#help-27
1 messages · Page 251 of 1
Its a derivative
^
oh ok
how long did it take you to memorize these all, it seems like a lot
or just do questions by looking at the sheet and eventually it will come?
I memorised all of the identities, inverses, and hyperbolic derivatives and integrals yesterday
I knew the others from before
what degree are you doing?
ok so du = 1 / sinx^2 dx
so du sinx^2 dx
that leaves e^u
so e^(-cotx)
Did you solve the integral?
+C
What work have you done so far
You need to know a half angle identity
do you think Pre Calculus Trigonometry or Calculus 1 re do would help me know all this
Trigonometry would, but you wouldn't do derivatives in that course
ok so while doing Calculus 2 work just always have the trig identities and integrals formulas out to look at?
I would say yes but you need to try and remember them without a sheet
Do some basic problems with a sheet but otherwise try to memorise them
yea we do get a sheet for tests, that's whats tough
I don't remember any of this, there is so much
you don't think Calculus 1 would help me with any of this? other than the derivatives at the end? and you don't think I need to re do Pre Calculus either just the Trigonometry identities?
what half angle identity to use?
@atomic dome what half angle identity should I use?
@atomic dome I only see when it's divided by 2
can someone help me with logic math?
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ok guys, when it comes to graphing y= rise/run or slope (x) + b in which b is the y intercept,
does that mean x= run/rise(y)-b in which b is x intercept?
i hate how they never use x= in a graph and suddenly show us this shit in a question omg
yeah, so if you go rearrange $y = mx + b$
south
you do get $\frac{y - b}{m} = x$
south
so yeah that's x = 1/m * (y - b) or (run/rise) (y - b), your brackets are a bit off
we can also write it as x = (run/rise)y + c, where c is the x-intercept
lol do you have a question
but just to reiterate, all 4 of these statements say the same thing
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"Knowing the magnitude of a vector one could abstract only the circle in which the vector belongs."
i know this is more a grammar question
but does anyone know if they mean circle as in an actual circle
or
circle as in like radius (not mathmematical)
@fading ledge Has your question been resolved?
A vector is a mathematical object with two properties: magnitude and direction. To my understanding, with magnitude only, then the vector could point in any direction starting at an origin point. Hopefully that answers your question.
oh
" vector unlike a scalar has direction, and it is broken into two components: magnitude
and angle"
so is angle here
meant to be direction??
yes
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why is A and B not compatible
what is B and what do you mean by compatible
like if A is invertible
why does Ax=0 only ave one solution
you can multiply on both sides by A^-1
@lament schooner Has your question been resolved?
huh?
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does T have to be isomorphism?
@lost crag Has your question been resolved?
(assuming context, holds for any linear map, isomorphism or not)
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How do I do 79
Chain rule
By applying it
What does chain rule formula say?
like if x^3 = 3x^2
This is power rule, not chain rule
Chain rule involves composition of (two) functions
oh yeah like if (g(x))^3= 3g(x)^2 * g'(x)
Yes
ok... how do i apply it to this i dont get it
If this is f(g(x)) In this case f(x) is x^3
Let’s think about what u did
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Pls
Do you know how to complete a square?
$1 - cosx = \sqrt{2\sin^2\frac{x}{2}}$
That question is long closed
Thanks for telling me
mari
@quasi vortex you wanna use complete the square.
Say
Y=x^2+ax+b
Then:
Y=(x+a/2)^2-(a/2)^2+b
You can easily verify this by expanding the second equation
Bro im too high to do all that ima just take the %80 i hot
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here i think opt b&c both are correct can u check it
@reef solstice there is no way both b and c can be correct
B scales with n^4 and C scales with n^3
Which values?
but in anc c is correct only
let n=1 or 2 or any int u get same value
What if you put in n = 10?
,w 10*191/2
@reef solstice are 2860 and 955 equal?
Here is a hint: $\sum_{k=1}^n k = n(n+1)/2$
OmnipotentEntity
And $\sum_{k=1}^n k(k+1)/2 = k(k+1)(k+2)/6$
OmnipotentEntity
@reef solstice ^ the above should help you prove this
@reef solstice Has your question been resolved?
i know this
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Hello I’m having trouble with these two
so I know that
I know how to solve it normally
but I assume they want us to like
use some method that makes it simple
using the fact that like 2 columns are linearly independent
hmm if two columns are linearly independet
that means they span the entire space
is [1, 1, 2, 3] are in the space
then I can just use those two columns as the coefficient matrix to find vector T that works
is that right
so I should be able to use A = {c1, c5}
and solve the corresponding system
or do we pick two linearly independent rows
for our coefficient matrix
For e I’d find the solution of the null space and add it to my unique solution?
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Is this correct?
import numpy as np
import sympy as sp
from matplotlib import pyplot as plt
def g(x):
return (np.cos(x) - 1) / x
L = np.arange(1, 18, 1)
base_seq = 0.5**L
xLeft=-base_seq
yLeft = g(xLeft)
print("Left Seq = ", xLeft)
print("g(Left Seq) = ", yLeft)
fig = plt.figure()
plt.plot(xLeft, yLeft, marker="o", color='red', label='Approaching from Left')
x = sp.Symbol('x')
g = (sp.cos(x) - 1) / x
limit_value = sp.limit(g, x, 0, dir='-')
print("The limit of g(x) as x approaches 0 from the left is: ", limit_value)
This is funnily enough for my math class🥲
I don’t know why we have this lab
I’m just hoping someone knows python as well🫠
also try asking in python discord server
this looks kinda like math
I did but they don’t seem to know what’s going on either
It’s calculus, we need to show limits using python
By graphing it
And they didn’t teach us anything they just threw us into it
but wouldnt it just suffice to write
print(np.cos(-0.0001) - 1) / -0.0001)``` or sth like this
What did problem 1 say? As in "look at the graph and what it looks like" or something?
Problem 1 is a whole mess we had to write it in sections which is so annoying because for each small change you have to run each block separately to make the next block run properly
Like this
And we had to show what number it’s approaching
I didn’t know how to do any of this btw, I just copy pasted the code from the example sheet
But I don’t know if that’s correct🥲
What am I reading
I don’t know what any of that means
<@&268886789983436800> spam
how is that spam
this is a help channel, you're not supposed to send random stuff in help channels
@vast helmCan you explain why you're talking about US GDP (unrelated to the ops question) when replying
oh sorry
Anyway🤣
ill leave
What can I use to know what I’m supposed to do here🥲
I don’t know how to even approach that
@amber merlin Has your question been resolved?
seems fine
but the range in np.arrange()
1 to 18
your points will be too cluttered
,calc 0.5^15
Result:
3.0517578125e-5
,calc 0.5^16
Result:
1.52587890625e-5
,calc 0.5^17
Result:
7.62939453125e-6
The guy there wrote that I wrote 10 but he changed it to 18 to show the point 0.00
To show that it’s approaching 0
When it was 10 it only showed 0.05
,calc 0.5^10
Result:
9.765625e-4
hmm
What is that😅
keep it 18 then
im not entirely sure if the yLeft thing works out or not
but just change it to g(xLeft)
I really don’t understand why they throw us into that, the professor said it’s so we can see how the graph behaves, but why aren’t we just using desmos
Should the graph below show if it’s correct or not?
It generated that graph when I ran the block
which graph
In this question
import numpy as np
import sympy as sp
from matplotlib import pyplot as plt
def g(x):
if x < 2:
return x**2 - x - 1.5
else:
return np.cos(x) + np.sin(x)
x_left = np.arange(1.5, 2, 0.01)
x_right = np.arange(2, 2.5, 0.01)
y_left = [g(x) for x in x_left]
y_right = [g(x) for x in x_right]
plt.plot(x_left, y_left, label='x < 2')
plt.plot(x_right, y_right, label='x >= 2')
here's the code for convinience haha
that's the assignment itself btw
seems right
Thank you! Gemini is pretty good at this it seems lol
u still have to check it tho
What do I look for when checking if it’s correct or not?
But they are all different graphs🥲
i can explain if you want
approach is same
Yes please, although I don’t think I have enough context to understand
Here for example you need to do a part b to problem 3, and you need to guess using code, what does it even mean to guess using code it just shows the answer nothing is guessing here
np.arrange(start, stop, step), the interval is [start, stop), this basically gives u a list with values within the given interval and the step u put
So like what number it starts what number it stops and the gap in between?
for example np.arange(1, 2, 0.1) will return [1. 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9]
u start at 1 (the start is included, stop isnt)
then add the step 0.1
so next value is 1.1
then again +0.1
1.2
and so on
until you hit the stop
I see, that one makes sense thank you😄
and the .plot(), well it just plots what you give
first argument is list of x values
second argument is list of y values
you can also give it label
for markers u can add the marker="o" argument
and for color, color="<color you want>"
the lines arent thinner than before but if u want to increase their width then u can add linewidth= argument
u have to use float btw
like 2.3
4.0
I pasted it from the previous problem and it gave me an error
dont add extra .plot() line
modify the one u have above it
these
plt.plot(x_left, y_left, marker="o", color="red", label="approaching from left")
yeah
Wth🤣
Where do I see that?
they are also evenly split
so you will have a bead garland type line
if you want it to be more like the first one, like the closer u get the more points increase
then use the exponential thing u did before
Inside the function or outside?
And why is desmos showing me this
It doesn’t look like the thing I just coded lol
anywhere works, but do it outside
import numpy as np
import sympy as sp
from matplotlib import pyplot as plt
def g(x):
if x < 2:
return x**2 - x - 1.5
else:
return np.cos(x) + np.sin(x)
L=np.arange(1,10,1)
base_seq=0.7**L
x_left = np.arange(1.5, 2, 0.01)
x_right = np.arange(2, 2.5, 0.01)
y_left = [g(x) for x in x_left]
y_right = [g(x) for x in x_right]
plt.plot(x_left, y_left, marker="o", color='red', label='x < 2')
plt.plot(x_right, y_right, label='x >= 2')
here
It’s the same as you have it but still looks different how🤣
This whole assignment is so frustrating 🫠
Or are you supposed to just zoom in? And focus on the point where it meets?
its not different
i was just very zoomed
Ohh I see, my bad🤣
I miss my previous professor he actually explained things
The graph that I coded doesn’t really show that they are approaching different values🥲
import numpy as np
import sympy as sp
from matplotlib import pyplot as plt
def g(x):
if x < 2:
return x**2 - x - 1.5
else:
return np.cos(x) + np.sin(x)
L=np.arange(1,10,1)
base_seq=0.7**L
x_left = 2-(0.7**L)
x_right = 2+(0.7**L)
y_left = [g(x) for x in x_left]
y_right = [g(x) for x in x_right]
plt.plot(x_left, y_left, marker="o", color='red', label='x < 2')
plt.plot(x_right, y_right, label='x >= 2')
try this
This works but the blue is higher than the red
Shouldn’t it be the opposite?
In desmos it also approaches 0.5 but here it’s 0.4
yeah because desmos goes upto 2
Oh it includes the limit??
wut
btw change the blue line to include markers too
Like x<2 but it includes 2
replace the last line with plt.plot(x_right, y_right, marker="o", color='blue', label='x >= 2')
Yup this looks good thank you!
Hopefully it’s correct cuz I still don’t understand it fully lol
I understand b. Cuz that’s like me answer in human language I guess
can you temporarily try changing if x < 2: to if x < 2.1: and check how it affects graph
i dont have the packages installed so cant run the code
It did this
I’m using google collab
it is
Thank you🥲
ah
i had the all these packages installed (i use python) before but due to recent drive crash all got wiped and i havent managed to installed everything again
Oh damn.. how did it crash?
fr
Use a raid system for important data
I’m gonna build a storage system with UnRAID, it’s really useful
thankfully i had important things store in a pendrive (not everything but the most important so its fine)
That’s lucky, but I feel your pain having to install everything🥲
fr
it crashed once before too
but only c drive was wiped
had to install everything back but i didnt mind it at that time
should've replaced it at that time
but welp
yeah, learnt the hard way
Do you like vs code btw? I saw you are using it rn, I tried using it for c++ to learn it but when I got to classes it caused problems cuz it can’t run multiple ccp files in one project
So when you try to run a class from the main file it doesn’t run it then shows you an error that the class doesn’t exist
you mean like return it?
Yes to the manufacturer and they send you a new one
its out of warranty lol
and im yet to try getting its stuff recovered
so even if it was in warranty i wouldnt return it
Can it be recovered once it fails?
3-3.5 years i think
they said they could recover the stuff
Damn yea warranty is usually 2-3 years on new drives
Oo hopefully🙏🏻
as for the drive itself, idk, i wouldnt risk it anyways
Yea I wouldn’t use it for anything important
vs code ftw
How do you run multiple files🥲
wdym
It doesn’t work for me
Like when you create a class
You open a new file right?
For that class
Then you mention that class in the main file and press run, but that only runs the main file
Yes
you have to import the class file first
Like import “class name”
?
In cpp it’s include and I did that but it still doesn’t run the class file itself
it wouldnt run the class file, yeah
you use the functions from that class file as needed
And then it’ll give you an error, so what do you do haha
you dont import a file to run it with another file
Ohh, so you can’t work with multiple files then?🥲
You put your classes with the main code?
are all files seperate?
like whatever is inside them isnt being used anywhere outside of its own file
Damn
ill give you an example how you use it
That sucks, because for classes I need a .cpp file and a .h file then include it in the main code
This is what I mean haha
if you have a file named tools.py with the following code in it:
def sum(a,b):
return a+b
def multiply(a,b):
return a*b
now if you want to use the functions from tools.py in another file, lets say main.py
so what you do is: (in main.py)
import tools
print(sum(1,5))
what error does this give
Tasks.json not found something like that
Like the thing that runs the code
Because it’ll only run Classes.cpp but not Cat.cpp
Since it can’t run multiple ccp files
can you run it rn and send ss of the error
It doesn’t do anything haha
you have a launch.json in your directory?
It just sends me here
you sure you installed compiler, added it to PATH
and also installed c++ extension in vsc
Yes it works fine when it’s one file
The problem happens when it’s multiple files
Hold on it gave me another error
There you go, this should answer it haha
Nope I just press run
try to compile using the terminal
What should I write?😅
g++ -o main Cat.cpp Classes.cpp
It didn’t do anything this time haha
No it just didn’t do anything, I wrote it pressed enter and nothing
no error, nothing?
Now it won’t let me write in the terminal at all wth
open an actual command prompt in the directory
then write this there
I forgot which one but the windows one
press on the address bar (in the directory) and type cmd
Here?
not vsc
i meant windows explorer
where the Classes.cpp file is
go there and open cmd in that directory
Wait what’s a directory haha
like the place where the files are stored
It’s on a folder on my desktop
yeah open it
Yes I’m there
and type cmd
yes
Yes but it already had it
When I ran it through yes main
run it
And it doesn’t do anything, it opens and closes something quickly
type main.exe in this terminal
yeah
Wait so you have to go through all of this to use classes?🥲
the issue was in compiling, it wasnt linking the files together (Classes.cpp and Cats.cpp) thats why it was giving undefined reference error because for it the function isnt defined
Yes that’s the thing that runs it right?
can u send its ss
replace the ${file} with ${fileDirname}/*.cpp
Astar777
lmao
🤣🤣
and delete the main.exe that u built before
then try to build it from vsc
What language is even tasks.json??
Thank you haha I’d have never figured this out
So on every project I just replace file with that and it’ll run all the cpp files?
its not actually a "language"
its just a format for storing data
the tasks.json file stores info on how to build the project
by default it only buils 1 file
Oh so it’s like raw code?
so if you want to compile all files in your directory then you repalce the ${file} with what i told
Ohh I see, there should be a setting for it haha
its not "code"
its just a format for data storage
here its storing the settings and info to use while building
things like ${file} and other similar things in it are all processed by compiler, json isnt doing any of it
Ohh it compiles that al together that’s cool
It’s like instructions for the compiler basically
On how to do it
Or what to do
its not "instructions"
its basically just storing the settings
so that the compiler can use them
they could also use a text file for storing settings, but json is better for this use case
I see, that makes sense haha
Thank you for your help🥲 I’ll try to do the next math problems
oh yeah i totally forgot about that
Hahaha
Code is more fun but college won’t allow me to just do code
They shove all this bs just so I can “earn” learning cpp
Thank you!
I guess so haha
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how did they get (i-1) for the left endpoint and
i for the right endpoint
im trying to figure it out with intuition but i cant 😭
@gentle wagon Has your question been resolved?
@gentle wagon Has your question been resolved?
So ur using the limit def of an integral?
no, sum definition
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For a I think i kind of get it
Do I just first find t at 0
Then at 5
Theb subtract and divide by 5
And do the same for 10?
But what about b
I’m confused
Wait hold up is an instantaneous rate?
Bc it says at the end
Why not f’(5)?
Yeah that makes more sense
But how would I find the derivative
Like do I use chain rule or
You don’t need the chain rule
It’s basic differentiation power rule
Average flow rate would be what?
Ohhh so it would be 40t-1609
1600
Then I just plug in right
Ya I think I get it
And I’ll just use velocity formula for b right
mari
@finite stag Has your question been resolved?
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Why doesn’t the nested interval property hold for rationals?
@unreal linden Has your question been resolved?
@unreal linden Has your question been resolved?
amphx 
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I have been trying this question for a while but I still dont know what to do
I could just tryout the numbers from the choises but I want to know the actual solution
let it be x, then (x/52)+33, (x/78)+59 should be equal right?
lets say x = 52*78
78+33 = 52+59
it is
it is equal.
but I dont understand where it come from tho
hmm
how is it equal
i would say chinese remainder theorem but you probably dont know what that is
do not crt this
some notable things here
52, 78, 117 all divisible by 13
and 52 - 33 = 78 - 59 = 117 - 98
oh
52 - 33 = 19
78 - 59 = 19
117 - 98 = 19
it's a coincident?
or it's a property
it is neither
this is a deliberate way this problem was set up so it would have an easy solution
ah
if I know that they are all divisible by 13
and divisor - remainder = 19
(x/52)+33 means that..
i'm not sure
52a+33 =x
divisor = remainder + 19
33(a+19) = x
no
the equation should look like this
can you explain how this equation is formed?
just make those two equal and solve for x
(x/52)+33 = (x/78)+59
(78x-52x)/7852 = 26
26X = 26 78 * 52
X = 26 * 78 * 52 / 26
4056
i did something wrong
how you do that?
thats not how ya solve
wait why?
4056
?
I couldn't understand their solution either
they did include what you said
I see now
ok ok I understand
thank you
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For the love of God, how does this work, i understand this upto the derivation of a = x² - y² and b = 2xy but after that it falls apart for me.
Why are we computing x² + y² anyways? Like it's so random, like sure it probably helps later but how exactly, the textbook does not specify a reason as far as I know, second, what identity are they even using to reexpress x² + y²? This is probably a stupid question but I am sorry i just woke up and can't think straight
While we are there, also explain the later steps to me once someone is able to get this thing through my thick skull
sqrt(x^2 + y^2) is the modulus of sqrt(a+ib)
if root(x² + y²) = root( root( a² + b² ))
Then by squaring both sides
x ² + y ² = root(a² + b²)
Do i follow?
The a = x ² - y ² and b = 2xy so
x ² + y ² = root( (x² - y²)² + (2xy)² )
Oh yeah i think i get it
But why is root(x² + y²) equal to the mod of root(a+bi) again?
@supple knot
What's the modulus of a complex number z = c + id?
|z| = root(c² + d²) i think
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Yes
so
like am i responsible to memorize these conversions
becuz they are not telling me how many inch in yard
);
they literally wrote in a bracket
w0t\
we can't tell thats on your school
is it that hard to memorise
i want to memorize
but dont know what to search
like if there is a paper that has them all i want to find it
try chatgpt or google
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i dont know how to do the third and fourth question. Even chatgpt cant help me
try finding x such that the value for some expression of x is -12, and p(-12)
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I need to calculate solar panel energy generation, i have these parameters:
- sunlight 3d Vector
- sunlight intensity
- solar panel tilt angle
- solar panel azimuth angle
- solar panel size
chatgpt told me this:
Calculate solar panel normal vector based on tilt and azimuth angles
panel_normal = np.array([
np.sin(tilt_angle) * np.cos(azimuth_angle),
np.sin(tilt_angle) * np.sin(azimuth_angle),
np.cos(tilt_angle)
])
# Calculate the dot product between sunlight and panel normal
dot_product = np.dot(sunlight_vector, panel_normal)
angle_cosine = max(0, dot_product / (np.linalg.norm(sunlight_vector) * np.linalg.norm(panel_normal)))
# Effective sunlight intensity
effective_intensity = sunlight_intensity * angle_cosine
# Panel area
panel_area = panel_size[0] * panel_size[1]
# Energy generation
energy = effective_intensity * panel_area * panel_efficiency # in watts
Is this correct?
@proud palm Has your question been resolved?
<@&286206848099549185>
yo
hi
is it correct to calculate solar panel vector like this:
X = Math.sin(tilt) * Math.cos(azimuth)
Y = Math.cos(tilt)
Z = Math.sin(tilt) * Math.sin(azimuth)
yips👍
great thx
np
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((F \circ G)(x)) means (F(G(x)))
Invariance
yes I already know that,
wait i will show my wrok
i don't even know if this is correct
why would g(x) = 3(x-1)^3?
bc i simplified the original g(x)
it's 3 (x - 1)^(-3)
oh there wasnt a minus sign sorry
you wrote ^3
oh i forgot to write the minus there
the domain of (FoG)(x) would be a subset of the domain of F
so how do I find the domain?
have you found the domain of F and the range of G?
oh wait i mixed up some things mb
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Is the equation right?
from second step its wrong
no
Why
$x^2 -2x+4 \neq (x-2)^2$
convergence
also that yea
Wut
you can do this equation like this instead $2x^2-4x+4=2(x^2-2x+1)+2$
convergence
Why is there a 1 in the end tho?
2(1) =2
$2x^2-4x+4=2x^2-4x+2 +2\
2x^2-4x+4=2(x^2-2x+1) +2$
convergence
So you divided the 4 into +2+2?
yes
$(a + b)^2 = a^2 + 2ab + b^2$
decaf
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hi um let me get this straight i hate maths AA and this is about the binomial thorem
this is where i am so far
5N2 = 10
10 . (3x^2)^3 . -(2/x)^2
(please tag when you see this)
i am most confused about the -(2/x)^2, how do i write it? (- 2/x)^2 or like how
anyhow, 10 . 9 . -4 . x^6 ÷ x^2
-364 x^4?
$(A + B)^5 = {5 \choose 0} A^5 + {5 \choose 1} A^4B + {5 \choose 2} A^3B^2 + {5 \choose 3} A^2B^3 + {5 \choose 4} AB^4 + {5 \choose 5} B^5}$
Alberto Z.
Compile Error! Click the
reaction for more information.
(You may edit your message to recompile.)
oh my God what the hell is that
The formula you should use 😅
I've expanded it for you, if you want I can write it with the summation symbol
you will make me end my life samarth
i just want to know where do i place the negative 😭
year 11 IBDP
Im in 11th btw :) [16yr old]
same here
alberto sounds like a very intelligent human being
foundation college year
Originally german, living in bahrain
waht
(never heard of that thingy)
IBDP?
what is happening
$$(A + B)^5 = \sum_{k = 0}^{5} {5 \choose k}A^{5-k}B^k$$
is K = R?
but not sooo much
Alberto Z.
i am getting scared
How do you expect to solve that exercise without this? 🤔
i am doing 40 pages assignment for the binomial theorem
i solve it very easily
like
(yeah i never learnt it)
and like i missed classes of binomial yk
Ahnn you have to use the Pascal triangle?
rah
I NEED TO BE TAUGHT TO TEACH FIRST, MY TABLE IS FILLED WITH TEARS FROM THIS ASSIGNMENT
so you're in 11th and you're a mum?
im curious but ok
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✅
Nope, it's 1080
I wrote the steps, if you want
The most important thing is knowing the Newton's binomial formula
$${\left(A + B\right)}^n= \sum_{k = 0}^{n} {n \choose k}A^{n-k}B^k$$
It is the binomial theroem
Alberto Z.
This one @fervent hinge
if you want something more visual, its like this, but this is only assuming its like (x+1)^n
ive never seen this in my life
i've never seen this either
1
11
121
1331
this is the one that i know
goes more
depending on the n
yeah thats pascals
you can see they are technichally the same just in a diffrent form
i am crying what the hell is this
ill just not solve it all around
i dont care if i lose marks
If you want I can paste my process anyway
So that you can see how it could be solved
In your exercise, the formula here becomes the following: $${\left(3x^2 + \frac{-2}{x}\right)}^5 = \sum_{k = 0}^{5} {5 \choose k}{\left(3x^2\right)}^{5-k}{\left(\frac{-2}{x}\right)}^k =$$
Alberto Z.
i found another answer
$$= \sum_{k = 0}^{5} {5 \choose k} \cdot 3^{5-k} \cdot {\left(-2\right)}^k \cdot {\left(x^2\right)}^{5-k} \cdot {\left(x^{-1}\right)}^k =$$
-1080x^4
Alberto Z.
wait
the negative
is it inside or outside the ()
believe me i dont understand what i am looking at
just hear me out
alberto
Alberto Z.
$$= \sum_{k = 0}^{5} {5 \choose k} \cdot 3^{5-k} \cdot {\left(-2\right)}^k \cdot x^{10-3k}$$
Alberto Z.
$10 - 3k = 4 \iff k = 2$
Alberto Z.
So the coefficient of x⁴ will be the number in the sum when k = 2
Hence, it's $$\text{coeff} = {5 \choose 2} \cdot 3^{5-2} \cdot {\left(-2\right)}^2 = 10 \cdot 27 \cdot 4 = \textbf{1080}$$
Alberto Z.
Yeah that's it
$${\left(-\text{stuff}\right)}^2 \neq -{\left(\text{stuff}\right)}^2$$
Alberto Z.
what
That's (very) basic algebra
i know
thats
why i am asking
because they're not the same
and the answer might be -1080x^4 or 1080x^4
This is the general formula, therefore in your case A = 3x² and B = **-**2/x
its only 6 marks
Please explain parts d and e to me
By the way the square of these two is the same
!occupied
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how are they the same?
one gives me -4x^2 and the other 4x^2 (after ^2)
Nope
$${\left(-\text{stuff}\right)}^2 = {\left(-{\left(\text{stuff}\right)}\right)}^2$$
Alberto Z.
how the hell did you () the ()
wdym
we're discussing
how
-(2/x)^2 is NOT the same as (- 2/x)^2
because one will give me a positive and the other a negative
yeah its not, what is there to discuss about
EXACTLY
.
ill resend it
my question is
for this one
main question showing its -(2/x) and not (- 2/x)
im missing the context to answer that
sent it
i am questioning my sanity
that doesnt make any sense
2/x is one term
and the minus is in front of this term
$-\frac{1}{2} = \frac{-1}{2} = \frac{1}{-2}$
Astar777
i am okay with losing 6 marks.
i will jsut
close this chat
i wont do this question
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💀
I think you won't be able to do other exercises as well, unless you understand the minus sign thing...
i am willing to lost all marks, my sanity is more important
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Parts d and e please
@swift valve Has your question been resolved?
<@&286206848099549185>
welp... im not sure if this is how you're meant to do it, but for (d), notice how 1/(4^2) is smaller than 1/(3^2) and in general 1/[(n+1)^2] is smaller than 1/n^2
so the avg btw 1/(3^2) and 1/(4^2) is smaller than 1/(3^2)
do this for the rest of the integers and (d) should be fine
for (e), you can manipulate the other inequalities to show it, what have you tried?
God I love you
This entire question has been driving me nuts
part a and part d have made me question my existence
np, it do be like that some times, i take it that (e) is ok?
Yes as long as I have d and c
Thank you kind woman
np have a nice day
I lied idk how to do part e
Could you help me again 🙏
haha... well i would try prove 3/2 < sum and sum > 7/4 cases separately
hint: use the red one for 3/2 < sum and the blue ones for sum > 7/4
Do I need both for blue one?
uhhhhh, the way i did it, yeah, there might be other ways
