in this case it happens to be a constant number
#help-10
daring tide
proper relic
Oh so in this case z would be the x and (x+y) is the a value.
daring tide
yeah the varialbes are arbitrary
daring tide
$$log_e(e) = 1 = ln(e)$$
warm shaleBOT
MattDog_222
proper relic
Yeah that.
So you got rid of the ln and the e through that.
And that leaves us with a^x from the previous law?
daring tide
$$e^{ln(x)} = x$$
warm shaleBOT
MattDog_222
daring tide
you just need to know that they cancel and that it "brings the inside down"
proper relic
Sorry if this is a dumb question but what is the inside?
daring tide
the value inside the parenthesis of the ln
ln(the inside)
$$e^{ln(1x+2x^2+3x^3+4x^4)} = 1x+2x^2+3x^3+4x^4$$
warm shaleBOT
MattDog_222
proper relic
Oh ok.
So combining ln with e removes it by cancellation and then it just brings the entire exponent down.
daring tide
in essense, yes
proper relic
Ok ok thank you.
warm shaleBOT
MattDog_222
daring tide
not sure when something like that might come into play with exponent rules
proper relic
Oh yeah laws of exponents.
Multiplication of numbers with same base becomes addition of the powers but base stays the same.
warm shaleBOT
MattDog_222
daring tide
$$ln(\frac{x}{y}) = ln(x) - ln(y)$$
proper relic
Multiplication of logs becomes addition.
warm shaleBOT
MattDog_222
daring tide
yeah theres a whole bunch of various laws you could get into
proper relic
Oh ok.
daring tide
$$ln(1) = 0$$
warm shaleBOT
MattDog_222
proper relic
I'm taking IB AASL, do you think I need to remember all of them?
daring tide
but you need to be familiar with the basic exponent and log rules
proper relic
Oh yeah because anything raised to the 0 is 1.
daring tide
yes, e^0 = 1
proper relic
Ok so I just have to play around and substitute differnt numbers for x.
To kind of get a feel.
daring tide
sorta? I mean the x and y intercepts are whats important
daring tide
wdym
proper relic
I was just wondering if for all values of x the intercept stays at 1.
warm shaleBOT
MattDog_222
daring tide
for ln(x)
proper relic
Oh ok.
daring tide
and the range is all reals
the inverse, e^x, has domain of all reals and range of positive reals
proper relic
Oh ok.
Thank you very much.
I will go practice more questions.
Thank you!
.close
flint kelp
obtuse pebbleBOT
flint kelp
can't figure out how to get to 5 as the answer here
obtuse pebbleBOT
@flint kelp Has your question been resolved?
flint kelp
<@&286206848099549185>
Yeah I just don't get this
I can think of more than 5 specific strings that match those requirements
1000000111111111
1100000011111111
1110000001111111
1111000000111111
1111100000011111
1111110000001111
1111111000000111
1111111100000011
1111111110000001
hushed moat
I can think of more than 5 specific strings that match those requirements
9 sounds like the right answer i think
hot hazel
.close
twilit forge
hi,
obtuse pebbleBOT
@twilit forge Has your question been resolved?
obtuse pebbleBOT
@twilit forge Has your question been resolved?
obtuse pebbleBOT
@twilit forge Has your question been resolved?
modern totem
math is so difficult
gloomy arrow
how do I solve this?
obtuse pebbleBOT
how do I solve this?
gloomy arrow
something I tried so far was letting u = x+1
but then I just end up with
(6*u-3) / u^(1/2)
to integrate
and I don't see how I can do that
am I missing something obvious? Cause this is supposed to be a pretty easy question
karmic grove
take 3 common.then split 2x+1 into x+x+1
now divide.u would get 1/root(1+x)+x/root(1+x)
obtuse pebbleBOT
@gloomy arrow Has your question been resolved?
gloomy arrow
now divide.u would get 1/root(1+x)+x/root(1+x)
wait how did you get that
gloomy arrow
root(1+x) will be in numerator
but still, I don't see how I can integrate that
root(x+1) + x/root(x+1)
timid silo
but still, I don't see how I can integrate that
root(x+1) + x/root(x+1)
(6*u-3) / u^(1/2) = 3(2u/ u^(1/2) - 1/u^(1/2)) maybe you know how to do now?
gloomy arrow
(6*u-3) / u^(1/2) = 3(2u/ u^(1/2) - 1/u^(1/2)) maybe you know how to do now?
actually yeah I think I got it, the right side turns into ln|root(u)| and the left turns into 2*root(u) which is easily integratable
lmao ok yeah that was pretty obvious
thx dude
.close
timid silo
help please
obtuse pebbleBOT
help please
timid silo
For how many values of $a$ is it true that the line $y=x+a$ passes through the vertex of parabola $y=x^2+a^2$?
warm shaleBOT
static
unique solstice
Setting them equal would be the first step
x+a=x^2+a^2
And theres one very important fact
About the x value of the vertex of this particular function thats true for all values of a
Where we're looking for intersections
What might that be
obtuse pebbleBOT
@timid silo Has your question been resolved?
timid silo
Whats the x value of the vertex of x^2+a^2
0?
unique solstice
Yes
And we're looking for the intersection to that vertex only
So what can you do to x+a=x^2+a^2 to make it easier to solve
Since we're only looking for intersections at x=0
obtuse pebbleBOT
unique solstice
@timid silo oh no
obtuse pebbleBOT
Available help channel!
winter jacinth
How many steps (20cm) would it take for a 20kg object to reach 6m/s when hitting the bottom of the supposed stairwell?
obtuse pebbleBOT
How many steps (20cm) would it take for a 20kg object to reach 6m/s when hitting...
winter jacinth
this is a part of another problem i am trying to solve, which includes human biology
unique solstice
What does falling down a staircase have to do with human biology
winter jacinth
i cant say too much, but the 20kg object is referencing to a 6 year old human
COMPLETELY HYPOTHETICAL
winter jacinth
my own research
i am bored
its something my friend thought of when sending me this image
jolly ginkgo
Mhh
obtuse pebbleBOT
@winter jacinth Has your question been resolved?
latent nova
You can try ask that in r/theydidthemath
obtuse pebbleBOT
Available help channel!
ebon holly
f:[-2,3]->[0,2], f(x) = (6-2x)/5;
I must study the bijectivity and ended up getting stuck with the intervals.
obtuse pebbleBOT
f:[-2,3]->[0,2], f(x) = (6-2x)/5;
I must study the bijectivity and ended up gett...
@ebon holly Has your question been resolved?
tight oasis
f:[-2,3]->[0,2], f(x) = (6-2x)/5;
I must study the bijectivity and ended up gett...
Is the function injective? Is it surjective?
ebon holly
I checked for the injectivity and it turned out true, as for surjectivity I simply did:
f(x) = y <=> x = (5y-6)/-2 [After some calculations]
celest sun
You need to take the domain and range into account. The equation is bijectice from R into R, but if you take a y in [0,2], will your x be in [-2,3]?
Does that make sense?
tight oasis
I checked for the injectivity and it turned out true, as for surjectivity I simp...
Nice! You proved surjectivity of the function f:R->R
But here f:[-2,3]->[0,2], so is it still surjective here in this special case?
ebon holly
You need to take the domain and range into account. The equation is bijectice fr...
Oh, I see what you mean!
celest sun
Try to take that into account then
Suppose y is between -2 and 3, then what is x between?
ebon holly
Hold on, let me calculate.
But wait, before that, isn't it between [0, 2] in theory? 🧐
tight oasis
Also trying to graph the function and the intervals of the domain and codomain of the function helps a lot!! 😁👌
ebon holly
Not sure, making an assumption here 'cause I am also thinking of the relation that x = (5y-6)/-2.
Indeed, you do have a point, however I was trying this method as well since it's.. safer in the case of an exam, hehe.
tight oasis
Hehehe
Oj
Ok
Then pick any y in [0,2]
Can you find some x in [-2,3] such that f(x)=y?
If you can for any y, then you won this little game and thus the function is indeed surjective!
If you found a y such that there are no x in [-2,3] such that f(x)=y, this means that the function is NOT surjective (don't worry, you also won the game :p)
ebon holly
I see what you mean, and this is theory if I am not mistaken. The problem here is, how am I supposed to write this in Mathematical terms, so that everything makes sense?
Since I tried something like:
x = -1 <=> f(-1) = y <=> 6-(-1)/-2 = y <=> 7/-2 = y, now what?
Wait, I got it!~
tight oasis
You're not doing it the right way
First pick some y, then challenge yourself to find an x
ebon holly
y = 0 <=> (6-2x)/-2 = 0, something like this?
tight oasis
Spoil: this is a nice function, so there is only ONE x such that f(x)=y, the only thing that you have to do is that your x is indeed in the interval [-2,3]
celest sun
But wait, before that, isn't it between [0, 2] in theory? 🧐
I swapped them yes.
I meant y between 0 and 2
tight oasis
y = 0 <=> (6-2x)/-2 = 0, something like this?
Here, you're doing it for only ONE value of y which is y=0
Now you need to do it for ANY value of y in [0,2]
(Don't to them one by one please, it takes a bit of time lol)
ebon holly
So that means the function is not surjective, thus resulting that it's not bijective, correct?
tight oasis
So that means the function is not surjective, thus resulting that it's not bijec...
Why is it not surjective?
If it is not surjective, it means that there is some y such that f(x) is NOT y for all x
Did you find such a y?
ebon holly
Yeah, y = 1.
{(6-2x)/-2 = 1;
6-2x = -2;
-2x = -8;
x = 4, which isn't part of [-2, 3].
}
tight oasis
PERFECT
tight oasis
The function is (6-2x)/5, no?
Good point lol
ebon holly
No, it's (6-2x)/5, perhaps I miswrote it in the beginning, and if that's the case, my apologies.
ebon holly
Oh, you are correct.
celest sun
I think you mixed it up with the x value you solved x = (5y-6)/-2
Hint: ||it is surjective, so don't look for a counterexample. Prove it.||
ebon holly
(6-2x)/5 = 3
6-2x = 15
-2x = 9
x= -2/9, which isn't part of [-2, 3].
celest sun
But 3 is also not part of [0,2]
ebon holly
.. Right.
celest sun
Read my hint lol
ebon holly
-2 <= (6-2x)/5 <= 3..?
celest sun
I think it's best with the formula you derived
Assume 0 <= y <= 2. Then show that -2 <= (5y-6)/-2 <= 3
ebon holly
I'll just go search it online, I am simply getting more lost h-
I swear that's the first formula I did, but ended up getting something weird.
-4 <= 5y-6 <= -6;
0 <= 5y <= 2;
0 <= y <= 10;
It's true then, since [0, 2] is part of [0, 10], right?
celest sun
Let 0 <= y <= 2.
Then _ <= 5y <= _.
Then _ <= 5y-6 <= _.
Then _ <= (5y-6)/-2 <= _.
Can you fill in the blanks?
celest sun
Other way around at the end
-2, 3
So for y in [0,2], (5y-6)/-2 is in [-2, 3] = dom(f) and f((5y-6)/-2) = y
This means f is surjective
ebon holly
Right, well one problem that I had, and is the reasoning behind why I asked, is because in the book it says it's not bijective.
Although we ended up proving that it's injective and surjective.
Thus concluding that the function is bijective.. 
celest sun
Book is wrong then
timid silo
obtuse pebbleBOT
@timid silo Has your question been resolved?
obtuse pebbleBOT
@timid silo Has your question been resolved?
timid silo
obtuse pebbleBOT
timid silo
Why is 4 not a local maximum
It changes from positive slope to negative
And its when f'(x) = 0
Because 6 is a local minimum, and it changes from negative to positive
Ugh
I think they mean positive y int to neg y int
But then why isnt -2 a local
obtuse pebbleBOT
@timid silo Has your question been resolved?
rugged barn
But then why isnt -2 a local
-2 is a local maximum
Since the slope change from positive to negative tho f increase then decrease
timid silo
obtuse pebbleBOT
rugged barn
It’s : M-5,3= (J+5,3) +10
So your first equation is ok but your second one is not correct
timid silo
So your first equation is ok but your second one is not correct
why
if M = 3J cant u just plug it in
rugged barn
It’s : M-5,3= (J+5,3) +10
Because of this
timid silo
where r u getting 30 from
rugged barn
It’s : M-5,3= (J+5,3) +10
Like this
timid silo
whats 5,3
rugged barn
Because Marcus gave 5,3 to José
timid silo
i think u mean 5.3
rugged barn
In France it’s the same but yes
warm shaleBOT
Alexis.d
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
timid silo
oh ya
rugged barn
So you need to add 5.3 to J un your second equation
rugged barn
Nop
timid silo
why not
rugged barn
You need to substrat J to erase the one in your right member of your equation
rugged barn
Eureka
rugged barn
Yup
timid silo
?
okay
i have a few mor
more
this is for practice for a test i have in a few weeks
is that fine
timid silo
a
timid silo
.close
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
ancient hornet
Quick question
if $L\in RE-R$ does that mean $L^C \in R$ ?
obtuse pebbleBOT
Quick question
if $L\in RE-R$ does that mean $L^C \in R$ ?
ancient hornet
just logically within groups 🤔
warm shaleBOT
blackmamba[they/them]
ancient hornet
not that I think about it might be false
So for example
$RE = {\set{1}, {2}, {3}, {4}, {5}, {6}} , R={{4}, {5}, {6}}$
$RE-R = {{1}, {2}, {3}}$
$ s \in RE-R => s = {1} or {2} or {3}$, but $s^c$ can contain everything else in the universe that's not these 3 groups
warm shaleBOT
blackmamba[they/them]
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
ancient hornet
smh
sweet trellis
Can I cancel a trig functions square when solving an equation? For example, if I have Sin^2(theta)=3, can I just square both sides or do I need to use the power reduction formula?
obtuse pebbleBOT
Can I cancel a trig functions square when solving an equation? For example, if I...
@sweet trellis Has your question been resolved?
spiral maple
Can I cancel a trig functions square when solving an equation? For example, if I...
yes you can square root, assuming you do it properly
however I can already tell you that your example has no solutions
sweet trellis
Yup just grabbed a random example for context
spiral maple
$\sin^2(t)=1\implies \sin(t)=\pm 1$
warm shaleBOT
Mosh
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
timid silo
H
timid silo
How do I write an equation in Ax + By = C with A > 0
I understand the Ax + By = C part, just not the A > 0 part
dapper bloom
I understand the Ax + By = C part, just not the A > 0 part
If you get A<0 you can just multiply both sides of the eqn by -1
And then rearrange it into the form Ax+By=C again with some algebra.
timid silo
I don't understand
timid silo
If you get A<0 you can just multiply both sides of the eqn by -1
What does that mean?
<@&286206848099549185>
dapper bloom
For ex if you had -2x+3y=5 you could do -(-2x+3y)=-5
By mutliplying both sides by -1
Then distribute the minus to get 2x-3y=-5 or whatever.
Do you see what I mean or nah?
timid silo
Why do I multiply both sides by -1
dapper bloom
So in my ex -2x+3y=5 so A=-2
So then A<0 right?
And we want A>0
You see how -A>0?
Well maybe that's a confusing way to say it.
You see how -1 * -2 > 0 right?
Since it is just 2.
timid silo
Yes
dapper bloom
Yah, in general you see how if A<0 then -A>0?
timid silo
No
dapper bloom
Ah I see. That's just a rule of algebra. Maybe you forgot it or haven't learned it yet. No worries though.
timid silo
Yes I haven't learned
dapper bloom
If you have Ax+By=C and A<0 then A is a negative number.
Negative numbers have a minus sign out front
If you multiply the whole thing by -1 and distribute the -1 to each term the minus sign in front of the number that was A will cancel out.
So in my example I had-2x+3y=5
So -1 (-2x+3y)=-5 then if you distribute the -1 you get (-1)(-2)x+(-1)(3)y=-5
But that simplifies to 2x-3y=-5
So now the number being multiplied by x is positive
timid silo
So what should I do for my problem
dapper bloom
So you see how if I gave you two pts you can come up with some equation of a line in the form Ax+By=C right?
timid silo
Yes
dapper bloom
Ok well if A is positive ur done
If A is negative (it has a minus sign) then multiply both sides by -1 like in my example.
Then do some algebra to write it in the right form ofc
timid silo
A is positive in 1.6x-y=2.6 though right
timid silo
It says its right but it's asking for a certain way of it
I don't know how to explain it
But I guess it needs to be more simplified??
dapper bloom
I double checked ur work and it seems fine.
timid silo
Yeah I'm not really sure why it won't accept it
dapper bloom
Maybe they want 1.6x+(-1)y=2.6?
timid silo
dapper bloom
Lmao wow
dapper bloom
It wouldn't hurt.
timid silo
It also said input error
dapper bloom
This doesn't seem like ur fault. You can even chk the pts on the eqn of the line ur giving
And it gives 2.6 on the lhs when you plug them in.
And A=1.6>0 like it's asking u for.
timid silo
Is there something I'm missing?
dapper bloom
Mathematically it seems like you understand what the question is asking you to do.
I just think the website has some mystery format it wants ur answer in that neither of us know. Is there some way to look at example versions of problems like this on the website? If so you might try copying the format.
If there's some sort of technical help service thru ur school or the website that might help.
If nothing works I would try contacting ur professor.
timid silo
I just think the website has some mystery format it wants ur answer in that neit...
I'm trying to look on the website but I can't find any examples. It usually gives us videos/examples on how to solve each problem but it doesn't have one for this problem
dapper bloom
I'd just bug ur professor I suppose.
Maybe show them what you are inputting. If we're doing it wrong they'll tell you, if not they'll tell you the right format or they should be able to make it so you don't lose pts over something that is not ur fault.
obtuse pebbleBOT
@timid silo Has your question been resolved?
dapper bloom
?
No
Try and plug in both of ur pts to see why 1.6(6)+7 is not 2.6
U can still chk if it will accept the answer tho
If somebody typed in the answer to the q by hand they could have mistyped
obtuse pebbleBOT
@timid silo Has your question been resolved?
lone bluff
Are you using that $y - y_1 = m(x - x_1)$ ?
warm shaleBOT
HrJonas
lone bluff
Maybe give this a read: https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:forms-of-linear-equations/x2f8bb11595b61c86:summary-forms-of-two-variable-linear-equations/a/forms-of-linear-equations-review
obtuse pebbleBOT
@timid silo Has your question been resolved?
obtuse pebbleBOT
@timid silo Has your question been resolved?
karmic whale
@timid silo if ur done can u do .close plz
young brook
hello :)) can someone help me with these problems? thanks :'))
obtuse pebbleBOT
hello :)) can someone help me with these problems? thanks :'))
obtuse pebbleBOT
@young brook Has your question been resolved?
young brook
hello :)) can someone help me with these problems? thanks :'))
<@&286206848099549185>
jolly ginkgo
hello :)) can someone help me with these problems? thanks :'))
If f is a quadratic function then what possibly will be its range
Btw for your problem check the change In y wrt x
young brook
If f is a quadratic function then what possibly will be its range
what-
frigid jasper
hello :)) can someone help me with these problems? thanks :'))
plot the points on a cartesian plane, it'll get you somewhere. or check whether there's a linear progression in values of y with respect to x
if you can plot it as a linear function, then it isnt quadratic
young brook
plot the points on a cartesian plane, it'll get you somewhere. or check whether ...
oh okay okay thanks
what about the other questions?
frigid jasper
you could do the same with this
frigid jasper
sum of roots is given by $\frac {-b}{a}$ and product is given by $\frac {c}{a}$
warm shaleBOT
Yash
frigid jasper
for question 12, equate it to 0 and solve for t. you could use the quadratic equation for it too
for 13, equate it to 320 and solve for t. subtracting 320 from the LHS would given you -16t^2+32t+(128-320). equate this function to 0 and solve it
young brook
for 13, equate it to 320 and solve for t. subtracting 320 from the LHS would giv...
what is an LHS
frigid jasper
left-hand side
an equation has two sides
for example, a = 2b. the LHS is a, and the RHS = 2b
young brook
for question 12, equate it to 0 and solve for t. you could use the quadratic equ...
is it the same with #14 and #15?
frigid jasper
yes
rugged nebula
.need
Given geometric sequence (un), u1=-61, d = 4 find minimum of n so that Sn > 0
rugged nebula
Given geometric sequence (un), u1=-61, d = 4 find minimum of n so that Sn > 0
Is the above the correct way of doing it?
zinc marsh
Given geometric sequence (un), u1=-61, d = 4 find minimum of n so that Sn > 0
If I am not wrong its an arithmetic sequence right?
jolly ginkgo
If I am not wrong its an arithmetic sequence right?
Yeah
rugged nebula
Yeah I mean arithmetic
jolly ginkgo
How did you get 31.5
zinc marsh
Yeah I mean arithmetic
what you thought was correct but its better to express it as an inequality
rugged nebula
2n-63=0
jolly ginkgo
Oh that's 63 not 6
rugged nebula
what you thought was correct but its better to express it as an inequality
Ahh I see
jolly ginkgo
The 3 looks like bracket
zinc marsh
2n^2-63n > 0
warm shaleBOT
it's Sam
zinc marsh
equating it to 0 might make the teacher who is correcting it confused
zinc marsh
yea
pastel obsidian
For the inequality E[{Y (t + τ ) + αX(t))}^2] ≥ 0,
Show that |R_xy (τ )| ≤ sqrt(R_xx(0) R_yy (0))
obtuse pebbleBOT
For the inequality E[{Y (t + τ ) + αX(t))}^2] ≥ 0,
Show that |R_xy (τ )| ≤ sqrt...
@pastel obsidian Has your question been resolved?
pastel obsidian
<@&286206848099549185>
obtuse pebbleBOT
@pastel obsidian Has your question been resolved?
fast kite
what are X & Y?
pastel obsidian
what are X & Y?
Just any generic probability distribution.
fast kite
Just any generic probability distribution.
i'm confused, you're taking Expectation of a squared quantity, won't it always be >= 0?
pastel obsidian
i'm confused, you're taking Expectation of a squared quantity, won't it always b...
Sorry! I picked the wrong part by mistake. I'll edit the question to fix it.
Show that |R_xy (τ )| ≤ sqrt(R_xx(0) R_yy (0))
fixed
obtuse pebbleBOT
@pastel obsidian Has your question been resolved?
obtuse pebbleBOT
@pastel obsidian Has your question been resolved?
warm shaleBOT
timid silo
Thank you
obtuse pebbleBOT
@pastel obsidian Has your question been resolved?
frosty jungle
obtuse pebbleBOT
frosty jungle
why is there 3 relative extrema
obtuse pebbleBOT
@frosty jungle Has your question been resolved?
obtuse pebbleBOT
@frosty jungle Has your question been resolved?
timid silo
Hello
timid silo
24 971 cm2 to the nearest 100cm2
jolly ginkgo
What's the closest number to 24971 divisible by 100
timid silo
250?
wise garden
timid silo
Ok
obtuse pebbleBOT
Available help channel!
gilded olive
obtuse pebbleBOT
https://gyazo.com/09df300e91d91f05ae2ce383ceaf58dc
gilded olive
coonfused
gilded olive
nvm i got it
obtuse pebbleBOT
Available help channel!
stray ember
Hi could anyone explain to me the question in B? What do I have to do??? Btw LE stands for Linear Equation
obtuse pebbleBOT
Hi could anyone explain to me the question in B? What do I have to do??? Btw LE ...
stray ember
<@&286206848099549185>
obtuse pebbleBOT
@stray ember Has your question been resolved?
obtuse pebbleBOT
@stray ember Has your question been resolved?
junior inlet
Hi could anyone explain to me the question in B? What do I have to do??? Btw LE ...
just
write the system?
keen ingot
I need help figuring the function the yeilds to the output
obtuse pebbleBOT
I need help figuring the function the yeilds to the output
keen ingot
a=1 b=1 c=3 d=2 output=0
a=0 b=0 c=3 d=2 output=0
a=2 b=2 c=3 d=2 output=2
a=3 b=2 c=3 d=2 output=3
a=2 b=1 c=3 d=2 output=1
a=0 b=1 c=3 d=2 output=0
a=0 b=2 c=3 d=2 output=0```I am pretty sure the max() is involced
you don't have to use all 4 vars
tired shell
max(a+b-2,0) 
dusk nexus
how would you solve things like sin^2(x) = 1/4
obtuse pebbleBOT
how would you solve things like sin^2(x) = 1/4
dusk nexus
for x?
like
is there a function
like for just normal sin
you would do
arcsin ( sin(x)) = x
is there smth
like
arcsin^2 (sin^2(x)) = x
bc i tried
and found
like
nothing
jolly ginkgo
First thing you do is (sin²x - 1/4) = (sinx -1/2) (sinx + 1/2) = 0
jolly ginkgo
You can just do sin²x = 1/4 implies sinx = ± sqrt(1/4) too
Whichever method you prefer
jolly ginkgo
It doesn't exactly work like that
dusk nexus
how does it work
jolly ginkgo
Do you know sqrt(x²)=?
dusk nexus
well i think about it in the difference of squares method then
dusk nexus
Do you know sqrt(x²)=?
wouldnt that just be +-x
jolly ginkgo
What do you think about absolute value
jolly ginkgo
Depends on value of x
If I say x=1 and you say sqrt(x²)= ±1
How will that be right
If I give x=-1 then sqrt(x²)= ±1 again according to it
But it's actually sqrt(x²)= |-1| =-(-1)= 1
warm shaleBOT
it's Sam
dusk nexus
oh ok
so
whenever x is greater or equal to 0, the answer to sqrt(x^2) is x, and vice versa?
dusk nexus
wait why not though
jolly ginkgo
Answer me this
dusk nexus
sorry im just a bit confused rn
jolly ginkgo
What is sqrt(4)
dusk nexus
2
jolly ginkgo
Ok so if you put it the way you did then sqrt(4)= sqrt(2²)= ±2
jolly ginkgo
What u guys solving?
Nothing much just a trigonometric question
gloomy trellis
Oh
jolly ginkgo
Ok so if you put it the way you did then sqrt(4)= sqrt(2²)= ±2
But you see -2 is not right
dusk nexus
yeah i see that now
jolly ginkgo
Ok
dusk nexus
so
jolly ginkgo
sqrt(sin^2(x)) = +-sinx
So that's why the problem here is sqrt(sin²x)= |sin x|
jolly ginkgo
**it's Sam**
Or you can write this way
dusk nexus
so its only going to be +- for specific values
jolly ginkgo
so its only going to be +- for specific values
- for x>=0 and - for x<0
Like if I give x=-2
Then x²=4
Sqrt(4)=2
And 2=-x
That means sqrt(x²)=-x in this case
dusk nexus
oh i see
how do things like x^2 = 5 work then
like i would do sqrt(x^2) = +- sqrt(5)
jolly ginkgo
You need to understand this graphically
dusk nexus
x = +-sqrt(5)
warm shaleBOT
jolly ginkgo
,w graph sqrt(x)
warm shaleBOT
jolly ginkgo
Two points in x² give same output like x=1 gives x²=1
x=-1 gives x²=1
But that's not the case in sqrt
jolly ginkgo
like i would do sqrt(x^2) = +- sqrt(5)
So you better not write it this way directly because sqrt is not inverse of x²
dusk nexus
oh i see
jolly ginkgo
Rather just write directly that x²=5 then x= ±sqrt(5)
dusk nexus
and
a square
is not a one-one function
so it can have multiple solutions
while
a square root
cannot?
at most one solution right
jolly ginkgo
so it can have multiple solutions
More like multiple input gives same output
jolly ginkgo
at most one solution right
Yes sqrt give one output for one input
dusk nexus
right
jolly ginkgo
In your question if I tell you what happens exactly then it is this
Sin²x = 1/4
Sqrt(sin²x)= sqrt(1/4)
|sinx|= 1/2
That's why we are looking for x such that sinx = ±1/2
dusk nexus
ah i see now
so theres not actually multiple solutions in the function its actually basically just an absolute value that makes it into two different functions which are sinx = 1/2 and -sinx =1/2
jolly ginkgo
I wont go much into this but I would just say avoid writing ±sqrt on squared side just write directly
jolly ginkgo
so theres not actually multiple solutions in the function its actually basically...
Yes but again write ± on right for better understanding how actually we get there
jolly ginkgo
So anyway we are finding sinx = ± 1/2
dusk nexus
i think the difference of squares method gives me more intuition so ill use that to think about it and ill keep in mind that sqrt(x^2) = |x| not +-x
yeah i found sinx = +-1/2
jolly ginkgo
So for what x values do we get 1/2
dusk nexus
i found
0.52+nπ and 2.62+nπ
i used a calculator and just wrote to the nearest two decimals
warm shaleBOT
jolly ginkgo
And other
dusk nexus
its 5pi/6
dusk nexus
its 7pi/6 and 11pi/6 but i already considered that so i added +nπ to each solution to include every single solution
obtuse pebbleBOT
cerulean tartan
Is there an equation relating flux and flow(circulation)?
obtuse pebbleBOT
Is there an equation relating flux and flow(circulation)?
@cerulean tartan Has your question been resolved?
obtuse pebbleBOT
@cerulean tartan Has your question been resolved?
white nova
So this is more of a thing that confuses me more than anything. I have a task to do with a theoretical house. There are some rooms in that house with different areas but only two rooms are relevant for my question. The living room should be half the size of the area of the house and the kitchen should be 1/3 of the living rooms area. The kitchen is part of the house meaning it adds to area of the house. But the kitchens size is dependent on the living rooms size. Isn't that just a paradox?
obtuse pebbleBOT
So this is more of a thing that confuses me more than anything. I have a task to...
white nova
How so? Can you explain?
But if you increase the living room, you also have to increase the kitchen? And when you increase the kitchen, you have to increase the living room because it is a part of the house? Seems pretty paradoxical to me but maybe I just have a small math brain.
Ah, okay. Thanks for the help:)
.close
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
meager sorrel
obtuse pebbleBOT
meager sorrel
When I differentiate I get: (3x^2-6y)/(6x-3y^2)
but the mark scheme says:
this has happened before as well for another question
I am not sure where I am going wrong
undone geyser
._. its the same thing
meager sorrel
how
undone geyser
multiply top and bottom by -1
undone geyser
$\frac{3x^2-6y}{6x-3y^2}=\frac{-1(3x^2-6y)}{-1(6x-3y^2)}=\frac{6y-3x^2}{3y^2-6x}$
??
warm shaleBOT
kiid
subtle hearth
obtuse pebbleBOT
subtle hearth
I am wondering if i can do this transition from line 2 to line 3 by modus tollens
<@&286206848099549185>
obtuse pebbleBOT
@subtle hearth Has your question been resolved?
subtle hearth
I am wondering if i can do this transition from line 2 to line 3 by modus tollen...
someone?
just need someone to answer to this question, because im not sure about it
brisk matrix
i’m confused by the symbols. is -> implies? or is => implies
subtle hearth
need to prove this is taotology, dont pay attention for -> / =>
brisk matrix
i can’t understand the question then
subtle hearth
I mean I need to prove that something -> not(p)
need to get not(p) ~> not(p)
well i think i solved it, but im not sure if my transitions are allowed
brisk matrix
alright so now we have -> ~> and =>
subtle hearth
If you can just take a look to line 2 for line 3 if its allowed
brisk matrix
what does each of them mean
subtle hearth
=> means assemption and conclution . ~> is like cause and effect
=> means that the phrase is taotology( if im not wrong)
brisk matrix
does the first line at the end read “(p iff r) implies not q”
subtle hearth
what?
i just figure that (q~>r /\ r~> q ) /\ r => (q~>r) /\ q according to modus tollens **(p~>q) /\ p => q**
brisk matrix
the first line, is it (p XOR q) AND r AND (q <-> r) -> p?
there’s a symbol next to the p at the end there
is that meant to be “not p”?
subtle hearth
at the end is not p
subtle hearth
yea i want to develop the assemption and to get into this
to not p
i want to get (not)p => not(p)
subtle hearth
tried to use MT
subtle hearth
i just figure that (q~>r /\ r~> q ) /\ r => (q~>r) /\ q `` according to modus ...
.
this is my solution
tbh i am not sure about couple of things i did here lol
brisk matrix
i’m a little confused on 6 -> 7
subtle hearth
5~>6 is good?
brisk matrix
just redundant
you added an extra Q
you only need 1
oh
what about 3->4
how did q->r disappear
subtle hearth
i’m a little confused on 6 -> 7
6~~>7 yea it is confused i used disribution and Absorption laws
subtle hearth
what about 3->4
Modus tollens
brisk matrix
but if p is true and r is false then (p->r) /\ p is false, but p alone is true
so they’re not equivalent
or q instead of p but same thing
subtle hearth
Modus ponens***
subtle hearth
so it doesnt equivalent according to this?
ohh wait
nvm 🤦♂️
I will fix it, thank your for the help
.close
velvet flicker
how do you get x intercept and y?
obtuse pebbleBOT
how do you get x intercept and y?
naive fossil
how do you get x intercept and y?
I recommend simplifying f(x) first (it's usually easier to tackle things after simplifying). Then, imagine f(x) = y. With that, to find your x-intercept, simply let y = 0 and solve for x. To find your y-intercept, let x = 0 and solve for y.
velvet flicker
oh thats what I did before but my textbook says factorize, quadratic formula etc so i was confused
thank you
.close
restive osprey
help pls
restive osprey
timid silo
so if QS goes through the center of the circle then the angle QRS equals what?
vagrant basin
obtuse pebbleBOT
vagrant basin
yeah and if I were to find the diffrence I would subtract og and new
oh good point I forgot about the brackets
brave bramble
But part b) looks good!
vagrant basin
and I simplified/expanded b) and got 81x^2 + 18x - 15
so I would take that nd subtract it by (9x)^2 to find diffrence
brave bramble
The picture above is not asking for the difference. If you do want the difference, yeah you'd subtract (9x)²
Note (9x)² = 9²x² = 81x²
vagrant basin
alr I was jus wondering cuz previous worksheet questions asked for the diffrence
sso 18x-15
cuz they two cancel out
nvm figured it out
thx for the help
.close
idle galleon
hi! im solving quadratic inequaliites and im confused why this method (from khan academy) doesnt work for this question
obtuse pebbleBOT
hi! im solving quadratic inequaliites and im confused why this method (from khan...
naive fossil
MY ATTEMPT
https://cdn.discordapp.com/attachments/713778930867503114/90985533750...
Ok, so...what exactly is your question? Also, if you're trying to use the khan academy method (which absolutely makes sense and is easy to understand so I would if i were you), you must use test points
idle galleon
my question is why is my case thingo saying that there's a solution at both x<2 and x>2 when the original answer is x<=2 and 3<=x
but that's prob cause i didnt test points
in this case, where would you test points
unique solstice
Before and after or between all of the zeroes
idle galleon
right
idle galleon
okay lmfao ig the point checking thing wasnt made super clear in the viddy i watched (or i just missed it)
idle galleon
oh damn can you link the vid
naive fossil
Here's a bit of a visual to try and help.
idle galleon
i defo didnt watch that one
naive fossil
On the left of 2 you can pick any test point - I chose 0. If you plug 0 in to (x-3)(x-2) you get (0-3)(0-2), which is -3*-2 which is a positive number.
Since you're original equation has >= you know you're looking for positive numbers, so you want to "keep" this interval. So, at least part of your solution so far is (-∞, 2] <- notice we're including 2 because it's >=
naive fossil
Now, you want to test each other interval. Between 2 and 3, and then to the right of 3.
idle galleon
so if the oriignal equation is just a > in this case, then your solution would be (-∞, 2) instead?
naive fossil
so if the oriignal equation is just a > in this case, then your solution would b...
Exactly, because it's saying greater/less than 2, not = 2
idle galleon
yep yep
one last q
with the inequality notation, are the arrows always pointing left?
or does it not matter
tyvm for explaining the test point thing btw! i forgot about that
naive fossil
It can vary - you must know how to read > vs < and solve accordingly
timid silo
Now, you want to test each other interval. Between 2 and 3, and then to the righ...
you dont actually need to test intervals if you got just quadratic equation because it always will be +-+ but if you would have -(x-3)(x-2) it would be -+-
idle galleon
oh that makes sense, it's because of the graph of a parabola right?
naive fossil
oh that makes sense, it's because of the graph of a parabola right?
More or less - the test points allow you to get comfortable with concepts like this and check yourself to ensure correctness.
naive fossil
Happyto help!:)
idle galleon
.close
long basin
obtuse pebbleBOT
timid silo
substitute that in for x
signal mango
Write the quadratic as a product of linear factors using those roots
Algebra works out, b and c will be real numbers
obtuse pebbleBOT
@long basin Has your question been resolved?
long basin
Write the quadratic as a product of linear factors using those roots
Can u show me
?
On paper if u can
topaz bridge
you need only to mulitply out $(x+2-i)(x+2+i)$
warm shaleBOT
tomzizek
obtuse pebbleBOT
@long basin Has your question been resolved?
obtuse pebbleBOT
timid silo
Can u show me
vieta theorem b=-(x1+x2) c=x1x2 b+c=-(x1+x2)+x1x2
obtuse pebbleBOT
Available help channel!
timid silo
obtuse pebbleBOT
timid silo
My question is the following.
why is the discord ask query giving this
wtf is -3hits
Whilst wolfram alpha on the internet gives this:
I'm unsure. perhaps I am using the wrong format for a query via query function
warm shaleBOT
topaz bridge
,ask query 1+1
warm shaleBOT
topaz bridge
for some reason he always see "hit" at the beginning
timid silo
Hmm weird. I'll stick to using the wolfram alpha on the internet
do you know any decent free EXPLANATION solving internet programs
rather than soley the answer
(other than that. my questions are solved)
timid silo
.close
timid silo
diff xy-x²=4
obtuse pebbleBOT
diff xy-x²=4
timid silo
u can implict differentiate of course
but
what if u just rearrange for y?
y = 4-x^2/x
do u get
-4x^{-2} + 1
but thats different??
im having trouble for some reason
uh wth one sec lemme send it
where does that negative come from
zinc stump
I'm not sure what the question is. What is x, y ?
timid silo
nevermind
im good lmao i was just being sleepy
they equal each other
whether u implicit differntiate or not
wait actually no im still confuised
timid silo
I'm not sure what the question is. What is x, y ?
yo
where does the negative come from here
xy' + y - 2x = 0
xy' = 2x - y
y' = (2x - y)/x
oh nevermind
its the same thing lol
zinc stump
It is
timid silo
now it makes sense
dont understand why they did this dumb thing with the -
but thats online solvers for u
obtuse pebbleBOT
@timid silo Has your question been resolved?
upper laurel
I want a function that looks something like this. I has to have a high slope in the beginning and a low slope at the end. I couldn't come up with an answer. I guessed that you could say that the slope exponentially decreases, but if you plot "y = √(x)", the beginning is way to steep. Could anyone help with this?
obtuse pebbleBOT
I want a function that looks something like this. I has to have a high slope in ...
timid silo
$\sqrt{x}$
warm shaleBOT
Jadefalke
timid silo
Or $\log{x}$
warm shaleBOT
Jadefalke
obtuse pebbleBOT
@upper laurel Has your question been resolved?
upper laurel
Or $\log{x}$
Those are almost right, but the slope in the beginning of both options is way higher than what I want
timid silo
You can scale it of course
upper laurel
You can scale it of course
How would you do that?
upper laurel
I really suggest you open desmos and just play around
Yes I'm trying things on desmos, but my problem is that the beginning should be less steep relative to the ending
And multiplying doesn't really fix that
timid silo
What about shifting $-\frac{1}{x}$ up and left
warm shaleBOT
Jadefalke
obtuse pebbleBOT
@upper laurel Has your question been resolved?
marsh wharf
405 degrees is 9pi/4 radians right?
obtuse pebbleBOT
405 degrees is 9pi/4 radians right?
marsh wharf
and 13/12 rotations is 405 degrees?
ember rune
Find the center of the circle that you can circumscribe about ΔABC given points A(2,8),B(0,8),C(2,2).
obtuse pebbleBOT
Find the center of the circle that you can circumscribe about ΔABC given points ...
ember rune
how do you solve this?
obtuse pebbleBOT
@ember rune Has your question been resolved?
carmine hawk
Find the center of the circle that you can circumscribe about ΔABC given points ...
You need to recall a property of circumscribed circles. The center is located at the intersection of the perpendicular bisectors of each side of the triangle.
Does that help?
ember rune
a little, how do i find the intersection of the perpendicular bisectors?
carmine hawk
You could write the equation of the lines (two of them is enough), and then solve for the intersection