so im assuming that means the equation works right
#help-17
sharp swallow
gritty sage
Yes, the points are nice and close.
sharp swallow
thank you so much!
gritty sage
You're welcome.
sharp swallow
i have one question left related to this situation problem thing
im assumign i just set y=82
sharp swallow
so im assuming the answer is "I would expect the temperature in Phoenix to be 82 degrees during the middle of March and during the middle of October"?
gritty sage
That's good, but not exactly.
The temperatures given are the average temperatures.
If I had to guess when in the month they happened at, it would be in the middle of the month.
So, 3 would be in the middle of March.
sharp swallow
yes
gritty sage
So, 4 would be in the middle of April. And 3.5 would be halfway between the middle of March and the middle of April.
When is that?
sharp swallow
the beginning of april
gritty sage
Right. Make sure to put that you're assuming that the average temperature happens in the middle of the month, so 3 would be the middle of March and 3.55 would be the beginning of April.
sharp swallow
Should I just say "I would expect the temperature in Phoenix to be 82 degrees during 3.555 months and 10.445 months"
gritty sage
Not really, since those are durations rather than times.
Like if I say 5 seconds, that doesn't tell you when.
But if I say 5:13:05 AM, that tells you when.
So, I'd put it as a when, like "beginning of April".
sharp swallow
"I would expect the temperature in Phoenix to be 82 degrees during the beginning of April (3.555 months) and the end of October (10.445 months)"
gritty sage
I'd go with the end of October since 0.4 is less than 0.5.
But yes, that looks fine.
Just make sure to state why you're doing that.
So that they know why you came up with that.
sharp swallow
I'd go with the end of October since 0.4 is less than 0.5.
true ok
gritty sage
Say something about the average probably happening in the middle of the month, so 3 would be the middle of March and 3.55 would be somewhere in the beginning of April.
sharp swallow
Assuming that the average probably occurs in the middle of the month, I would expect the temperature in Phoenix to be 82 degrees during the beginning of April (3.555 months) and the end of October (10.445 months)
gritty sage
Yes, that looks fine.
gritty sage
You're welcome.
sharp swallow
sorry to take so much of ur time but will u be willing to help me with one last scenario? (it only has 2 sub questions)
gritty sage
Yes, but open a new channel.
mortal dust
vocal sleetBOT
mortal dust
Hello I have done till this much
AB²= (r1+r2)²-(r1-r2)² = 4r1r2
So inequality becomes
Substituting a² = r1 and so on
How do I continue?
mortal dust
The last one is CA
2nd is BC
$AB= 2\sqrt{r_1r_2}$,
$BC= 2\sqrt{r_2r_3}$,
$CA= 2\sqrt{r_3r_1}$
twin meteorBOT
dum
vocal sleetBOT
@mortal dust Has your question been resolved?
mortal dust
<@&286206848099549185>
vocal sleetBOT
@mortal dust Has your question been resolved?
mortal dust
<@&286206848099549185>
vocal sleetBOT
@mortal dust Has your question been resolved?
mortal dust
<@&286206848099549185>
twin meteorBOT
dum
mortal dust
Where are you stuck, and what do you need help on?
Now I'm stuck
vocal sleetBOT
@mortal dust Has your question been resolved?
dull maple
How are you getting CA equal to that ?
You realise that same procedure doesn't work for CA?
mortal dust
You realise that same procedure doesn't work for CA?
Why should it not
Could you elaborate
dull maple
try making a triangle of the same type the way you did for AB and BC.
mortal dust
Yep done
dull maple
Do you think that hypotenuse will still be sum of radii ?
mortal dust
Yes
dull maple
Lol. See again.
mortal dust
.
dull maple
Oh. Wait.
mortal dust
🧐
dull maple
Gosh. I'm so so sorry. 🤦♂️
mortal dust
Gosh. I'm so so sorry. 🤦♂️
It's ok
Happens
Any idea on what to do next
I think it is almost done
Some am gm thingy
dull maple
I'm seeing now. I stopped then at that point.
dull maple
Some am gm thingy
Yeah. Suspected so.
dull maple
If a^2 = r1, then shouldn't it be 16?
dull maple
OH so ab is not AB? I thought you just felt lazy to press shift. Lol
dull maple
**dum**
I mean... Why are you doing this though ?
You are manipulating what you have to achieve.
Shouldn't you be independently trying to get this inequality?
mortal dust
What should I do then?
dull maple
I think. You should be doing something over this. Try to manipulate this information.
twin meteorBOT
dum
vocal sleetBOT
@mortal dust Has your question been resolved?
glad python
How do I prove that inequality
Write r_2 in terms of r_1 snd r_3 and use standard inequality techniques
vocal sleetBOT
vocal sleetBOT
Available help channel!
primal shell
vocal sleetBOT
primal shell
how do u find the mode of this?
flat whale
Do you know what mode means
flat whale
most frequently occurring number
What is that in terms of the graph
primal shell
What is that in terms of the graph
ive got no idea to be honest
my guess would be 0.5
flat whale
Most frequent means where the probability is the highest
For densities that is
How do you find where the probability density is the highest ?
flat whale
Have you taken differential calculus?
sharp lynx
They don't want calculus
primal shell
im doing year 12 math
flat whale
is it at its highest at 1.5
Close, but you probably should be precise
primal shell
Close, but you probably should be precise
yeh so 1.4
primal shell
Most frequent means where the probability is the highest
so when its a graph it is when it is at its maximum?
which is the mode?
flat whale
so when its a graph it is when it is at its maximum?
Yes
hushed hamlet
What is understand solving equations as a process of reasoning and explain the reasoning, what’s is that supposed to mean
vocal sleetBOT
What is understand solving equations as a process of reasoning and explain the r...
hushed hamlet
When I look it up I can’t find any explanations
Is it the same as reasoning with linear equations?
mild flower
it's pretty vague but it might include like quadratic as well
hushed hamlet
Yes but I don’t really know how to get an explanation for this specific thing because I don’t know what to look up
vocal sleetBOT
hushed hamlet
.close
vocal sleetBOT
Available help channel!
river kettle
is part a, log_a(c) - log_a(b) which equals 1.3
vocal sleetBOT
is part a, log_a(c) - log_a(b) which equals 1.3
ornate ember
is part a, log_a(c) - log_a(b) which equals 1.3
yep!
paper depot
well, what's being raised to the third power in the original expression?
river kettle
a
paper depot
so
log_a(a^3), if you're so inclined, is 3 log_a(what)?
||but also, log_a(a^3) can be found directly via the definition of log, no log laws required: "a raised to what power gives a^3?"||
river kettle
oh its 3?
paper depot
ok that was my mistake asking two questions at the same time
no, $\log_a(a^3) \neq 3 \log_a(3)$
twin meteorBOT
Ann
paper depot
but $\log_a(a^3)$ does equal just 3 yes...
twin meteorBOT
Ann
river kettle
so a raised to the what power gives a cubed
do i use exponent laws here
7log_a(b)
paper depot
so a raised to the what power gives a cubed
that's what i wrote earlier, yes.
good job parroting it.
vocal sleetBOT
@river kettle Has your question been resolved?
magic bluff
Need help in a permutations question
vocal sleetBOT
Need help in a permutations question
magic bluff
This procedure is used to break ties in games in the championship round of the World Cup soccer tournament.
Each team selects five players in a prescribed order. Each of these players takes a penalty kick, with a player from the first team followed by a player from the second team and so on, following the order of players specified.
If the score is still tied at the end of the 10 penalty kicks, this procedure is repeated. If the score is still tied after 20 penalty kicks, a sudden-death shootout occurs, with the
first team scoring an unanswered goal victorious.
a) How many different scoring scenarios are possible if the game is settled in the first round of 10 penalty kicks, where the round ends once it is impossible for a team to equal the number of goals scored by the other team?
b) How many different scoring scenarios for the first and second groups of penalty kicks are possible if the game is settled in the second round of 10 penalty kicks?
c) How many scoring scenarios are possible for the full set of penalty kicks if the game is settled with no more than 10 total additional kicks after the two rounds of five kicks for each team?
vocal sleetBOT
@magic bluff Has your question been resolved?
vocal sleetBOT
@magic bluff Has your question been resolved?
maiden lagoon
vocal sleetBOT
paper depot
,rccw
twin meteorBOT
woeful cloak
your bounds for the integral wrt z is wrong
maiden lagoon
your bounds for the integral wrt z is wrong
wym
woeful cloak
z is not bounded by 0<=z<=1, it's 0<=z<=2
the planes intersect at z=2 on the z-axis
fluid sandal
how do i do question 12?
vocal sleetBOT
how do i do question 12?
outer warren
consider using the discriminant
fluid sandal
not sure if i did it wrong but here’s what i did so far
wait holdon
that’s meant to be a 16
16-4
paper depot
also (m^2 x + 8m)x is... unhelpful
fluid sandal
so + 12
paper depot
collect like terms by powers of x
fluid sandal
didnt i just do that?
paper depot
yeah so it should go with the other x^2 you've got...
paper depot
??? how did that happen
paper depot
you have the equation x^2 + m^2 x^2 + 8mx + 12 = 0
fluid sandal
wtf
paper depot
x^2 + m^2 x^2 = (1+m^2)x^2.....
grave falcon
yes but what was that
grave falcon
How did you write the equation that contains only m terms?
fluid sandal
what do you mean
fluid sandal
it is
i checked the book
but my answer was meant to be +- sqrt 3
not +- sqrt 4 or +- 2
grave falcon
only has one solution right?
fluid sandal
Im pretty sure?
crimson pumice
cambridge math book?
fluid sandal
it’s asking for values that make a line be a tangent to the circle
fluid sandal
cambridge math book?
yup
grave falcon
okay then its true for m
fluid sandal
i usually skip these questions but how do i do 15?
i solved it and got no solution
but how do i actually put it into words
grave falcon
i have no time for now see you later
civic otter
but how do i actually put it into words
Draw those curves and the answer will come afterwards
civic otter
y = 3 and x²+y²=4
civic otter
Nothing you can do here of course...
RHS is positive, LHS is 0 or negative
Hence there are no solutions (and so no intersections)
vocal sleetBOT
@fluid sandal Has your question been resolved?
fluid sandal
.close
vocal sleetBOT
vocal sleetBOT
Available help channel!
vast shale
vocal sleetBOT
dull maple
what did i do wrong?
That thing inside the red circle.
vast shale
That thing inside the red circle.
could you tell me what am i supposed to do instead?
dull maple
could you tell me what am i supposed to do instead?
Since power is only 2. One way could be to just solve quadratic to get values.
Other way would be to write it in form of e^(itheta) and then take root.
vast shale
Since power is only 2. One way could be to just solve quadratic to get values.
...
oh alright, i was able to get z = 3+2i, z= -3-2i
as my final answer
maiden lagoon
the planes intersect at z=2 on the z-axis
how were u able to tell this?
Im doing another practice problem
And this is what i have
woeful cloak
occupied channel, please open a new one
vast shale
cuz the person got the answer
not yet
maiden lagoon
but kk ill open a new one
woeful cloak
that's an assumption
maiden lagoon
my bad
vast shale
I think that similarly you can calculate w.
i got the values for w = 2-i, -2+i
dull maple
i got the values for w = 2-i, -2+i
Do you want them checked or what ?
vast shale
Do you want them checked or what ?
nah it's alright, i'll get them marked by my teacher
thank you : )
.close
fluid pumice
explain why in {2^(2x-1)} = {2^(x+3)} - k, if k > 32, no real solutions?
fluid pumice
im praying it isnt a discriminant question
dull maple
Well, if you are not doing higher level mathematics, discriminant is the standard way to go.
Alternatively, you can use derivatives to find minimum and graph to show it.
fluid pumice
if i use substitution of 2^x to do this, my discriminant is 64 - 2k, so do i just say if k > 32, disc is < 0, hence no real sols?
(if 64 - 2k = 0, k = 32)
peak scroll
vocal sleetBOT
lofty bridge
How do I do the last one
10th question ?
paper depot
,rccw
twin meteorBOT
vocal sleetBOT
@peak scroll Has your question been resolved?
boreal root
vocal sleetBOT
boreal root
.close
zenith wren
hello I've solved this exercise by moving the 1/x and 1/x+4 on the other side. This is supposed to have 3 solutions but I only got 2
vocal sleetBOT
hello I've solved this exercise by moving the 1/x and 1/x+4 on the other side. T...
paper depot
these are super low resolution
zenith wren
mb
paper depot
i for one cannot make out a thing
zenith wren
the first one is clear at least?
paper depot
i can see the equation $\frac{1}{x} + \frac{1}{x+1} + \frac{1}{x+3} + \frac{1}{x+4} = 0$ yes
twin meteorBOT
Ann
paper depot
when you divided both sides by (2x+4), you destroyed the solution x = -2.
also you should cross out the ENTIRE factor you're cancelling out and not just draw a flimsy diagonal stroke through a tiny part of it.
zenith wren
why can I not divide by 2x+4
paper depot
that is not what i said.
and even if you DID read what i said as forbidding the action, then i told you exactly why.
vocal sleetBOT
@zenith wren Has your question been resolved?
vast shale
vocal sleetBOT
vast shale
Help I don't understand the wording
So from what I understand Valentin
can pick up any coin he wants and then if he flips it
then he flips all the coins on the left
when he flips a coin, a coin can either black or white
so why can't all of them be correct
paper depot
flip means turn over, not toss.
a coin that gets flipped changes color from white to black or vice versa.
it doesn't get tossed for a fifty-fifty chance of either one
vast shale
vocal sleetBOT
mild flower
kind of
my mind is thinking of base 3 notation
there are some things you know are a waste of moves, like subtracting 1 three times in a row
vast shale
there are some things you know are a waste of moves, like subtracting 1 three ti...
sure
I wrote all the powers 3 until it > 2023
mild flower
so you need to remove 2187-2023 from that which is 164
by putting -1 at strategic locations
what happens to the answer if you put it at each spot?
vast shale
what happens to the answer if you put it at each spot?
so when I put -1 after 9
to make 8
I got up to 1944
mild flower
,calc 2187-1944
twin meteorBOT
Result:
243
vast shale
which is 3^5
mild flower
yes
mild flower
we do, yes
vast shale
,calc 8 * (3^5)
twin meteorBOT
Result:
1944
,calc 26 * (3*4)
twin meteorBOT
Result:
312
,calc 26 * (3^4)
twin meteorBOT
Result:
2106
,calc 25 * (3^4)
twin meteorBOT
Result:
2025
yes, the way you're doing it seems like trial and error
perhaps you might find a pattern in how the output changes when you insert -1 at each location
start at the end
vast shale
I guess the argument would be that when you subtract 1 from 9, and then multiply that by 3^5 you get something way lower than 2023, and when you do 27-1 you get something way above 2023, so the only logical thing to do would be to subtract 1 more since going up to 81 would be pointless as subtracting 27 by 1 already creates too large a number, and playing around with 9 would be pointless as it's too little and you can't add 1
mild flower
yes, as we established you shouldn't subtract 3 and all you can do is subtract so it's really just a matter of making it smaller
vast shale
i'm wondering if there is something more concrete
mild flower
yes, sort of
vast shale
something quicker
mild flower
you've already kind of identified it
if you -1 at the end what happens to the output?
it goes down by 1 ok great
what if you -1 before the last multiplication?
vast shale
,calc 8 * (3^3)
twin meteorBOT
Result:
216
,calc 25 * (3^3)
twin meteorBOT
Result:
675
Result:
2022
no
mild flower
before multiplying by 3 the final time, subtract 1 first
what do you get? what's the difference between that and 2187?
vast shale
,calc 728 * 3
twin meteorBOT
Result:
2184
Result:
164
,calc 243-164
twin meteorBOT
Result:
79
2184 - 1944
mild flower
what about if you put it in the third-last position instead?
vast shale
,calc 2184 - 1944
twin meteorBOT
Result:
240
,calc 242 * (3^3)
twin meteorBOT
Result:
6534
,calc 242 * 9
twin meteorBOT
Result:
2178
2184-2178
mild flower
you are like a kid that won't put down their new toy
vast shale
,calc 2187 - 2178
twin meteorBOT
Result:
9
ok I see
I'm starting to see the link
if you put -1 at the before last position
it creates an effect -3
mild flower
without calculating it, what do you think the next one would be?
vast shale
without calculating it, what do you think the next one would be?
like 4th last position?
mild flower
yeah
vast shale
,calc 2187 - (3*4)
twin meteorBOT
Result:
2175
so far we've had -1, -3, -9
vast shale
,calc 80 * (3^4)
twin meteorBOT
Result:
6480
,calc 80 * (27)
twin meteorBOT
Result:
2160
yesss
vast shale
soit would be 2187 - (3^3)
mild flower
okay, we could do that
twin meteorBOT
Result:
2106
but
vast shale
okay, we could do that
what would you do
vast shale
yes
mild flower
so we are charged with subtracting any number of powers of 3 from 2187 to arrive at 2023
how much do we need to subtract in total?
twin meteorBOT
Result:
164
yes good
mild flower
yes very good
vast shale
so we go to the 4th last position
mild flower
,w 164 in base 3
vast shale
and subtract that by -2 since we have to double
vast shale
,w 164 in base 3
what is this
mild flower
it's a breakdown of the powers of 3 that fit into 164 but if you haven't done number bases don't worry about it
mild flower
and subtract that by -2 since we have to double
okay so we've subtracted 81 twice, what are we left with?
vast shale
2025
mild flower
we started with 164
vast shale
subtract another 2
vast shale
ah I see this method is a lot more practical
mild flower
so how many subtractions did we do? and how many multiplications?
native kiln
i
native kiln
can i get some help with calculuc
green hinge
Sure
native kiln
i dont know how to evulute
limits
this is my equation
\lim _{x\to 2}\left(\frac{\left(x^3-8\right)}{x-2}\right)
green hinge
$\lim _{x\to 2}\left(\frac{\left(x^3-8\right)}{x-2}\right)$
twin meteorBOT
FuriousChocolate
green hinge
Alright
So here you can't just plug in 2 since 2-2 = 0 and you can't divide by zero right
native kiln
yah
green hinge
But if you can simplify the fraction
Then you can get rid of the x -2
And plug 2 in directly
native kiln
$\lim _{x\to 2}\left(\frac{\left(x^2-8\right)}{-2}\right)
green hinge
You need dollar signs on both sides
twin meteorBOT
Sunset
green hinge
Is this the limit you're working with?
Or the first one
Because they are different
green hinge
**FuriousChocolate**
This one has an x in the denominator
native kiln
to get this equation
green hinge
You can't cancel like that
native kiln
oh
green hinge
You have to factor the top expression
In order to make it so the top and bottom both have (x -2) in them
And then you can cancel (x-2) from both sides
One thing you might notice is that the top equation is the difference of two cubes:
$x^3 - 2^3$
twin meteorBOT
FuriousChocolate
green hinge
Do you know the formula for factoring a difference of two cubes?
native kiln
i know but i forgot what it looks like
green hinge
Do you know polynomial division?
green hinge
**FuriousChocolate**
Since the fraction means the top divided by the bottom, try dividing x^3 - 8 by x - 2
native kiln
$(a-b)(a^2+ab+b^2)$
twin meteorBOT
Sunset
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
green hinge
$(a-b)(a^2+ab+b^2)$
twin meteorBOT
FuriousChocolate
green hinge
$\lim _{x\to 2}\left(\frac{\left(x^3-8\right)}{x-2}\right)$
twin meteorBOT
FuriousChocolate
native kiln
(x-2)(x^2+2x+4)
green hinge
Right
native kiln
the x-2 cancel each other
green hinge
Yep!
native kiln
then i add the 2
green hinge
$\lim _{x\to 2}\left(x^2 + 2x + 4\right)$
twin meteorBOT
FuriousChocolate
green hinge
This is what you get after the (x-2)s cancel
What happens if you try to plug in x = 2 now?
native kiln
12
green hinge
Nice!
twin meteorBOT
Sunset
green hinge
So what happens when you try to plug in x = 3
karmic imp
$\lim _{x\to 3}\left(\frac{x^2-8x+16}{x^2-x-12}\right)$
Factor the numerator and denominator
green hinge
Are you sure the limit wasn't to -3?
native kiln
is positive 3
karmic imp
Then you don't need to factor it
green hinge
And its x^2 - x - 12 in the denominator?
karmic imp
$\lim _{x\to 3}\left(\frac{x^2-8x+16}{x^2-x-12}\right)$
Just plug in x if this is the proper question
karmic imp
You should simplify when is it results in an indeterminate form or divided by 0
native kiln
oh
karmic imp
Because that problem did not have divided 0, you didn't need to simplify
karmic imp
**FuriousChocolate**
This resulted in division by 0 so you needed to simlify it
native kiln
okey i understand now
i will probably have more question latern on
but i want to try solve for myself
i will be back
green hinge
Alright
karmic imp
i will be back
Don't forget to .close the channel if you are done, you can always open a new one when needed
native kiln
how do i do that
karmic imp
.close
native kiln
.close
open spade
Have I made a mistake thus far? I used the angle subtraction formula for tan(x-y) and got a finished product. I'm wondering if I either made a mistake or have not simplified enough.
vocal sleetBOT
Have I made a mistake thus far? I used the angle subtraction formula for tan(x-y...
whole oasis
your answer is correct
but you have to manipulate it a bit
to make it one of form they have
notice that it's > 0, so you can reject answers A. and D.
since they're < 0
open spade
Mhh ok I'll give it a shot. Tysm
whole oasis
what's more
notice that, the numerator is less than the denominator
so you're sure answer is less than 1
so the only one which fits to the conditionds is B.
open spade
Lol I never thought about it like that 
whole oasis
open spade
Thanks again :)
whole oasis
yea, deduction can save us some work
open spade
True, I'll still give it a shot at manipulating my answer just incase I can't use deduction in further questions.
.close
vast shale
vocal sleetBOT
paper depot
the tables are going to form a polygonal approximation to an annulus
in particular the inner sides of the tables will form a regular n-gon
find from the diagram this n-gon's interior angle
also who's karlo
vast shale
@vast shale this guy
vast shale
the tables are going to form a polygonal approximation to an annulus
sorry I don't understand
paper depot
don't mind that one
in particular the inner sides of the tables will form a regular n-gon
this is more important
vast shale
> in particular the inner sides of the tables will form a regular n-gon
ok so the inner sides of the table
this I take it
paper depot
no
i meant the sides that are facing inward once the ring of tables is complete
those green ones aren't gonna be sides at all
the side at the top, which is adjacent to both 99° angles, will become an inner side
the one at the bottom will become an outer side, accordingly
vast shale
paper depot
yes
paper depot
yes
you can do it
as in it is possible and i am 100% certain you specifically are able to
vast shale
won't it be 360 - 99 - 99
paper depot
of course, what else
vast shale
,calc 360 - 99 - 99
twin meteorBOT
Result:
162
we do not have the number 162 suspended in midair
and there is no such thing as a "polygon formula" unless you are a fan of vagueness
vast shale
these are 4 different things
wouldn't we able to do $162 = \frac{(n-2)*180}{n}$
twin meteorBOT
Mushaar
paper depot
do you just doubt yourself that much
vast shale
yes
paper depot
what room is there for doubt
vast shale
Geometry is the main reason i stopped like math before I became interested in it again
vast shale
what room is there for doubt
well none now after you asked that
paper depot
you have that your polygon's interior angle is 162°
you have that the interior angle of an n-gon is 180(n-2)/n °
put two and two together
\cdot in latex btw
vast shale
,calc 162 - 180
twin meteorBOT
Result:
-18
.close
vocal sleetBOT
open spade
,calc 162 - 180
#bots would be a better place to do that
vocal sleetBOT
Available help channel!
naive notch
I can't figure out a)
vocal sleetBOT
I can't figure out a)
naive notch
can anyone walk me through it?
paper depot
do you know how to solve linear equations like these but without fractions
naive notch
yeah
paper depot
ok
then multiply both sides of (a) by 35 and you will end up with an equation just like that
paper depot
show what you did.
naive notch
I'm an idiot
paper depot
i didn't ask you what you are. i asked to show what you did...
or should i read this as "i realized where i messed up"
vast shale
vocal sleetBOT
vast shale
I think it’s true
paper depot
well it is awfully typeset but yes it is true
vocal sleetBOT
vast shale
Is this false?
civic otter
,w derivative of (x+1)/(3x-2)
twin meteorBOT
vast shale
Because this makes it seem true
harsh canopy
yes it is false
vocal sleetBOT
Available help channel!
swift night
help
terse warren
First try to identify common parallelograms (rectangle, triangle,etc..)
terse warren
nice, well done
now what is the area of a rectangle ?
and what is the are of a triangle ?
swift night
square is 40
rectangle its like
1 long line
so i dont really know
im gonna split it in have and just assume its 5
i think the sqaure is 40, idk rectangle, and triangle 30?
terse warren
remember that the area of a triangle is (basis * height)/2
swift night
so 15 then
terse warren
i have to divided both then divide?
nono, its (basis+height)/2
sturdy viper
( P ∨ Q ∨ ( ¬ P ∧ ¬ Q ∧ R)) ⇔ P ∨ Q ∨ R prove that they are logicaally equivalent
please help
terse warren
( P ∨ Q ∨ ( ¬ P ∧ ¬ Q ∧ R)) ⇔ P ∨ Q ∨ R prove that they are logicaally equivalen...
Please use a not ocuppied channel
sturdy viper
how do yk if its not occupied?
vocal sleetBOT
( P ∨ Q ∨ ( ¬ P ∧ ¬ Q ∧ R)) ⇔ P ∨ Q ∨ R prove that they are logicaally equivalen...
Please read #❓how-to-get-help
swift night
nono, its (basis+height)/2
so 3 + 10 = 13 divided by 2?
terse warren
no, because i was trying to say that 3 foot is a SEGMENT of that side not the entire side
swift night
ohhhh
terse warren
look, you identify the right parallelograms
now can you tell me the length of the colored lines ?
swift night
which ones red yellow or green or all
terse warren
all}
the red ones will help you to calculate the area of the first rectangle, the yellow ones with the second rectangle and finally the green ones will help you to calculate the area of the triangle
swift night
the red say 8 and 5
yellow are 6 and 8 (i say 6 bc on the bottom it is 6 so im pretty sure they are equivalent)
green are 6 and 10 (i say 6 because the 3 doesnt cover the full thing so if i doubled the line it would cover)
terse warren
Almost right but be extremly cautious with this "i say 6 because the 3 doesnt cover the full thing so if i doubled the line it would cover"
swift night
hm
terse warren
the image can be missleading, the orange line may looks like its the same length that the purple one
swift night
it looks like 1 extra
terse warren
But you have to infer it only with calculations, this is a extremely common math trap
swift night
But you have to infer it only with calculations, this is a extremely common math...
how do i do that
terse warren
what i'm trying to say, do not relie on the scale of the figure
only the numbers
like this
swift night
ok
but i’m still confused on how to find out how long the line would be bc i don’t see anything equivalent to it on the shape
terse warren
did you see it right? the length of the pink one?
swift night
so it’s the whole line on the other one 13 feet?
swift night
And it only shows the 10 feet which means the missing piece is 3 feet
swift night
so yellow + pink = ?
13 feet
terse warren
EXACTLY
now back, to the original figure
you said to me that the red ones are, 5 and 8, the yellow one 8 and 6, and the green ones are x and 10
what is x equal to ?
using the previous information
swift night
x = 6
terse warren
correct
now you have that the first rectangle (the red one) have basis=8 feet and height of 5 so the area is ?
swift night
8 x 5 = 40
swift night
the yellow is 48
swift night
dont i have to divide it by 2 after i multiply it u said
terse warren
yes only the triangles tho
swift night
so 6 x 10 = 60 divided by 2 = 30
terse warren
correct, now 40+48+30 its the total area
terse warren
there you are
swift night
it was correct
the was only the first question so tis prolly gonna get more complex later on
im gonna use the info u helped me get n try to use it to solve later problems
ill message again if i need help or if i finish
thank u!
terse warren
not problem, good luck on your remaining questions.
swift night
ty
terse warren
.close
vocal sleetBOT
Available help channel!
dull scaffold
Help plssss
dull scaffold
<@&286206848099549185>
gilded copper
u just need to combine number of boys and girls and the 1 adult is included and use the calculator for combination the following formula nCr
dull scaffold
I don't understand why it's 7C4
vast shale
<@&286206848099549185>
its (mc ^ 2 * pi / e) / i
dull scaffold
its (mc ^ 2 * pi / e) / i
Wtf
vast shale
what
vast shale
what language
dull scaffold
its (mc ^ 2 * pi / e) / i
This
west raptor
its (mc ^ 2 * pi / e) / i
Is that even relevant?
vast shale
Is that even relevant?
idk
gilded copper
I don't understand why it's 7C4
it's 7C4 cuz we have already 1 adult chosen so we should count only boys and girls and remove 1 from 5 of people cuz 1 adult is already selected
west raptor
Shouldn't it be 5 times 7C4 because there's 5 adults?
opal ridge
I need to find the number of 6-digit numbers where the sum of the first three digits is equal to the sum of the last three digits, but I don't know where to start
vocal sleetBOT
I need to find the number of 6-digit numbers where the sum of the first three di...
opal ridge
a+b+c=d+e+f and their sum should be in the set of numbers {0,1,2,...,27}
rugged orchid
a+b+c=d+e+f and their sum should be in the set of numbers {0,1,2,...,27}
That restriction doesn’t work
opal ridge
Why
rugged orchid
You want to restrict each letter to 0->9
If you use this restriction instead you’ll automatically get the max of 27 restriction
opal ridge
from (0,0,0),(0,0,1),(0,1,0),(1,0,0) combinations to (9,9,9)
that was my initial idea
rugged orchid
a+b+c=d+e+f and their sum should be in the set of numbers {0,1,2,...,27}
Yes I understand but there’s no point doing this
There’s no point restricting the sum in this way
opal ridge
okay
rugged orchid
Since it doesn’t do the “letters must be 0->9” part
But if you do this restriction it automatically also applies “sum mustn’t exceed 27”
Now we pretty much have 2 sets to play with
We just want to iterate through each possible sum and choose 2 ways to construct such a sum
Then do the permutation stuff for different numbers
Then remember you can’t have 0’s at the front
rugged orchid
So 6 digits or less?
rugged orchid
Right ok
opal ridge
because the original question is to look at arrays of length 6, and to answer if we have more arrays that the sum of all digits is 27 or that the sum of first three digits is equal to the sum of last three
and the first answer is binomial coefficient 6 + 27 - 1 choose 27
and i'm trying to solve the other part now
it's too much, i tried writing the numbers to try and see some kind of logic but just from 0,0,0 to 1,1,1 we have like 38 combinations so that won't work
or more
maybe i missed some
.close
pale siren
vocal sleetBOT
pale siren
Could someone please give me a hand on this
vocal sleetBOT
@pale siren Has your question been resolved?
pale siren
<@&286206848099549185>
vocal sleetBOT
@pale siren Has your question been resolved?
warm prawn
Hello, I need help with some trig. Here is the question I am struggling with right now.
vocal sleetBOT
Hello, I need help with some trig. Here is the question I am struggling with rig...
warm prawn
ornate ember
hi there, what have you tried so far?
warm prawn
So I have set it up using the difference identity but my answer is not matching up with any of the choices.
I tried to first find cos(t) and arrive at radical15/4
And I found sin(s) to be 2radical2/3.
ornate ember
so you have the right idea, but those values aren't right
let's start with sin(s) since that's in Q1
how did you get to that answer?
ornate ember
okay so I think I see what's going on.
So remember SOH CAH TOA: sin = opp/hyp, cos = adj/hyp, tan = opp/adj.
in other words, these are our lengths to a right triangle!
give me one second I'm drawing up something
forgive the crude drawing lol
so how would we find x here?
this triangle represents sin(theta) = 1/3
ornate ember
yep exactly!
so in other words: 3^2 = 1^2 + x^2. So just solve for x 🙂
and you can throw away the negative answer since it won't apply
ornate ember
bingo!
so then we get that sin will be (2sqrt(2))/3
oh wait
damn my b I misread
warm prawn
lol
ornate ember
yeah that's what you got
ok sorry lol
the only one that's incorrect then is cos(t), and that's a sign issue
remember that in q2, only sin is positive
which implies that cos is negative
(so it tan)
warm prawn
It came out to -.67080947
ornate ember
oh this part you want to give the exact answer in the form they're asking above
aka
$sin(s - t) = sin(s)cos(t) - sin(t)cos(s)$
$\\ = \frac{2\sqrt{2}}{3}\frac{1}{3} - \frac{1}{4}\frac{\sqrt{15}}{4}$
twin meteorBOT
MellowDramaLlama
ornate ember
then simplify
warm prawn
I see what I did wrong, thank you!
ornate ember
yep np!
pallid forge
$sin(s - t) = sin(s)cos(t) - sin(t)cos(s)$
$\\\\ = \frac{2\sqrt{2}}{3}\frac{1}{3...
shouldn't sin(t)cos(s) be 1/3*1/4
ornate ember
shouldn't sin(t)cos(s) be 1/3*1/4
oh yeah you're right. Typo on my end, my b
ornate ember
and sin(s)cos(t) should be the radical factors
yeah I just goofed up. Latex is wonky sometimes lol
pallid forge
yea i do the same. just wanted to point it out
ornate ember
np, thanks for the catch 🙂
warm prawn
Awesome, I appreciate the help.
ornate ember
$= \frac{2\sqrt{2}}{3}\frac{-\sqrt{15}}{5} - \frac{1}{4}\frac{1}{3}$
warm prawn
.close
vast shale
$log_{75}(x^2)$
vocal sleetBOT
$log_{75}(x^2)$
twin meteorBOT
fdasghwwds
vast shale
so far I have $\frac{2b}{2 + log_5 3}$
twin meteorBOT
fdasghwwds
vast shale
but i don't know how to change $log_5 3$ in terms of a and b
twin meteorBOT
fdasghwwds
vocal sleetBOT
@vast shale Has your question been resolved?
river kettle
What does OR mean in probability I keep getting it mixed up with and
river kettle
like if I have numbers 1 to 20 and I have to find the probability of a number being divisible by 3 OR being less than 10
How would I go about that
pallid zenith
usually means non-exclusive
the problem is that you tend to double count
so youd use inclusion-exclusion
pallid zenith
How would I go about that
consider you have a group of people
you wanna know how many people are male, OR named sam
so you go through and count all the people named sam
and you go through and count all the males
problem is, if a male is named sam, you double counted them
so you subtract the intersection of the two
river kettle
problem is, if a male is named sam, you double counted them
So in this case I would add the number of males and number of people named Sam and subtract the number of people who are both male and named Sam
pallid zenith
So in this case I would add the number of males and number of people named Sam a...
right
river kettle
like if I have numbers 1 to 20 and I have to find the probability of a number be...
But wait
For numbers divisible by 3
Does that also refer to those numbers being less than 10
Or applies to 1-20 in general
round plover
river kettle
There are 6 numbers divisible by 3 from 1-20 in that case
And 9 numbers less than 10
So do I do 6 + 9 - 3
-3 from the numbers that are both divisible by 3 and less than 10
vocal sleetBOT
@river kettle Has your question been resolved?
vocal sleetBOT
vocal sleetBOT
Available help channel!
tidal trail
f'(2) = 2
g'(2) = 2
(fg)' (2)= ?