#help-27
1 messages · Page 301 of 1
so u js told me to do it but it was wrong
It has nothing at all to do with part b
Well yeah, it's how to check your answer to part a
ON your exam
It's how you prevent silly mistakes
ikik
would that not just equal 2.4
just by looking at it
ur multiplying everything by 0 exept for 2.4
Exactly
alr perfect
and as for the max height the baloon reaches
i can take my too zeroes, add them and divide by
2
to get axis of symetry
then plug that into my equation?
Yep, that's one way of doing it
shorter way?
Because quadratics are symmetric equations
It won't work in general but it works for quadratics
General way involves calculus
nah bro its only avaialbe in gr12
It's nothing to do with how simple they are, it's just a special case for quadratics
It doesn't matter because you're dealing with quadratics
i need advanced functions, calculus, chemistry, physics and english to become an architectural engineer
and im trying to get into waterloo with a 5-15% aceptance rate
no bro
ill get it marked wrong
its fine
maybe next year when im doing calculus
oh wait
yo
@toxic grove
can i not just complete the square to get it into vertex form then take the c vaalue and thats my maximum
That's another way of doing it
im mad creative
on sum real shit
do u have the answer
im about to say it and i want u to tell me if its right
Yes I've had it written down for nearly an hour now
No
Yes
plug it in
So what's h(2.7)?
62.07
No
You did the same thing as last time
Exactl ythe same thing
You forgot the square
It's vitally ipmortant
Much better
alright bro
its bed time ngl i got an exam 9am tmr
gota say bro w out u ion think id know shit for these last couple units
Good luck then
preciate u bare man i added u if u ever tryna tutor me for sum cash or smth id appreciate it
u rlly helped on sum real shit
3 straight hours
word bro
Also thx
Godspeed
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how to solve ii
i have factorized it already
x^3-y^3?
how do i proceed
multiply numerator and denominator with (sqrt(7)+sqrt(5))
why
For that crazy fraction, let x=7^1/2 and y=5^1/2
because the denominator is (sqrt7 - sqrt5)
or to put it differently, say the denominator is a-b
if you multiply num and denom with a+b
you get in the denominator:
(a-b)*(a+b) = a²-b²
and since in our case a=sqrt(7) and b=sqrt(5)
you can easily just do a²=7 and b²=5
Then you can simplify using that formula like how the problem want you to do
oh ok lemme try
what about the power dont we need to consider that
I think op already did the factorization and cancellation?
yes i did factorize it
oh did I miss it
lol
Yes, but @strange arch is suggesting rationalizing the denom for some reason
If you're done factorizing and cancelling, you're only left to evaluate (√7)² + (√7)(√5) + (√5)² are you not?
that's just 12 + √35 ?
oh so i just sub the values like @graceful stone told
because this doesn't look like the factorized version
result needs to have form c+dsqrt(35)
my school suggested this solution but idk
that's the same thing
ah right top contains the factor @graceful stone makes sense
You factorized, you cancelled the (x - y), and proceed to evaluate (x² + xy + y²)
is it not the same thing?
Pretty much
The problem was dropping hints about factoring so i was trying to suggest that
But rationalizing the denom should give the same answer
wait so can 1 person type out the easiest method and explain it
cuz im confused with so much stuff here
$\frac{x^3 - y^3}{x-y} = x^2 + xy + y^2 = (\sqrt{7})^2 + (\sqrt{7})(\sqrt{5}) + (\sqrt{5})^2 = 12 + \sqrt{35}$
ok ig also the question says $\frac{x^3/2 - y^3/2}{x^1/2-y^1/2}
So @slender mirage is taking $x=7^{1/2}$ and $y=5^{1/2}$
Cris
||why ping
||
To give credit sry
then it should be 0.5 right?
huhhhh
im actually so confused could you please explain it simpler
yes
Similar if y=5^1/2 then y^3=5^3/2
yes
So you can rewrite the original fraction as $\frac{x^3-y^3}{x-y}$
Cris
And follow this solution
Np
should we leave the help channel open just in case? :]
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So say we have 5 points on one side, 7 points on another and 6 points on another
The amount of triangles we can make is 5x6x7 right
How is that different from adding 5+6+7=18 and taking the combination of that
Aka 18x17x16/6
816
no
?
Yes
so to explain why the calc isnt 5x6x7
Sorry about the delay discord mobile app for some reason is shitting itself
Yes
right
so while theres probs a few ways to do it
i'd do it using casework
so split it into calculations for each case then add them up all together
- each point is on a seperate side
So take one point on the 5th side
As a case
- one point is on a side while the other 2 points is on a second side
And use the other two sides as two points??
Ohhh
Ah i get ot
Btw i gotta ask
Do you have any book recommendations/class recs for permutation and combination and stuff
This case would be 5x6x7 right?
yeah
Wait would there be double counting
wdym
Ok nah we're good
Like if a,b,c is a triangle, b,a,c is same triangle
So we'd have to divide by the number of those repeated counts
Doesnt seem like thats a prob here tho
na should be all good
Ye
then for the next case
In this case we have to split it in to two more cases I think
if we pick the side with 5 points first
Take 1 point on 5 side as starting
yep
Take the point with 7 side
then you can pick any of the remaining 13 poinst
Pick another point
I was gonna do it like
Pick a another side
In this case 7
Pick a point on there
6 points remaining
So 5x7x6 for that side
Then pick the 6 point side
Take one point, 5 points remaining
5x6x5
Add those two cases together?
This would include both the 7point and 6 points and we'd end up with the prev case coming me thinks
wait
im gonna label it
for case 2, one from one side and the other 2 from another
Ye
2 from AB) 5C2 * 13
thats what i was trying to get to sorry
you pick one side
and then you pick the 2 from that side
to prevent the double counting
2 from BC) 6C2 * 12
2 from AC) 7C2 * 11
Im ngl i havent studied combinations yet 💀
the C just represents choose
Sorry
Will this case have double counting?
Doesnt seem like it
Wait cant we just group the 6 point and 7 point side together as 13 points
ya
Take one point from the 5 side
but thats the one that you'd pick one from
no because then you'll
double count
itll clash with choosing 1 point from each side
because from the 13 remaining points
theyre not all on the same side
if you wanted to group the rest
you'd have to only pick 1 from the 13
and pick 2 on the same side
Huh
which is essentially this
cuz
Lemme think
say you pick a point from the side with 5 points right
as you said, you'll have 13 points remaining
Yes
our first case was that
On the other side yes
each point is on seperate sides
your suggesting to pick 2 points from the remaining 13 correct?
Yep
Regardless of seperate side
if you pick 2 that aren't on the same side
then you end up picking 1 from each side
which clashes with your first case
No im uh saying cant we do the whole thing via that one operation without splitting it into cases
there is another way to do it, which is to find the total combinations then subtract the cases where the points chosen are all on the same line
oh without the first case
It doesn't it would to be too large
From what i think the 5x13x12 should double count prob
You have to find the multiplier for each angle
So we need to divide it by 6
yeah 5 x 13 x 12 is 780
Yes
too high
But u need the multiplier
Sine if we have 3 points a,b,c they can be arranged in 6 ways
So there would be 6 times repeated count
No
So just divide by 6?
Its not
i'm getting 751 i think 780/6 is too low
Damn
no
i think if you do it that way
Wait wait
youll have to also make an operation where you do 6
I dont think we need to divide by six
and one from 7
We are only selecting 1 point from 5 in the first place
So 5 wouldnt factor into the repeated count
So itd be 780/2 maybe?
Oh right
because your assuming you start with the 5
That 5 thing just takes from the 5 side
when you can also start with the 6 and the 7
And doesnt consider just 6 and 7
yep
The angle doesn't contain just 6 and 7
What
How?
Or where it repeats
Uh
Can u show where it repeats?
Normally you have a rule to know these Angles
You need to find the rule
What the fuck do angles have to do with this
....
Where is the question?
Huh?
Find the matrices
This
Is this guy trolling or what
if you study combinations this question becomes much easier
i cant think of a way to do it without it
Oh
its definitely possible i just can't work it out
Well im trying to do this question
I just made that other question up to get a base
That question doesnt use combinations
717?
Its 717 yes
Howd you get it
Consider a rectangle ABCD having 5,7,6,9 points in the interior of the line segments AB, CD, BC, DA respectively. Let a be the number of triangles having these points from different sides as vertices and B be the number of quadrilaterals having these points from different sides as verticies then (B -a) is equal to:
-
795
-
1173
-
1890
-
717
It is 4, 717
What
you can calculate alpha
Ok bruh
by
so you have the sides AB BC CD and DA yes
all the possible combinations is just
AB * BC * CD
AB * CD * DA
AB * BC * DA
CD * BC * DA
wait
i think that's right
so
ab has 5, bc has 7, cd has 6 and da has 9 points
so
yes
Ok uh mind if i ask one more question
Yes
Got it
beta is just AB * BC * CD* DA = 1890
Correct
then subtract ya
I assume for the rectangle we just do the other
Yes
yep
Okay thanks
Correct
Mind if i ask another question
Yes go ahead
i can try ya
90
Idk where to start on that one
Is it 90?
hang on
u need to choose f(1) and f(2)
so if the set is {1, 2, 3, 4, 5, 6}
6 choicse for f(1) and 6 choices for f(2)
not sure if i'm going the right way with this
The total numbers of functions f : {1,2,3,4} > {1,2,3,4,5,6} such that f(1) + f(2) = f(3), is equal to
-
60
-
90
-
108
-
126
ERM, 90
How do you get 6 choices
Oh is it because of the range?
the codomain is 1,2,3,4,5,6
If this is the question
so the possible outputs is 6
Got it so
so the total combinations is 36 for f(1) and f(2)
f(3) has 6 values it can take
yep
Wait no
5
F(3) cant take the value 1
Cause theres no zero in domain
The minimum is 2
1+1
i was more thinking to count the solns such that f(1) + f(2) is smaller or equal to 6
Yeah but see
f(3) can take values 2,4,5,6
So if f(3)=2 then that means f(1)=f(2)=1
are u a learner like me
Hmmm
yes that's correct
Yes
My bad
Wouldnt that just be one comb
yep
which is theyre both 1
for f(1) + f(2) = 3, you have 2 combinations
which is (1,2) and (2,1)
Hm no
Uh dude
The solution says theyre 6cases
Which is where im stumped
f(3)=2,(f(1),f(2))→(1,1)→6 cases
Yes
Thats what the solution says
Its right
Not sure where tf that came from
Any idea?
I can help you learn but you need to take the structure
no im not too sure with that sorry, if @restive river understands then they can help out
That dude is just trolling lol
But i think we kinda screwed up
I'm not
Because uh
f(1) should also have the codomain value from 1 to 6
Same with f(2)
Its only f(3) that doesnt have the value 1
Oh shit dude
I think i got it
So we have a singular case for f(3) right
f(1)=f(2)=1
mhm
Im thinking that f(1) can have 4 values that give 1
Since the domain is 1,2,3,4
Wait no its already specified which number we take thats 1
Okay bruh
@lilac mango should i close this or leave it open for help
not rlly sure where to go with this, leave it open for help and if you still cant get it ping helpers role
Aight got it
Tysm for your help so far
Uh thanks for your help?
The question is above
....
How the fuck they got 90
Whats the point of knowing the answer if you dont know how they got it
@toxic spire Has your question been resolved?
@toxic spire Has your question been resolved?
So uh any new help
@toxic spire Has your question been resolved?
@toxic spire Has your question been resolved?
just check for f(1) = 1
then f(1) = 2
and so on
posting again so dont have to scroll
and f(4) has no constraints on it so it can be any of the 6 values in the co-domain
rough idea
so that f(3) = f(1) + f(2) doesn't go beyond 6
@toxic spire
Yes i got that
But how does that help with the question
Because the question has the relation f(1)+f(2)=f(3)
f(4) isnt a part of that
check this
you still need to map 4 to something in the co-domain
you have 6 options
so 15 * 6
Yeah i got that
Ugh
It still isnt clicking
Why are we using that as the conditions for it
Yes
and you still have to map 4
Yep
Why are we multiplying it by the values of f(1) tho
Sorry not the values
The amount of options f(1) has
What does that have to do with f(4)
Wait
write down all possible funcs for f(1) = 5
Since it has to satisfy the f(1)+f(2=f(3) thing
its not much
no
Wut
do this, you'll see
f(1) = 5 then f(2) can only be 1
right?
Yes
so f(3) is 6
Yes because every domain element needs a codomain ye
1,2,3,4,5,6 ye
so what's the doubt?
Why are we multiplying six by the numbers of options of f(1)?
Here f(3) can be 2,3,4,5,6
f(3)=2,(f(1),f(2))→(1,1)→6 cases
f(3)=3,(f(1),f(2))→(1,2),(2,1)
→2×6=12 cases
Here
We are multiplying the 2 cases of f(3)=3 which is (1,2) (2,1) by 6
To get 12
Why do we need to do that dont we already have which cases satisfy the relation
f(4) can be any of 1 to 6 though
1,2,3,(1 to 6)
and
2,1,3,(1 to 6)
Yes so f(4) has 6 options i get that, but what does that have to do with us finding which cases satisfy f(1)+f(2)=f(3)
it doesn't have anything to do when finding the cases
but you still haveto consider f4 when finding how many fns there are
Then why are we multiplying the number of cases we get by 6
.
Cant a function only have 1 value
does this help? 😅
Wait so is it like when the relation f(1)+f(2)=f(3) is satisfied f(4) can take any of the 6 values?
Oooh
AH
I think i get it
So if like
We take it in a bracket(just for clarity) as {f(1),f(2),f(3),f(4)}
In the case of f(3)=2, then f(1) and f(2) will both be 1
And f(3) is 2
yeah
yess
Thank you so much man
no prob
haha, happens
Got it
^
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Hello, I'm confused how adding "b" to both sides equals 5/3b can anybody provide insight?
$$\frac{2}{3} + 1 = \frac{2}{3} + \frac{3}{3} = \frac{2 + 3}{3} = \frac{5}{3}$$
Alberto Z.
If this is not clear, then I strongly suggest you back up a little bit and revise basic algebra
Because it's fundamental for understanding equations with letters appearing besides numbers
got it thank you! writing it out makes it clear
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sketch
I need final speed on contact and time of position being 0
I wrote the function as
f(y) = 39 + 21t - 0.5gt^2
where g is the magnitude of acceleration due to gravity, 9.80 m/s^2
ok I got it
I messed up with my function in my calculator

and the program didn't want the velocity as negative
(I put 32 instead of 39 somehow)
But it didn't specify that it was the magnitude only :(
Thank you for coming to help :)
.close
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Trying to figure out where/how I need to start to get #17 done (finding the value of a,b,c,d)
Ignore #16
@arctic mist Has your question been resolved?
<@&286206848099549185>
start with a: 3/6 = 9/(9 + a)
then angle bisector theorem: 5/7.5 = (6 + 6 + c)/(9 + a + 4.5) gives c
then d/5 = 6/(6 + 6 + c) gives d
and lastly, I hope you can find b
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why are they using synthetic division if you can factor it normally
you can't perform partial fraction decomposition unless the degree of the denominator is higher than the degree of the numerator
so, you have to perform the polynomial division to fix that first
or equal?
no, it has to be strictly higher
Well you're doing partial fractions, that term isn't part of the fraction anymore
the 2x is pressent on both sides of the equation, so you can ignore it for the purposes of solving for A and B
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i have no idea how to do this on a calculator
,rotate
thanks
First is to rearrange the equation in terms of x
okay how
i apologize my skills with the calculator are equivalent to an average grandma on a nintendo switch
Basically x = arccos (2/5)
got it
Now in your calculator you may have arccos or cos^(-1)
it does yes
What did you get?
66.4218
,calc (2/5)
Result:
0.4
,calc arccos (2/5)
The following error occured while calculating:
Error: Undefined function arccos
Maybe it's set in degrees ?
you got it in degrees insted of radians
Instead of radians?
i’m sorry i have to go
yes
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Hi, could anyone enlight me what characters should I use if I wanna denote coefficients or weights in an expression? Thanks
This may depend entirely on the context of the problem and conventions persuant to the topic
But in general, variables can be represented with whatever you'd like as long as you communicate what they represent
The standard form linear equation is generally Ax+By=C, but A, B, and C are only chosen because they're the first three letters in the English alphabet and generally easy to remember
The equation itself could just as easily be represented by Wx+θy=ξ and mean the same thing, as long as someone working with it understands what W, theta, and epsilon mean
If not using characters at the head of the alphabetic order, could you suggest some characters that are used the most?
Any particular subject?
Bcoz I saw ppl using x, y in italy style.
Oh yes the subject
I'm doing a financial cost expression
Algebra tends to use x, y, and z for coordinates/experimental variables and i, k, m, and n for some formulas
Probability and stats use n, p, q, and some others
by adding weighted components, I need some characters to denote the weights
should they be in italic style?
They don't need to be
It helps if x is if it's a more rudimentary field of math because younger students may confuse "x" for a multiplication symbol
But letters in formulas are generally understood to be variables
Subscript is used a lot to denote distinct variables of the same type
For example, the slope formula is $m = \frac{y_2 - y_1}{x_2 - x_1}$, where $x_1$ and $x_2$ are both x-component coordinates, but are different from each other (same for y)
al-jebruh
Ah the subscript is so good. Also, if i'm hoping to do this financial expression in a linear regression style, which means the weights would be used as coefficients, is there suggestions abt which characters to use instead?
Nevermind, I think I just need to follow your explanation above 🙂
Weights afaik are usually just multuplied to variables
you are right.
Thanks for the help, I think I'm much clearer abt the character usages.
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Here is my attempted solution for the question: Suppose f and g are boh functions that are uniformly continuous on the set A and both f and g are bounded on A. Show that the product of the 2 functions fg is also uniformly continuous on A. Is my attempt correct?
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!status
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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
Apply trigonometry.
So for ∆RQM, you get
RQ/MQ = cos(60°)
Or RQ = MQ/2
So, RQ is half of MQ.
Now, look at ∆PQM, you get the angle(MQP) as 15° because angle(PQR) is 75° and angle(MQR) is 60°.
Now let's look at ∆PQR, it's a right angled triangle, so angle(RPQ) should be 15 degree.
So from this we get that PM = MQ = 18.
And hence, RQ = 9
thank you but we did not learn any of that yet
we only learned basic geometry
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What's wrong here
it is not clear ma dude
It split the integral
And I'm supposed to get alpha as 30 and beta like -25
Here i got 10 and -12
@remote charm Has your question been resolved?
$\int_{n-1}^n \frac{{x} + 2[x]}{e^{{x}}} = \int_{n-1}^n \frac{{x}}{e^{{x}}} + \int_{n-1}^n \frac{2(n-1)}{e^{{x}}}$
Left side is just (e - 2)/e, Right side is 2(n - 1)(e - 1)/e
Yes I saw
Bruh silly mistake
Add them and you should get: 5(e - 2)/e + 20(e - 1)/e
I took x for [x]
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In our math book it is claimed that: "If the function 𝑓 has 3 roots, the inequality ax^3+4x^2+x>0 cannot have a solution for x>0 because in that case, the values of the function 𝑓 are necessarily positive in two intervals."
How can one intuitively determine the number of positive and negative intervals in a polynomial function of degree n? I kinda figured out it was true after considering the roots of the derivative. I realized that the roots of the first derivative aren't enough for the "direction changes" required for the function to only have a single positive interval. The textbook, on the other hand, doesn't offer a reasoning process. it just states this as a fact.
So my question remains: How can one intuitively determine the number of positive and negative intervals in a polynomial function of degree n?
@shut tendon Has your question been resolved?
<@&286206848099549185>
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$(P \land Q) \lor (P \land \neg Q)$
What a wonderful world!
yes
Thanks
P
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The annual salaries of six employees of a company are as follows
$22,000, $35000, $22,000, $46,000, $57,000, $90000
get the standard dev
is my calculation correct im doubting my calculator
@distant wave Has your question been resolved?
Result:
272
,calc 272/6
Result:
45.333333333333
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560
236
236
340
590
248
248
360
620
260
260
380
655
275
275
400
690
290
290
420
725
305
305
440
760
320
320
460
795
335
335
480
835
352
352
510
875
369
369
540
915
386
386
570
955
403
403
600
995
420
420
630
1040
439
439
670
what is the pattern scheme for these numbers
what kind of pattern do you desire
i cant find anything in oeis but you could definitely do a polynomial interpolation
Look at it in sets of 4
i think that
There's a resemblance that continues till 4 sets of 4
it just says pattern scheme to me too
Then the middle number starts incrementing by 15 instead of 12 and again after a few sets it increases by 7 !
hm let me s
wdym by incrementing?
have u tried yet
polynomial interpolation? hell nah i aint interpolating like 70 numbers
what do y think the pattern is by looking at the first few terms
i agree with what xor said that you can group it into 4 and then theres a pattern
ty
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Y'all could someone check my integration?
err ignore the last step on the second page, should pi/2 instead of pi
<@&286206848099549185>
anyone???
neonnnn!
jee 25?
i am the kraken guy..
@rugged sparrow
<@&286206848099549185>
<@&286206848099549185>
Okay so I messed up somewhere
Because my integrals finally diverged xd
Just a sec. What is it that I should be seeing?
@finite briar Has your question been resolved?
check 2nd page its in reverse order
Holy.. what are these 
Okay wait am stupid..why am I evaluating Jk at 0
Mse integral xd
I:ll send link to u on dm if u want tmrw
I don't have it on phone
cannot , the order is not very well organized and readable :c at least for me
perhaps someone knowledgable might be able to differentiate between the steps better
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I'd like to check my work
ty!
working on another one like that rn)
i got this for the first two but im not sure if thats correct?
it's one of those im not sure which one
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True or false? If the slope of the line is -3, then the line goes down 1 and right 3
I thought i had a grasp on simple algebra but apparently not
what does the "the line goes down 1 and right 3" mean??
quite the opposite.
If the slope of a line is -3, then the line goes down 3 and right 1.
what you have just described is a line of slope -1/3.
a kind of "interval of steps".
here, let me show you.
recall that $\text{slope}=\frac{y_2-y_1}{x_2-x_1}$, where $(x_1,y_1)$ and $(x_2,y_2)$ are points on the line.
;(
for some reason I was thinking it had something to do with transformation which made zero sense

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Slightly confused on what is happening here>
Why do we have a (-1/x) and why do in the second step there is an extra negative
The -1/x is just from differentiating ln
But why is that -1/x instead of positive?
I think it's going the chain rule so
dy/du is -1/x
du/dx is d/dx (-x)
But then is the u just -x?
The derivative of ln(x) is 1/x as you probably know. It just takes the reciprocal of whatever function is inside
so d/dx(ln(-x)) = 1/(-x)
Yeah but why would 1/(-x) = 1/x in this case?
Ok nvm it's the second step
Right remember you need to apply the chain rule
and the derivative of -x is -1
so there's an extra negative
Ok yeah I got the hang of it
Remind me again though, is the u just -1?
So g(u) where u=f(x)
In(u) where u = -x ?
Yes
So dy/du becomes 1/(-x) and du/dx becomes -1
OH ok
Yeah great make sense now
Thanks

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hi, how does radiation work? is it like invisible fire or smthing?
seems beyond the scope of a math server, have you tried the physics server? see #old-network
I dunno maybe it is
ill ask there, thanks
Its the emission and propagation of energy in the form of waves or particles
Usually by photon
Sometime other particles
But mainly photon
that's not what they meant
huh?
people mean ionising radiation
are this
invisible fire is probably how they explained it back in the 1400s
you;re explaining "invisible light" but they could mean nuclear decay type of thing
but it's not clear which one maybe you guessed right
it's invisible fire, it happens to the atoms, so you can't stop it, but it gets more intense if you bring it in a pile
just from the amount of material close together
ok i continued this in the physics server so i didnt check in on this channel
im watching chernobyl series from hbo
apparently electricity was referred to as the "subtle fire" well into the 1700s
and i dont understand how radiation works, so i just want to learn out of curiousity
Its just usually photon which high energy
Shoot the eletron of the atom
Which excitize it, and ionize it
Death basically (if ionize enough)
ah wow interesting
some atoms are unstable, they have a chance to blow up every moment, it's literally random
one example is potassium-40, the stuff in bananas
if you put alot of it close together it intensifies like i said
nuclear batteries work like that
it's just a tiny thing that's constantly hot, you use heat to make electricity, in space probes and rovers and stuff like that
apart from heat it shoots ionizing radiation, in 3 types
Anyone has those old ceramic like plate and bowl with paint made from uranium compounds
makes sense, yeah this is to my understanding basically how nuclear plants work as well, but im more curious about how it affects us and other things, like it can burn us, it can destroy our cells from within, it penetrates most things, it gets carried in the wind and clouds, it contaminates things like clothes and metals, how does all that work?
it's like microwave oven type of thing, but for some reason it gets absorbed by dna at low energy
like if you point a microwave at yourself you will die from burns before you get dna damage
but not with this thing
the thing that gets carried and contaminates is the dust, the actual chemicals
the radiation can't do that
