@crimson sedge Has your question been resolved?
#help-13
cedar kilnBOT
Available help channel!
sharp jetty
Hello, I have a high school math contest question. If there's another place I should be asking this question, please let me know. I've posted the question and my attempt, which doesn't go very far because I'm missing a key piece of logic I need I'm sure. Any suggestions would be appreciated. Thanks!
sharp jetty
My feeling is that I need to use the relationship between area and sides of similar triangles, but I can't seem to put that to use
cedar kilnBOT
@sharp jetty Has your question been resolved?
sharp jetty
<@&286206848099549185>
night forum
I have no idea how to solve it with proof and shit, but the answer is A
sharp jetty
haha, ok, well thanks for looking at it
cedar kilnBOT
@sharp jetty Has your question been resolved?
gleaming spindle
Why 5x^2/(5x)^2 is = 1/5 ?
cedar kilnBOT
Why 5x^2/(5x)^2 is = 1/5 ?
dire geode
Why 5x^2/(5x)^2 is = 1/5 ?
,tex .exp rules
wraith daggerBOT
riemann
wintry pier
what is 5/25 and x^2/x^2 ?
primal socket
x^0 ≠ 0
gleaming spindle
Huh?
wintry pier
x^0=1
primal socket
That's what you are saying
x^2 / x^2 ≠ 0
Let x^2 = 2
You have 2/2
And you said that isn't 0
Then why would x^2 / x^2 = 0
Anything raised to the 0th power is 1.
gleaming spindle
We cancelingout factors
primal socket
Yes
wintry pier
Anything raised to the 0th power is 1.
exept 0^0
gleaming spindle
Yeah i know
primal socket
That means you are dividing the numerator and the denominator by the thing you are "cancelling"
primal socket
Which just ends with 1/1, which is 1
primal socket
Yes.
primal socket
anything
gleaming spindle
Okay
wintry pier
so going back to 5/25=?
wintry pier
no pb take your time
gleaming spindle
Can gou explain the rules again? @wintry pier
Everything until 5/25 make sense to me
But them becoming 1/5
I dont get
wintry pier
$\frac{5}{25}=\frac{5}{5^2}$ right ?
wraith daggerBOT
Joseph.P
wintry pier
you can apply $\frac{a^m}{a^n}=a^{m-n}$
gleaming spindle
Not meaning something else
wraith daggerBOT
Joseph.P
gleaming spindle
How?
gleaming spindle
How to apply
wintry pier
take m=1 and n=2
gleaming spindle
-1?
wintry pier
because $a^1=a$
wraith daggerBOT
Joseph.P
gleaming spindle
**Joseph.P**
Should i do this to the numerator or denominator?
wraith daggerBOT
Joseph.P
wintry pier
Should i do this to the numerator or denominator?
it's always the exponent of the numerator minus the exponent of the denominator
wintry pier
**Joseph.P**
see this
wintry pier
so the exponent of the numerator is 1
gleaming spindle
**Joseph.P**
I just don't get how AND why those guys r turning to 1/5
What are you doing so the 5 becomes 1, amd 25 becomes 5
wintry pier
$\frac{5}{5^2}=\frac{5}{5\cdot5}$
wraith daggerBOT
Joseph.P
gleaming spindle
Yeah i know this
primal socket
Well, 5 can be written as 5^1 because anything raised to the first power is itself
wintry pier
$\frac{5}{5^2}=\frac{5}{5\cdot5}=\frac 15 \cdot \frac 55$
wraith daggerBOT
Joseph.P
primal socket
The exponent tells you how many times the number is multiplied by itself
5^2 means twice, so 5 * 5
gleaming spindle
I know
primal socket
And 5^1 means once, so it's just 5
gleaming spindle
Well, 5 can be written as 5^1 because anything raised to the first power is itse...
What is "first power"?
primal socket
Exponent = 1
x^1
x is raised to the first power
This is when you can use this: $\frac{a^m}{a^n}=a^{m-n}$
wraith daggerBOT
USS-Enterprise
gleaming spindle
$Hi$
wraith daggerBOT
Jizzy
primal socket
You've got $\frac{5^1}{5^2}$
wraith daggerBOT
USS-Enterprise
gleaming spindle
5/25
primal socket
Yes
gleaming spindle
And?
primal socket
But it also equals $5^{1-2}$
wintry pier
$5^{1-2}$
primal socket
What
wraith daggerBOT
Joseph.P
wraith daggerBOT
USS-Enterprise
primal socket
Wrong brackets
gleaming spindle
**USS\-Enterprise**
So 1-2 is -1, and -1 means 0, and 0 means 1 right??
primal socket
What
gleaming spindle
Anything powerd to 0 is 1
Nah dude answer to this
primal socket
What
primal socket
No
gleaming spindle
Or 1
primal socket
Yes
wintry pier
$5^{-1}=\frac 15$
wraith daggerBOT
Joseph.P
primal socket
$5^{-n}=\frac {1}{5^n}$
wraith daggerBOT
USS-Enterprise
primal socket
We literally told you above
wintry pier
do you understand @gleaming spindle ?
wintry pier
U could just say shorten it
?
gleaming spindle
Both are equals to 0.2
wintry pier
1/5 is same as 5/25
that's what you didn't understand
gleaming spindle
But if i only write 5/25
It is still correct right?
Since shortening is not necessary
wintry pier
ye still correct
gleaming spindle
Damn man
wintry pier
it's the same as $\frac ab=\frac{k\cdot a}{k\cdot b}$ can you see why ?
wraith daggerBOT
Joseph.P
wintry pier
no
gleaming spindle
17
wintry pier
ok thanks
latent smelt
The equation x^2 = 9 has the solution x= ±√9 i.e. x= ±3.
• The number √9 is not equal to ±3. According to the definition, square root of √9 = 3
latent smelt
I don't understand the second line, why isn't √9 = ±3?
obsidian coral
Because x^2, the domain is all real numbers but sqrt(x), the domain is all real numbers greater than 0
obsidian coral
I don't quite understand what you mean
x^2, the domain is all real numbers, meaning you can have positive and negative solutions but sqrt(x), the domain is numbers greater than 0, meaning that you would only have positive solutions
latent smelt
from what I understand it's a fixed solution that can only have one answer and the domain for sqrt(x) is numbers greater than 0 because that's what mathematicians agreed on
.close
olive briar
how would we do this question
i understand to prove a basis it needs to be in the span and linearly independent but not sure what they are meaning by the equation and hyperplane part
olive briar
<@&268886789983436800>
<@&286206848099549185>
is this correct
.cise
.ckise
.close
tranquil lotus
Let $\phi: \bF_{1} \to \bF_{2}$ be an isomorphism. Show that $\phi(\b0_{\bF_{1}}) = \b0_{\bF_{2}}$.
tranquil lotus
.close
exotic verge
Can <@&286206848099549185> assist?
cedar kilnBOT
@exotic verge Has your question been resolved?
exotic verge
.close
crude scroll
cedar kilnBOT
cobalt parrot
Express the following matrix
$\begin{bmatrix}
1&2&3&8 \
-2&-5&1&-8\
0&1&7&8
\end{bmatrix}$
in the form A = EFR where E and F are elemntary matrices and R is in row-echelon form
wraith daggerBOT
farmer
cobalt parrot
don't understand how this is possible. How can I express this matrix which has no inverse in this form.
I can't obtain this matrix from an elemntary matrix
How do I do it
All I know if that the reduced row echeolon form for that matrix is
$ \begin{bmatrix}
1&3&3&8 \
0&1&7&8 \
0&0&0&0\
\end{bmatrix}$
wraith daggerBOT
farmer
cobalt parrot
which was the result of this two elementary row operations: 2R1 + R2, and -R2+R3
So the answer above is R. So we found R
maybe I can apply inverse operation on I3
so, I am performing a row operation on E which is producing I. But, i can find E by multiplying the I3 matrix by those inverse row operations
and F by multiplying the matrix by the inverse of the 2nd row operation
am I on the right track?
!status
cedar kilnBOT
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
@cobalt parrot Has your question been resolved?
snow elbow
find f
snow elbow
whoops i meant
find f' of f(t)=3tsin(pi t)
so using the chain rule my inner function here is pi t and my outer is 3tsin(t)
correct?
Then I derive each
After deriving each now I can go back to applying the chain rule
Outer deriv eval at the inner times the inner
lucid topaz
but like why are they split up
livid hound
if you wanted to do it like that, you'd want to introduce another variable
also not actually ideal
snow elbow
What do you mean? Idk what's going on here ive been on this for a hour
lucid topaz
for reference d/dx sin(kx)=k cos(kx)
snow elbow
for reference d/dx sin(kx)=k cos(kx)
Have I made a mistake somewhere?
livid hound
if you wanted to apply chain rule that way, that require something like
u = pi * t
f(u) = 3u/pi * sin(u)
differentiate that wrt u (requiring product rule)
then multiply by that by du/dt
lucid topaz
well more like: whats your reasoning to need to split it into 2 equations? thats the part that confused me
snow elbow
Could this one just be done using the product rule alone?
livid hound
no
snow elbow
well more like: whats your reasoning to need to split it into 2 equations? thats...
Oh well i didnt really split it up I was just showing myself the derivative of the inner and the outer on different lines
livid hound
chain rule would be required in some way to differentiate the sin(pi t) component,
although it's possible to approach it in the way you described (if done properly).personally I'd apply product rule first
snow elbow
So, so far is my work incorrect anywhere?
livid hound
yes, it's incorrect
snow elbow
Could you expand? I'm not picking up on this one
livid hound
if you wanted to apply chain rule that way, that require something like
u = pi * t
f(u) = 3u/pi * sin(u)
differentiate that wrt u (requiring product rule)
then multiply by that by du/dt
snow elbow
Ok so step 1 can we identify the inner and outer function
livid hound
substitution makes it clearer
snow elbow
I'll assign the inner function as pi t
livid hound
note that you have t elsewhere in your function as well
snow elbow
so i cannot double up on the same variable is my issue than?
I cannot use pi t and 3tsin(t)
that would be incorrect
livid hound
yeh
livid hound
you could still use pi * t for the inner function of you want, going through the steps I've outlined
but as mentioned, it's better to first apply product rule here
and chain rule after that for the components that require it
snow elbow
Could you please type it out for me
Again
This isn't sticking I don't understand
So we apply the product rule first to 2 terms here
3t and sin(pi t)?
livid hound
not terms, but yes, you could even factor out the 3 first
from constant multiple rule
$$f'(t) = 3 \cdot \dv{t} \blue{t}\red{\sin(\pi t)}$$
wraith daggerBOT
ℝam()n()v
snow elbow
Using the product rule I got 3tcos(pi t) + 3sin(pi t)
if we want we can do 3 [tcos(pi t) + sin(pi t)]
livid hound
missing chain rule in that first term
for $3t \red{\dv{t} \sin( \pi t)}$, \
differentiating that requires chain rule
wraith daggerBOT
ℝam()n()v
snow elbow
in tcos(pi t) there is 2 functions right t and cos(pi t)
so now while applying the chain rule our inner function is cos(pi t) and our outer is t
livid hound
skipping ahead
snow elbow
oh?
snow elbow
3[tcos(pi t) + sin(pi t)]
livid hound
no
snow elbow
damn....
livid hound
$$3t \red{\br{\dv{t} \sin( \pi t)}}+ \blue{3\sin(\pi t) \dv{t}{t}}$$
wraith daggerBOT
ℝam()n()v
livid hound
that's what's you have after applying only product rule (no simplification)
livid hound
you simplified the blue part correctly
your first term is incorrect because the d/dt of sin(pi t) isn't just cos(pi t)
snow elbow
I appreciate your help I'm gonna walk around clear my head for a few mins here before I get back into this one
Been on it for a hour +
Think im brain locked or something
livid hound
if the question was just
$$\red{\dv{t} \sin( \pi t)}$$
what would you do?
wraith daggerBOT
ℝam()n()v
livid hound
wdym by separate pi and t
snow elbow
So like I could use the product rule here i think
snow elbow
the derivative of that should just be cos pi?
livid hound
no
snow elbow
zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz
snow elbow
The inner function in that I guess can be t
livid hound
no
livid hound
yeh
snow elbow
Then I would fund the derivative of the outer which is cos
and then eval at the inner
so cos(pi t) times it all by pi
livid hound
no
you're forgetting one of the key components of chain rule
specifically multiplying by the derivative of the inner function
livid hound
don't use x for multiplication but yes
wraith daggerBOT
ℝam()n()v
livid hound
overall you'll end up with that
snow elbow
And that is our final solution?
Jesus I don't understand how the hell I was stuck for so long
Thanks for the help man
(Lot of practice is needed for me)
lone ferry
how to paproahc
lone ferry
bold anchor
You can set up a system of equations for this 🙂
lone ferry
the rectangles overlap thouhg
bold anchor
Indeed, but we're not super interested in the rectangles so much as the fence.
lone ferry
im not sure how to set up the second equation
ohh
i know the first equatoin is 3w + 4l = 1200
but i dont know how to continu efrom there
bold anchor
I believe it would be more like 4l+2w, there's only 2 sides of the fence.
royal loom
bold anchor
Hey Austin, been a while.
royal loom
Hey!
lone ferry
<:this:489575168314966026>
sorry 😓
lone ferry
I believe it would be more like 4l+2w, there's only 2 sides of the fence.
ohh ok i split it into 3 parts
how would i contineu from there?
bold anchor
how would i contineu from there?
What are we trying to maximize here?
nimble veldt
i know the first equatoin is 3w + 4l = 1200
it depends on what you call w and what you call h but i am pretty sure the factor for w should be even.
lone ferry
What are we trying to maximize here?
area
bold anchor
Do you know the formula for the area of a rectangle?
lone ferry
l * w
bold anchor
Yup! So now you have a system of equations with length and width
lone ferry
but what would l*w be equal to
bold anchor
Area
Area is what you're trying to maximize so you can treat it like a function since it's variable
lone ferry
so i did
l * w = a
4l+2w = 1200
2w = 1200-4l
w = 600-2l
(600-2l)l = a
-2l^2+600l = a
oh nevermind i did it wrong
bold anchor
Looks correct thus far.
lone ferry
im not sure how to continue
bold anchor
im not sure how to continue
You can take the derivative and then set it to 0 to find L
lone ferry
You can take the derivative and then set it to 0 to find L
i dont know how to take the derivative
i havent learned that yet 😓 sorry
bold anchor
Wait what class is this for?
lone ferry
precalculus
ive learned the difference quotient but i dont thikn its necessary for this problem
bold anchor
I thought this was a Calc 1 the whole time although to be fair I never took precal so I don't actually know what they teach in there.
In precalc terms, you can find the vertex of this parabola using l=-b/2a
b and a, as in the quadratic formula ax^2+bx+c
cedar kilnBOT
@lone ferry Has your question been resolved?
lone ferry
I thought this was a Calc 1 the whole time although to be fair I never took prec...
ohh okok
thank you!
.close
cedar kilnBOT
cedar kilnBOT
Available help channel!
crimson sedge
Please help me I have 9 minutes to do this
cedar kilnBOT
Please help me I have 9 minutes to do this
crimson sedge
And I don't know jow
cedar kilnBOT
@crimson sedge Has your question been resolved?
crimson sedge
no one can even read the question the quality is terrible
crimson sedge
cedar kilnBOT
crimson sedge
not sure how i would start this
cedar kilnBOT
@crimson sedge Has your question been resolved?
tidal kernel
not sure how i would start this
Hi! You can try rewriting the expression using geometric series
crimson sedge
Hi! You can try rewriting the expression using geometric series
(w-wr^9)/1-r ?
tidal kernel
(w-wr^9)/1-r ?
Well, more like a sum of geometric series
i.e. (w + ... + w^9) + (w^2 + ... + w^9) + ... + (w^9)
Now write each term using the formula and see if you can simplify
crimson sedge
formula as in a(1-r^n)/1-r?
tidal kernel
formula as in a(1-r^n)/1-r?
Yup!
crimson sedge
hmm wouldnt it be (w + w^2 + w^3... + w^8 - w^10 - w^11... - w^17)/1-w?
tidal kernel
how come that is 1
It's easier to work with radians
40 degrees is 2pi/9 radians
Do you know about roots of unity
crimson sedge
nope i dont think so
tidal kernel
No matter. But you know radians yes?
crimson sedge
yeah
tidal kernel
Ok so for example $w^2+...+w^9 = w^2\left(\frac{w^8-1}{w-1}\right)$
wraith daggerBOT
Wheeler
tidal kernel
i.e. (w + ... + w^9) + (w^2 + ... + w^9) + ... + (w^9)
Can you use this to simplify this expression?
crimson sedge
wait how does that work
is that the roots of unity thing you were talking about
lemme see
tidal kernel
wait how does that work
no that's just geometric series
crimson sedge
also just something that i noticed, would i be able to use demoivres to solve this problem
tidal kernel
The ratio r is w
tidal kernel
also just something that i noticed, would i be able to use demoivres to solve th...
I don't think de moivres is very useful here
Since |w|=1
crimson sedge
no that's just geometric series
is it not w^2 - w^10 over 1 - w
since in a(1-r^n)/1-r a is w^2 and r is w n is 8 w^2(1-w^8)/1-w
w^2 - w^10 / 1 - w
crimson sedge
since in a(1-r^n)/1-r a is w^2 and r is w n is 8 w^2(1-w^8)/1-w
w^2 - w^10 / 1 ...
when i factor out a w^2 from this though its w^2 (1-w^8/1-w) not what you wrote
or was it just a typo
tidal kernel
Just multiply top and bottom by -1
crimson sedge
oh ok im stupid give me a sec to simplify the other expression
tidal kernel
And remember w^9 = 1, and so w^10 = w etc
tidal kernel
oh ok im stupid give me a sec to simplify the other expression
Nws take your time and keep at it but I need to go somewhere soon
crimson sedge
Nws take your time and keep at it but I need to go somewhere soon
oh ok can i just close this and then dm you when finish
since i gotta sleep soon as well
tidal kernel
Sure
crimson sedge
.close
cedar kilnBOT
cedar kilnBOT
Available help channel!
upbeat moth
hi , any ideas?
cedar kilnBOT
hi , any ideas?
upbeat moth
bro i have
it doesnt help
i think the expressions inside the brackets should be a common factor
all i managed to get to is
$$3cos(2x)(6sin(2x)-1) + 2\sqrt{2}(3sin(2x)-2)$$
wraith daggerBOT
exterminated
upbeat moth
i factored $cos(2x)$ out
wraith daggerBOT
exterminated
foggy merlin
the goal is to solve for x ?
upbeat moth
yes
I have also tried to change $cos(2x)$ in the first brackets to $2cos^{2}(x)-1$ etc but it didnt help either
wraith daggerBOT
exterminated
foggy merlin
hmm i am struggling fr
tropic oxide
\cos and \sin btw
tropic oxide
I have also tried to change $cos(2x)$ in the first brackets to $2cos^{2}(x)-1$ ...
definitely counterproductive -- all angles involved here are multiples of 2x
upbeat moth
`\cos` and `\sin` btw
oh hey Ann
thanks for the tip
You got any ideas on the problem?
wraith daggerBOT
Ann
upbeat moth
show us
tropic oxide
ok so some prep work first
$18 \sin(2x) \cos(2x) - 3 \cos(2x) + 6 \sqrt{2} \sin(2x) - 4 \sqrt{2} = 0$
wraith daggerBOT
Ann
upbeat moth
yeah i''ve done that
tropic oxide
$\sin(2x) \cos(2x) - \frac{1}{6} \cos(2x) + \frac{\sqrt{2}}{3} \sin(2x) - \frac{2\sqrt{2}}{9} = 0$
wraith daggerBOT
Ann
tropic oxide
ok so far so good by the looks of it
$\sin(2x) \cos(2x) - \frac16 \cos(2x) + \frac{\sqrt{2}}{3} \sin(2x) = \frac{2\sqrt{2}}{9}$
wraith daggerBOT
Ann
tropic oxide
simple shit so far
now i subtract sqrt(2)/18 from both sides.
$\paren{\sin(2x) - \frac{1}{6}} \paren{\cos(2x) + \frac{\sqrt{2}}{3}} = \frac{2 \sqrt{2}}{9} - \frac{\sqrt{2}}{18}$
wraith daggerBOT
Ann
upbeat moth
okay?
foggy merlin
ah i see :/
upbeat moth
ok the right side is $\frac{3\sqrt{2}}{18}$
wraith daggerBOT
exterminated
upbeat moth
what do we do now?
foggy merlin
pretty much nothing, we cant continue from there
or if you can find the angle for which cos or sin of that angle is 3√2/18
upbeat moth
why would i
tropic oxide
pretty much nothing, we cant continue from there
or can we?
tropic oxide
now comes the real pro gamer move
foggy merlin
tropic oxide
let $u := \cos(2x) + \frac{\sqrt{2}}{3}, v := \sin(2x) - \frac16$
wraith daggerBOT
Ann
upbeat moth
what for
foggy merlin
let her do 
tropic oxide
then we have: $\begin{cases} uv = \frac{\sqrt{3}}{6} \ (u - \sqrt{2}/3)^2 + (v + 1/6)^2 = 1 \end{cases}$
wraith daggerBOT
Ann
foggy merlin
upbeat moth
hw
foggy merlin
upbeat moth
i've solved 12/14 so far
sooo @tropic oxide what do you think
<@&286206848099549185>
cedar kilnBOT
@upbeat moth Has your question been resolved?
cedar kilnBOT
@upbeat moth Has your question been resolved?
upbeat moth
@somber lantern what do u think?
upbeat moth
yes i am
thorn lotus
It's still 14 right?
upbeat moth
yeah
thorn lotus
ok
so have you gotten to a point where you have everything in terms of either sin(2x) and cos(2x)?
upbeat moth
uhmm yes
thorn lotus
good. let t = sin(2x)
upbeat moth
aight
thorn lotus
that should give you a quartic that is actually factorable. just use rational root theorem
upbeat moth
that should give you a quartic that is actually factorable. just use rational ro...
is that the theorem about the rational factors of the last term?
let me check
thorn lotus
the one where you guess the roots by using the factors of the constant term and the coefficient of the leading term
upbeat moth
yeh thats what i meant
thorn lotus
ok, yeah
upbeat moth
okay gimme a sec i will try to figure it out
thorn lotus
there's only a couple of combinations so it's pretty easy to check
thorn lotus
so that should give you a sqrt(1-t^2)
upbeat moth
but its not always that
thorn lotus
manipulate the equation to get rid of it by squaring it
upbeat moth
it can also be -sqrt(1-t^2)
thorn lotus
you can deal with that when you get there
try doing the positive root for now and see what you get
upbeat moth
i don't think that's a good idea
looks like an overkill to me
that would be a quartic equation with like 10 terms
thorn lotus
let me say this then
it doesn't matter
because of the squaring part
how about you just try instead of complaining
somber lantern
hi , any ideas?
The identity $\sin(4x) = 2\sin(2x)\cos(2x)$ gives $$18\sin(2x)\cos(2x) - 3\cos(2x) + 6\sqrt{2}\sin(2x) - 4\sqrt{2} = 0.$$
Substituting $u = \sin(2x)$ and $v = \cos(2x)$ gives $$18uv - 3v + 6\sqrt{2}u - 4\sqrt{2} = 0.$$
$$\frac{6\sqrt{2}u - 4\sqrt{2}}{18u - 3} = v = \sqrt{1-u^2}.$$
Squaring both sides will result in a quartic in $u$ that I do not want to even calculate but may be solvable.
wraith daggerBOT
chencking
somber lantern
That's pretty much the only way to approach it from what I see.
You can't work directly with offsets and the polynomial simply doesn't factorize in u and v.
thorn lotus
You should get to a point where you have $-324t^4 + 108t^3 + 243t^2 + \cdots$
wraith daggerBOT
TooManyCooks
thorn lotus
I'm not giving you the whole thing, but that should give you an idea of whether you're doing it right
upbeat moth
**chencking**
i can't see how y'all think that $\cos(\alpha) $ is always the same as $\sqrt{1-\sin^{2}{\alpha}}$
wraith daggerBOT
exterminated
upbeat moth
I'm not giving you the whole thing, but that should give you an idea of whether ...
yeah its okay
somber lantern
Technically it's plus minus that
thorn lotus
Like I said, it doesn't matter if you use +-
upbeat moth
yes
somber lantern
but you're squaring either way
thorn lotus
the problem will sort itself out
thorn lotus
but you're squaring either way
this
upbeat moth
aight let me try
somber lantern
Honestly, this is a horrible problem and you should make your teacher do it as well.
thorn lotus
It's not that bad
somber lantern
I doubt there's a quicker way to do it. But if there is, it's worth seeing.
thorn lotus
It's tedious, but all the principles involved are simple
somber lantern
Factorizations with 3 digit coefficients are difficult.
thorn lotus
just keep dividing by 2, and 3 
somber lantern
even then
upbeat moth
I doubt there's a quicker way to do it. But if there is, it's worth seeing.
there for sure is , but my teacher just always gives us really advanced problems
thorn lotus
it is annoying to do the synthetic division part, but you gotta do what you gotta do
upbeat moth
~~Honestly, this is a horrible problem and you should make your teacher do it as...
I will
upbeat moth
You should get to a point where you have $-324t^4 + 108t^3 + 243t^2 + \cdots$
yeah i got to this point
let me try to solve that
my brain hurts
to factor the polynomial that i got i need to solve a system of equations with 6 variables
i know the solution but it just brings me down
cedar kilnBOT
@upbeat moth Has your question been resolved?
thorn lotus
my brain hurts
to factor the polynomial that i got i need to solve a system of e...
uh, what?
upbeat moth
uh, what?
using the method of undetermined coefficients
$(ax^{2}+bx+c)(dx^{2}+ex+f)=324x^{4}-108x^{3}-243x^{2}+12x+23$
wraith daggerBOT
exterminated
thorn lotus
why are you doing it that way?
upbeat moth
what are the other ways? there are no whole roots
thorn lotus
who said anything about whole roots?
upbeat moth
no one
thorn lotus
do a prime factorization of the constant and leading coefficient
upbeat moth
**exterminated**
this is just how you generally solve a rational quartic+ equation
thorn lotus
you'll see easy test cases there
upbeat moth
do a prime factorization of the constant and leading coefficient
but how do i find the roots using factorisation if they are not whole
thorn lotus
pretty sure i mentioned that earlier
that or do it graphically
it just so happen that there are roots here you can easily identify in a graph
well, 2 of them are. but once you identify those two, you are left with a quadratic equation and you can go crazy on that however you want
upbeat moth
it just so happen that there are roots here you can easily identify in a graph
yes but i can't just use a graphing tool during a test can i?
thorn lotus
this is not a test right?
hw are supposed to be more difficult
not the other way around
upbeat moth
i am just telling you that i am not allowed to use desmos etc. to solve my problems
thorn lotus
besides, i already pointed out the rational roots thing
thorn lotus
i am just telling you that i am not allowed to use desmos etc. to solve my probl...
and i'm giving you options. I'm not saying you do it that way, i'm just suggesting you can do it if you want
upbeat moth
@thorn lotus ohhh man i got it thank you a lot
the thing is they don't teach us the full rational root theorem at school so i was really confused in the beginning
thorn lotus
so you got the first two roots now?
upbeat moth
yes 1/3 and -1/3 but it doesnt matter since i now understand what you meant by "rational root theorem"
thorn lotus
okay. then the rest should be straightforward
cedar kilnBOT
@upbeat moth Has your question been resolved?
pallid vapor
Can someone do question 10 and tell me what they get. I think my textbook is wrong
pallid vapor
My working is on calculator but I just substituted the dimensions into the volume of cone and cylinder formulas and half of the volume of a sphere
livid hound
,rotate
wraith daggerBOT
livid hound
show the actual calculations you made
and what you got
do the work on paper
clearly write out what you're doing
livid hound
mmk, seems there's an issue with the book
pallid vapor
Thank you
errant wasp
I have a kinda really stupid question
cedar kilnBOT
I have a kinda really stupid question
errant wasp
im trying to prove something
i think its actually simple
but i might be over thinking it
im trying to prove that for all inhabited sets X there exists A, B in P(x) such that A neq to B
but by definition of a powerset,
P(X) = {y | y in X}
if A and B were equal, they would be the same element
can i just prove this by contradiction
sleek totem
Why (sen(x))²+(cos(x))² is equals one?
cedar kilnBOT
Why (sen(x))²+(cos(x))² is equals one?
scenic smelt
what is sen?
scenic smelt
oh
scenic smelt
Why (sen(x))²+(cos(x))² is equals one?
ya you jus need to kno that
usually they're given on a test
at least in my case
drifting marlin
It's easy enough to show, since it just follows from pythagoras and a unit circle
drifting marlin
Indeed
sleek totem
basically (sen(x))²+(cos(x))² is 1
drifting marlin
Again it's pretty straightforward if you look at a unit circle
drifting marlin
Take the unit circle, pick any point on it, and call the angle between the terminal arm and the x-axis t
You should hopefully know that if the point is some coordinate (x, y) then sin(t) = y and cos(t) = x
sleek totem
Interesting
drifting marlin
Since the radius of the circle is 1, applying pythagoras immediately gives the identity
drifting marlin
That also works
idle raft
hi could anyone help me with this
cedar kilnBOT
hi could anyone help me with this
idle raft
🥹
idle raft
could u explain
final meteor
this basically
idle raft
so qpt is equal to prq?
final meteor
yes
thats the first part
for the second part, do u know the properties of parallel lines and transversals?
idle raft
nope
idle raft
yes
final meteor
so its basically that here
RQ and PT are parallel
RQP and QPT make alternate angles
so they're equal
idle raft
so thats also 51?
final meteor
yea, since they're alternate interior angles
idle raft
oh okay
final meteor
for the third part, you can use the angle sum property
idle raft
is it okay if i ask more questions?
idle raft
this one
final meteor
so angle SPT is 10° right
tan10°=ST/PT
= 5/(PR+8)
from here you can solve for PR
once you have PR you can find PT aswell, and also find QR using the same method
idle raft
= 5/(PR+8)
im confused at this part
idle raft
how do i find the pr when idk pt
final meteor
PT=PR+RT
we can rewrite it like that
and we already know RT=8
or yk what just directly find PT
we can do this in step 2
idle raft
oh okay
final meteor
yeah so now u have PT
now rewrite PT as PR+RT
28.356=PR+RT
we are given RT to be 8
here u can find PR
idle raft
20.356?
idle raft
how do i find pq?
final meteor
you don't need to
idle raft
oh
final meteor
you have to find QR
and you have PR and angle QPR
what t-ratio would u wanna use in this scenario
idle raft
sin?
final meteor
QR, PR, and angle QPR
you have only the perpendicular and the base
sin uses hypotenuse
final meteor
.. cos uses hypotenuse aswell
idle raft
oh
final meteor
what did we use in the first part
final meteor
we had ST, PT and angle SPT
the third part is exactly the same scenario
so we'll use tan
idle raft
qr is 7.409?
final meteor
yea that's what im getting too
final meteor
oh mate i haven't really studied functions yet so can't do that sorry
final meteor
ok so
there's this one property of tangents
the radius is perpendicular to the tangent at the point of contact
do u know about this
idle raft
yea
final meteor
so in quadrilateral SPTO
you have OSP and OTP to be 90°
and SPT to be 20°
can u find x through angle sum now?
final meteor
yea
and now there's another neat property of circles
the angle subtended by an arc at centre is double of what is subtended at a point on circumference
do u know about this one
idle raft
so 160 times 2?
final meteor
yeah if sqt is 200 then sot would be 400
in this case we have sot to be 160
so what would sqt be
idle raft
80?
idle raft
x/2?
idle raft
ah this one
idle raft
wait so thats it?
idle raft
where do i go from here 2π(x/2)^2 + x(2π(x/2))
final meteor
2×π×(x²/4) + x×2×π×(x/2)
you can cancel out a 2 in both the terms
what do u get after that?
final meteor
i can't seem to understand what u wrote? we already opened up the (x/2)^2?
idle raft
nope
final meteor
look its 2πx²/4 + x2π*x/2
which is 2πx²/4 + 2πx²/2
2/4 is 1/2 and 2/2 is 1
so itd be πx²/2 + πx²
take the lcm from here and that should be the final answer
idle raft
so i multiple the 1 with 2
idle raft
ohhh
(oh and is it okay if i add u just in case i needa ask more in the future , i understand if thats not possible)
final meteor
yeah sure why not
so here you gotta use a property of exponents
(a^b)^c = a^(bc)
here 4^(x+1) is the same as (2^2)^(x+1)
final meteor
(a^b)^c = a^(bc)
what do suppose thatd be given this property
idle raft
so can i do 2^8-4^x+1 = 0?
final meteor
That'll become more complicated
idle raft
oh
final meteor
(a^b)^c = a^(bc)
its best if we just use this property
final meteor
can u understand how 4^a would be 2^(2a)?
idle raft
4^(x+1)=0
0/4 =x+1?
final meteor
4^(x+1)=0
0/4 =x+1?
That's.. not what i meant by simplify
idle raft
😭
final meteor
you understand that 4^(a) is 2^(2a) right
idle raft
yes
final meteor
and can u write 2(x+1) as 2x+2?
idle raft
no?
final meteor
have u done the distributive property
idle raft
nope
final meteor
the one where u multiply each term seperately?
you multiply 2 with x and then multiply 2 with 1
to get 2x and 2
so 2x+2?
idle raft
yea
final meteor
4^(x+1) is 2^(2(x+1))
yeah so this would then be 2^(2x+2) right?
idle raft
yesss
final meteor
alr so we finally wrote 4^(x+1) as 2^(2x+2)
replace that in the equation
2⁸ = 2^(2x+2)
what do you suppose we do from here?
idle raft
so 8-2 = 2x
6=2x
3=x
final meteor
yeah you got it
x is 3
that's the answer
alr imma get going now
good luck with your math mate
dull cairn
I'm trying to solve cathesis B when A is 5.581 and angle is 59.6 degrees, i keep getting 9.51 with 5.581*(tan 59.6) which is wrong
dull cairn
modern compass
you're trying to solve for side b? or angle B?
dull cairn
side B i think, i was told the answer is 3,27 but i've no clue how to get to that conclusion
the picture is not on scale btw
modern compass
tan(alpha) = a/b, your solution has it backwards
dull cairn
im trying to figure out how to type it in my calculator
modern compass
b = a / tan(alpha)
autumn wolf
struggling with writing the other expressions idk how to do it
autumn wolf
actually ignore the one i wrote i have no how to write them
frank bronze
hu
cedar kilnBOT
Available help channel!
frank bronze
i was wondering how to do this, why we take these steps to solve this
frank bronze
i got lost at the first step
plug in x+h for x it seems but how do we have a/a-3
<@&286206848099549185>
cedar kilnBOT
@frank bronze Has your question been resolved?
karmic horizon
$f(a) = \frac{a}{a-3}$
wraith daggerBOT
Scerball
kind magnet
how do I find the derivative of f(x) if all I have is y = f(x)
kind magnet
Every problem I find online I see them beginning with an equation where they take the derivative to find the slope of that equation
but all I am given is 2 points, which are (2,f(2)), which is really (2,78)
I am not given an equation
would I have to use my secant line to find the tangent line?
<@&286206848099549185> can someone explain how to find the equation here?
cedar kilnBOT
@kind magnet Has your question been resolved?
cedar kilnBOT
@kind magnet Has your question been resolved?
cedar kilnBOT
@kind magnet Has your question been resolved?
vernal shell
but all I am given is 2 points, which are (2,f(2)), which is really (2,78)
Yeah you have that point. In order to find the tanget line, you need the slope at that point too🤔
Are you sure the problem doesn't give you that slope?
kind magnet
Are you sure the problem doesn't give you that slope?
This is all the problem gives me
Every video I search online, they begin with a slope and a point or a function and a point in which they derive
Unless I have to do something with the other information such as using secant line? Because the tangent line is also at the secant line. They share the same point of (2,f(2)) or (2,78)
kind magnet
<@&286206848099549185> Can someone explain how to find the tangent line without the slope?
faint perch
Hi, do you know the general equation of the tangent line at a fixed point?
kind magnet
Hi, do you know the general equation of the tangent line at a fixed point?
My understanding is lim f(a+h)-f(a)/h
or are you asking for the equation?
faint perch
Ok, that's the value of the derivative at a, which is the slope of the tangent line, but I ask about the equation of the tangent.
kind magnet
because I've been stuck on this problem for a good while and I found the average velocity or slope of secant line
faint perch
Ok, that's the value of the derivative at a, which is the slope of the tangent l...
So it's a part
kind magnet
I have no idea how to find that
faint perch
Do you know how to find the equation of a straight line?
You know already its slope: the value of the derivative, and you need one point of the line, so just take the point where it's tangent to f.