high fiber
oh
timid silo
Wait
high fiber
length of the interval is 4 units
timid silo
No I'm really tripping today
high fiber
or width
timid silo
You already answered earlier
high fiber
oh
high fiber
yeah
timid silo
So what is the average value
high fiber
So what is the average value
2
timid silo
No
timid silo
Again, I just said I was mistaken
high fiber
ohh
timid silo
The integral is 4
high fiber
yeah
timid silo
So what is the average
high fiber
so e would be 4
timid silo
No
timid silo
No again
high fiber
4
high fiber
12 units
high fiber
I thought you meant of the function
timid silo
Nah just the length of the interval
high fiber
ah
high fiber
yeah
high fiber
4
high fiber
is it the same formula you put?
timid silo
Yes
high fiber
then it would be 1
timid silo
There you go
high fiber
There you go
ahh
timid silo
There's a couple ways to interpret it geometrically but honestly I think if you're in hs you should prob just try to remember this fact
high fiber
yeah
timid silo
Ok so 2 and 4 implies that at x=0 y= whatever you want other than 1
Because if it equals 1 then it would be continuous
high fiber
oh, that makes sense
high fiber
would that be a curve
timid silo
Could you try to draw one now that fulfills the conditions on the right side of the y-axis?
timid silo
Interestingly, this fulfils none of the conditions we went through 💀
high fiber
bruh
timid silo
Start with making sure your graph goes through the points it's supposed to man
high fiber
Start with making sure your graph goes through the points it's supposed to man
i know that there will be a break in the graph at x=2
so
timid silo
Brother
You have point
That you drew
Draw the line
Through the points
We have a start
high fiber
I see
timid silo
Now, draw a vertical asymptote at x=2
Just erase a bit around it
And redraw it so it's a vertical that goes up to infinity and comes back down to rejoin the line
We'll go with that
high fiber
okay
timid silo
In the same way, work your way through the other half of the graph
high fiber
like a mirror image
timid silo
And check to make sure it fulfils the requirements as you go along
No
Absolutely not
Read the question
high fiber
ahh I see
timid silo
The stuff on the left looks different from the stuff on the right
I need to go for now
Good luck
high fiber
How's this
flat pier
Hi
obtuse pebbleBOT
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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above
1
flat pier
Yes
deft crescent
and you know that f(100)-f(120) = 40
flat pier
Yes
deft crescent
which is f(100) - f(120) = f(80)/2
deft crescent
got it now?
flat pier
Yes
deft crescent
nice
deft crescent
all good
flat pier
Are you able to help me in another question
deft crescent
sure
flat pier
flat pier
Portuguese
deft crescent
ah
flat pier
I can translate if you need
deft crescent
yeah translate
flat pier
Let f be a function where Df=...
And D'f=...
Let g be a function in which Dg= Real positive numbers (without 0)
As it is shown on the picture
g touches the Ox axe in a positive value of x
Dh= R+
h(x)= (x-1)f(x)g(x)
Which is the minimal amount of roots that h has
?
I don't know if I had explained myself
deft crescent
i think i get it
flat pier
Ok
deft crescent
what is $D_f$
warm shaleBOT
Zouni
deft crescent
ah ok weird notation
flat pier
The value the x can be
So I'm sure we are talking about the same thing
D'f is the same thing but for y
deft crescent
pretty sure answer is one
flat pier
Ok
deft crescent
and (x-1)(-log(x))f(x)
considering f(x) is kind of an abstract function and exists throughout the specified domain and range
we can solve for the x intercept of the function (x-1)(-log(x))
so, 0 = (x-1)(-log(x))
which you should be able to solve very easily
deft crescent
cool
deft crescent
you too
flat pier
.close
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
crisp lichen
Suppose the measure of angle 6 can be represented by (2x -5)º.
What equation can be written to solve for the value of x? How do you know?
obtuse pebbleBOT
Suppose the measure of angle 6 can be represented by (2x -5)º.
What equation ca...
tardy epoch
!15m
obtuse pebbleBOT
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.
tardy epoch
i need help rn
there are 30 help channels open. you're not more important than them
crisp lichen
yall only have 30 helpers
static beacon
Suppose the measure of angle 6 can be represented by (2x -5)º.
What equation ca...
so the angle is size 6?
or did u forget to paste the image?
static beacon
and the full question
static beacon
???
crisp lichen
thats all the question was
static beacon
Oh
i mean you can
call the missing angle angle 8
we know 5 + 6 + 7 + 8 = 360 correct?
static beacon
no?
crisp lichen
static beacon
Yea ok
clam down
we know 6 = 2x-5
and so therefore using the fact that oppsotie angles r the same
angle 8 is also 2x-5
crisp lichen
so 4x-10
crisp lichen
ok
crisp lichen
yes
crisp lichen
yes
crisp lichen
yes