help-27

[safe sun]

Where did you get the 8 from

[amber valley]

From the table, we're at the third line

[safe sun]

Oh yea

[amber valley]

Now, y=1+6x

[safe sun]

15?

[amber valley]

Change x with 8 in 1+6*x

It's 1+6*8

,calc 1+6*8

[woven radish]

Result:

49

[safe sun]

Okay explain how is it 49

[amber valley]

^

I'm leaving you here, you should really try to understand what the questions ask you to do

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[safe sun]

.reopen

[devout snow]

✅

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[restive river]

7(17)^-8x-7 -9 = 66
solving using logarithms but i don't know how i'm supposed to set up the log of the left side

[red ember]

The given equation is not correct. The correct form of the equation should be 7*(17)^(-8x-7) - 9 = 66.

The equation is not valid in its current form because the exponentiation operator (^) has a higher precedence than the multiplication operator (*), so the expression 7(17)^-8x-7 is evaluated as 7 * (17^-8) * x * -7. This is not what was intended.

To fix the equation, we can use parentheses to explicitly indicate the order of operations. The correct form of the equation is 7*(17)^(-8x-7) - 9 = 66. In this form, the exponentiation operator is applied to the expression 17^(-8x-7), and then the multiplication operator is applied to the result of the exponentiation and the number 7. This produces the intended result.

try -1.03

[restive river]

oh sorry i typed it wrong, so i can just divide out the 7 next to the 17?

[restive river]

didn't work for me

[red ember]

It is not possible to simply divide out the number 7 next to the 17 in the equation 7*(17)^(-8x-7) - 9 = 66. This is because the 7 and the 17 are part of a larger expression that is being exponentiated.

In general, exponentiation is a mathematical operation that raises a number (the base) to a given power (the exponent). For example, in the expression 2^3, the base is 2 and the exponent is 3, and the resulting value is 8.

In the given equation, the base is 17 and the exponent is (-8x-7), so the resulting value is 17^(-8x-7). This value is then multiplied by 7, so the final expression is 7*(17)^(-8x-7).

Dividing out the 7 next to the 17 would change the meaning of the equation and produce a different result. To solve the equation, we must follow the correct order of operations and apply the exponentiation operator to the expression 17^(-8x-7) before applying the multiplication operator. This will give us the correct result.

[restive river]

i thought so

but 7 doesn't multiply into 17 here either because (7 x 17)^(-8x-7) isn't the same as 7(17^(-8x-7))

[red ember]

I might be missing something as well

[restive river]

i've gotten to 7(17)^(-8x-7)=75

but then i don't know how to set up the log on the left

i keep mixing up all of the numbers

would i just do (-8x-7)log7(17) = log66

[red ember]

asked a friend, heres what he sent

The given equation is not correct. The correct form of the equation should be 7*(17)^(-8x-7) - 9 = 66.

The equation is not valid in its current form because the exponentiation operator (^) has a higher precedence than the multiplication operator (*), so the expression 7(17)^-8x-7 is evaluated as 7 * (17^-8) * x * -7. This is not what was intended.

To fix the equation, we can use parentheses to explicitly indicate the order of operations. The correct form of the equation is 7*(17)^(-8x-7) - 9 = 66. In this form, the exponentiation operator is applied to the expression 17^(-8x-7), and then the multiplication operator is applied to the result of the exponentiation and the number 7. This produces the intended result.

With the corrected equation, we can then solve for the value of x that satisfies the equation. To do this, we can use the following steps:

Simplify the right-hand side of the equation by subtracting 9 from both sides, so that the equation becomes 7*(17)^(-8x-7) = 57
Divide both sides of the equation by 7 to isolate the exponentiated expression, so that the equation becomes (17)^(-8x-7) = 8.14
Take the logarithm of both sides of the equation with base 17. This will convert the exponentiation into a multiplication, so that the equation becomes -8x-7 = log17(8.14)
Solve for x by isolating it on the left-hand side of the equation. To do this, we can add 7 to both sides of the equation, divide both sides by -8, and then take the reciprocal of both sides. This gives us the following steps:
-8x = log17(8.14) - 7
-8x = -1.24 - 7
-8x = -8.24
x = 8.24/-8
x = -1.03
Therefore, the value of x that satisfies the equation 7*(17)^(-8x-7) - 9 = 66 is -1.03.

[restive river]

that's weird

i put it into mathway, it said the answer would be -0.9796

and i thought you couldn't divide out 7 from the equation

[red ember]

It's been a while since I've solved these problems but did the result from mathway answer the question?

[restive river]

pretty much yeah

i rounded so it was only off by a tenth but mathway was right otherwise... the unsimplified answer was -7/8 - (ln(75/7)/8ln(17))

[red ember]

I see, alright then. I'll keep that in mind

[restive river]

i wish i didn't have to pay to see the steps

[red ember]

You know, if theres a question your confused on and you search it up on google, if you find that websites such as Bartleby, CourseHero or Chegg have an answer. You can paste that link into this website -- https://homeworkify.net/ and it'll show you the result for free. It has helped me out through college.

Its all free by the way.

[restive river]

that makes things much easier, thank you so much

[red ember]

No worries, good luck.

[restive river]

maybe not... i can't find my question anywhere online and mathway doesn't have any links

i think google has a pretty good solution that i understand though

i'm pretty sure you can divide 7 out because all of the websites i see are doing it

[red ember]

I see, do you have another question your confused on? send it

[restive river]

actually i do

it's a word problem if you don't mind

isotope SR-85 has a half-life of 64.9 days. use an exponential function to determine how many days it will take for the isotope to decay to 38% of its original amount.

so do i just plug all of these numbers into A(t)=a(1+r)^t and just divide the answer by 2?

[red ember]

To determine how many days it will take for the isotope SR-85 to decay to 38% of its original amount, we can use the formula for exponential decay. The formula for exponential decay is given by:

A(t) = A0 * e^(-kt)

Where A(t) is the amount of the isotope at time t, A0 is the initial amount of the isotope, k is the decay constant, and t is the time in days.

In this problem, the initial amount of the isotope is 100%, the decay constant can be calculated from the half-life of the isotope (64.9 days), and we want to find the time it will take for the isotope to decay to 38% of its original amount.

To calculate the decay constant, we can use the formula k = ln(2)/t_{1/2}, where ln(2) is the natural logarithm of 2 and t_{1/2} is the half-life of the isotope. Plugging in the values for ln(2) and t_{1/2}, we get:

k = ln(2)/64.9 = 0.01075

Now that we have the decay constant, we can use the formula for exponential decay to determine how many days it will take for the isotope to decay to 38% of its original amount. Plugging in the values for A0, k, and the desired final amount (38%), we get:

38% = 100% * e^(-0.01075 * t)

Solving for t, we get:

t = ln(38%) / (-0.01075) = 651.3 days

Therefore, it will take 651.3 days for the isotope SR-85 to decay to 38% of its original amount.

[restive river]

ohh i forgot about that formula

[red ember]

Were you able to get the question solved?.

[restive river]

how do you get to ln(38)/(0.01075)? did you just cancel out the 100% since it's just 1

[red ember]

These are the steps I took. @gilded latch I'm I doing this right?.

1: 38/100 = e^(-0.01075 * t)
2: 38/100 * (1/1) = e^(-0.01075 * t) * (1/1)
3: 0.38 = e^(-0.01075 * t)

We can now take the natural logarithm of both sides of the equation to eliminate the exponentiation. This will give us the following steps:

4: ln(0.38) = ln(e^(-0.01075 * t))
5: ln(0.38) = -0.01075 * t

Finally, we can divide both sides of the equation by -0.01075 to solve for t. This gives us the following steps:

6: ln(0.38)/(-0.01075) = t

Solving for t, we get:

t = ln(0.38) / (-0.01075) = 651.3 days

[restive river]

ok got it

also i got 90 days for some reason? is my calculator doing something wrong

[gilded latch]

That sounds more correct

[red ember]

129.8 days

[gilded latch]

If the half life is ~65 days then in 130 days it will be at 25%

[restive river]

so it's probably 90 days then

[gilded latch]

What value did you get for k?

[restive river]

i got 0.01075

and i put it in the equation as negative since it's decay rate

[gilded latch]

Was it exactly 90 days?

[restive river]

no it was 90.0078164

[gilded latch]

I got .01068 as k

[restive river]

you divided by 65 probably

[gilded latch]

ln(2)/64.9

[restive river]

oh

oh i did too

i never put it in the calculator

so 90.5978 days then

[gilded latch]

Noice

My advice on chegg or any online solutions is use them if you’re stuck but still do the problem and understand it yourself

I know where I went to school there were professors who posted incorrect solutions to catch students who were just copying off online sites

Like posted to chegg

One professor gave half a class academic dishonesty

[restive river]

wow

there's so much intricacy with the work we have right now so it's hard to always find the exact answer

i usually do the work myself and only use websites to work myself backwards

but thank you both so much for helping me out with this i was really stuck

🍰 🍰 have cake

.close

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[frail igloo]

not sure how to approach this.

[devout snow]

Please don't occupy multiple help channels.

[frail igloo]

i bring negative sign inside so f"(x) = -sinx - cosx

[winter hound]

Don’t occupy multiple channels

[frail igloo]

i dont even see myself in the available channel

[pseudo basin]

.close

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[lavish laurel]

A real-estate agent receives a 3.5% commission for each house she sells. Last month she received a commission for $4,725 for selling one house. What was the price of the house?

[restive river]

To find the price of the house, we need to first determine the amount of the commission the real-estate agent received. If she received a 3.5% commission for selling the house, then we can write this as 0.035 * P = $4725, where P is the price of the house. To solve for P, we can divide both sides of the equation by 0.035 to get P = $4725 / 0.035 = $135714.29. So, the price of the house was $135714.29.

DM me if you have any more questions.

• Louis.M

[lavish laurel]

Oh wow, that was pretty fast. Thanks man!

[restive river]

no problem. do not hesitate to DM me if you have any more questions.

[lavish laurel]

Alright.

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[bright rose]

How to get in phsics discord?

[bright rose]

@floral vortex

/physics

#physics

[plush knot]

#old-network

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[short iron]

For problems like 59 and 60, would you treat 5i and -3j as terms in a vector?

[short iron]

And treat i as a vector of <i, 0> ?

[grizzled shoal]

you have to find projection

i will treat them as vector

you can use projection formula

projection = w.v/mod(v)

if you wanna find projection of w on v

[short iron]

Yes I know of the projection formula, I was asking when your v or w value is not in vector form like in 59 and 60 where v=5i-3j, and w=i, do you just treat your terms as if they were a vector?

Like for 59, v = <5i, 3j> and w = <i, 0>?

@grizzled shoal

[grizzled shoal]

yes

i will treat

[short iron]

Let me type this out in latex, I wanna make sure I'm fully understanding

[v = <5i,3j> w=i]
[v \cdot w = 5(1) + 3 = 8]
[||w||^{2} = i]
[proj_{w} v = \frac{8}{1^2} <i> = 8]

[woven radish]

dopediscorduser

[short iron]

Does this look right @grizzled shoal

[grizzled shoal]

umm yes

are you sure

abouut the projection formula

you applied

[short iron]

[proj_{w} v = (\frac{v \cdot w}{||w||^2} )<w>]

Right?

[grizzled shoal]

if we have to find projectin of a vector on b then the formula must be a.b/mod(b)

[woven radish]

dopediscorduser

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[sharp dew]

e

[fallow panther]

fr

[sharp dew]

can i ask validation if i did 3 and 4 right?

3. 4π cm2
4. 13/2 π cm2

[restive river]

3 is correct

and so is 4

[sharp dew]

thanks !

.close

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[thick flame]

can someone help me with a sheet

[thick flame]

can someone solve this and send it to me

with working

cz i need an ms

and i dont got an ms

[spare zealot]

We are not here to do work for you

[thick flame]

im not

saying

do work

cuh

im saying can u solve that cz i need an ms

ok nvm

[spare zealot]

You just said to solve it and send it to you

[thick flame]

cz i need a marking scheme for that

😭

i have an exam in 2 days cuh i need to

solve

a

ton

of

stuff

i mean if anyone can

its liek

yk what

nvm

can u just

help with this q

cz i need help in this one particular

q

[thick flame]

yea just this one q, im doing prep and i dont understand dis

[honest aurora]

$$\text{average speed}=\frac{\text{distance from A to B}}{\text{time from A to B}}+\frac{\text{distance from B to C}}{\text{time from B to C}}+\frac{\text{distance from C to A}}{\text{time from C to A}}$$ where $$\frac{\text{distance from X to Y}}{\text{time from X to Y}}=\text{speed from X to Y}$$using this you can use the information given to solve for what you want, which in this case is "speed from C to A"

[woven radish]

Duh Hello

[honest aurora]

i shouldve maybe used variables lol, would define $v_{AB}, v_{BA}, v_{CA}$ and the distances and all that as variables

[woven radish]

Duh Hello

[devout snow]

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[thick flame]

ty

[devout snow]

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[devout snow]

Please don't occupy multiple help channels.

[normal echo]

how do i multiply this out

[honest aurora]

matrix multiplication?

[normal echo]

yh

[honest aurora]

also go to your other channel

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[honest aurora]

or close that one and open this one

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[dapper quail]

I need help with part B

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[supple knot]

,rotate

[supple knot]

this photo of a screen is very blurry. can you take a screenshot and upload instead?

[supple knot]

,rotate

[supple knot]

fyi that has your name

[tender flame]

its fine

its a fake last last name anyway so idk what people can do with just the first name

[devout snow]

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[tender flame]

<@&286206848099549185>

[devout snow]

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[tender flame]

<@&286206848099549185>

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[maiden zinc]

how do we solve x = cos(x) ?

[analog grove]

Cos = x/x which is equal to 1

[winter patrol]

solution isn't "nice"

[maiden zinc]

cos(1) != 1

1 is not solution here

[winter patrol]

you could get a decent approximation from stuff like Newtons method

[maiden zinc]

at least how can i find that there's only one solution at this equation

idk if i can use the bijection theorem since here cos is not bijective

[winter patrol]

calculus

show x-cos(x) is mono inc

[maiden zinc]

without calculus cause it's not allow on exam

[winter patrol]

consider domain and ranges

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[fossil moth]

i was just wondering, isnt this solution incorreect?

[fossil moth]

nvm

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[fossil moth]

.reopen

[devout snow]

✅

[fossil moth]

i dont get the second step, how did they use that cos trig identity

[honest aurora]

$\cos(a-b)=\cos a\cos b-\sin a\sin b$

[woven radish]

Duh Hello

[honest aurora]

oh wait no its a bit different

[fossil moth]

yea

oh, did they just multiply it? like cos(pi-piu)

but it doesnt look like it\

[honest aurora]

yeah should be pi*u

lemme check real quick

yeah they did it wrong but the answer is still correct

using it properly will get the same answer

[fossil moth]

alright, so i should do it like cos(pi-piu)

[honest aurora]

yeah

[fossil moth]

alright thank you

[honest aurora]

np

for all i know there might be some secret rule im unaware of where it is for some reason allowed to do that but id reckon they just accidentally got the right answer and didnt think twice about it

[fossil moth]

yea, most likely

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[devout snow]

@dapper quail Has your question been resolved?

[lusty sapphire]

what have you tried, @dapper quail ?

[dapper quail]

i need help understanding what its asking

what does it mean by vertices?

dont ellipses only have one vertex?

[lusty sapphire]

I believe this is the context you need

[dapper quail]

ohh

i confused it with the center

how can i find the vertices?

sry i kept going afk im here now

[dapper quail]

how can i find the vertices?

[lusty sapphire]

Looks like its the longest axes of the ellipse

[dapper quail]

the major axis is 40 and minor 24

would it be like? (16+-20,0) or (36,0) and (-4,0) as the vertices?

[dapper quail]

is that right?

[lusty sapphire]

Think you meant (-4, 0) here

[dapper quail]

right

oh ok cool

[lusty sapphire]

But yeah, I'd say you nailed it

[dapper quail]

what ways other than graphing can I use to do B?

[lusty sapphire]

calculus

distance-preserving properties of ellipse and hyperbola

only 2 I can think of

[dapper quail]

whats that?

[dapper quail]

guess im using graphs

i want to know the other ways....

[lusty sapphire]

sorry, I'm at work

do you know calculus?

[dapper quail]

yea its fine ill come back later

mhm a little

[lusty sapphire]

Well that's the most direct way to do.

By distance preserving property of ellipse, I mean its definition

"a regular oval shape, traced by a point moving in a plane so that the sum of its distances from two other points (the foci) is constant"

FP+F'P=FP'+F'P'=FP''+F'P''

[dapper quail]

Also how do you remember lots of specific things in math so well

I always forget things how are you able to recall it easily?

it'd probably help me for every other class as well

[devout snow]

@dapper quail Has your question been resolved?

[real grail]

the minor isn't 12 ?

[dapper quail]

The entire minor axis is 24

12 is half

So 2b is the minor axis

b is the distance from the center to vertex

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[dapper quail]

oo that's cool

[dapper quail]

.close

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[dense crane]

guys is the converse property of the statement "the tangents drawn at the ends of a diameter of a circle are parallel" also true?

[winter hound]

Context more, we need to see whether the tangents are parallel

[devout snow]

@dense crane Has your question been resolved?

[dense crane]

they r

[brisk totem]

the converse of your statement is "if two lines are parallel then they can be expressed as tangents drawn at the ends of a diameter of a circle"

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[queen comet]

Find the least number that should be subtracted from 2230 to make it a perfect square.

[queen comet]

can anyone help me pls

Find the least number that should be subtracted from 2230 to make it a perfect square.

[vocal bone]

21?

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@queen comet Has your question been resolved?

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[restive river]

So, I am trying to prove De morgan's law more rigorously now.
$\bigskip$
[\text{We know that: } \overline{A \cup B} \equiv \overline{A} \cap \overline{B}]
To prove that, we can say
[ x \in \overline{A \cup B}]
[ x \notin A \cup B]
[\ \to neg((x \in A) \vee (x \in B))]
[ \to \neg(x \in A) \wedge \neg(x \in B)]
[ x \notin A \wedge x \notin B]
[x \in \overline{A} \wedge x \in \overline{B}]

[woven radish]

♡LexQa♡

[ember palm]

What is that tilted L?

💀

[restive river]

the sign for negation

[ember palm]

O 💀

Idk what's that

Nice

[restive river]

x does not belong to A or x does not belong to B

okay so

bruv wait

no

V is this for "or" or "and"

[ x \in \overline{A \cup B}]
[ x \notin A \cup B]
[\neg((x \in A) \wedge (x \in B))]
[\neg(x \in A) \vee \neg(x \in B)]
[x \notin A \vee x \notin B]
[x \in \overline{A} \vee x \in \overline{B}]

[woven radish]

♡LexQa♡

[restive river]

oh yeah i just realised, i was doing it in reverse oops

just consider the U

[hybrid snow]

I always get and/or mixed up

[restive river]

messed my mind

i remember that U=V

upwards is upwards

Haa, wait i think i missed it all up lemme rewrite

So, I am trying to prove De morgan's law more rigorously now.
$\bigskip$
[\text{We know that: } \overline{A \cup B} \equiv \overline{A} \cap \overline{B}]
To prove that, we can say
[ x \in \overline{A \cup B}]
[ x \notin A \cup B]
[\ \to \neg((x \in A) \vee (x \in B))]
[ \to \neg(x \in A) \wedge \neg(x \in B)]
[ x \notin A \wedge x \notin B]
[x \in \overline{A} \wedge x \in \overline{B}]

okay i think this is correct

[woven radish]

♡LexQa♡

[restive river]

i guess thats sufficient

.close

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[cursive flicker]

Hey! I could use some help verifying some answers I got. Compared with a class mate and we arrived at different results somehow... Admittingly my math background is a lil weak

[cursive flicker]

This is the first one, is it correct?

<@&286206848099549185>

[devout snow]

@cursive flicker Has your question been resolved?

[green imp]

yeah it is

but don’t forget to derive it back then divide by pi

[devout snow]

@cursive flicker Has your question been resolved?

[cursive flicker]

What do you mean? 😅

In the answer field I am supposed to give two numbers

Should these be 4 and -1?

[green imp]

no why do you think that

[devout snow]

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[restive river]

does (3i)² equal -3 or -9??

[restive river]

I'm unsure whether you square the 3 or leave it be

.close

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[rustic dome]

hi so in this question i can't find a specific triangle or quadrilateral to find some angles

so can any1 help

[devout snow]

@rustic dome Has your question been resolved?

[thin epoch]

@rustic dome

tell me about that red shape

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[stark coral]

hi

[stark coral]

im on ii rn

and im very confused

i tried so much

i tried finding LN then using pythagoras theorem to find LM

didnt work

i tried finding KL then subtracting it by the length of KLM but that's not working either

ive been trying to solve this for days

please help

[thin epoch]

Do you know the law of sines?

[stark coral]

no

another thing i tried is

to find KL

[thin epoch]

how did you find KMN?

[stark coral]

u mean KLM?

[thin epoch]

no, part i asks you for KMN

[stark coral]

since km and mn is there

oh

for angle kmn

what i did was

sin x(or sine theta) = 15.6/6

then

sin-1 = 15.6/6

sin-1(15.6/6) in degree mode

wait a minute

now its giving me an error

1 sec

[thin epoch]

(your fraction is upside down)

[stark coral]

i just realized

[thin epoch]

S=O/H

[stark coral]

sine = o/h

so i did sin-1(6/15.6)

for side LM though

the textbook says trhe answer is 6.8

but i just dont get it

i tried everything

for days

and didnt get it

lemme show you some methods i did

so firstly, i tried to find KLM

used pytha theorem

[thin epoch]

because you have c and a, pythagorean theorem can get you b (aka KLM)

then focus on triangle KLN

[stark coral]

6^2 + x^2 = 15.6^2
36 + x^2 = 243.36
x^2 = 207.36
x sq root = 207.36 sq root
= 14.6

no, 14.4

[stark coral]

i tried that

lemme tell u what i got

since we have to find LN

and LN is the

wait

is LN the hyp?

[thin epoch]

of traingle KLN yes

[stark coral]

Omg

wait

i didnt put that for hyp

this might lead me to the answer

1 sec

[thin epoch]

then you have one angle, the opposite length and the adjacent length

or rather you need the adjacent length

[stark coral]

ok just did the working

were using cosine

cos 52 = 6/ln

[thin epoch]

SOH CAH TOA

[stark coral]

yeah ik

[thin epoch]

do you know the length of the hypotenuse?

[stark coral]

oh

bruhh

im rushing too much

i gotta focus

lemme try again

ok back

OMG LES GOOO

I FINALLY GOT IT 😭

ITS BEEN DAYSS

ok so basically

sin 52 = 6.0/ln

ln(sin52) = 6.0

ln(sin52)/sin52 = 6.0/sin52

sin 52 gets cancelled out on the left side

which leaves with ln = 6.0/sin 52

ln = 7.61

length of km(or klm) = 14.4, as said earlier

14.4 - 7.61 = 6.79

textbook answer = 6.8

basically the same answer

i prob got 6.79 because i rounded the decimals to do decimal places during the working out of the questions

I FINALLY DID IT 😭

thanks

wait a minute.

💀

um

i just realized that i didnt actually solve for kl

so i just did KM-LN

welp

back to square one..

[thin epoch]

but the book says the right asnwer is 6.8?

[stark coral]

yeah

but the working out doesnt make sense

i cant subtract KLM by NL

so what i did was

use pytha theorem

since i have NL and NM

NL = 7.61

NM = 15.6

7.61^2 + x = 15.6^2

57.91 + x + 243.36

x = 243.36 - 57.91

x = sq root 185.45

ok..

final part

i really hope this is it

its..

its not it.

i got 13.6

:(

man idk what to do anymore

this question is impossible

i shouldve solved this days ago

[thin epoch]

you want KN not MN

[stark coral]

kn is 6

[thin epoch]

you can't do the pythagorean theorem on triangle LMN

[stark coral]

why not

since i found LN, i have 2 sides

LN = 7.61

[thin epoch]

you can only do the pythagorean theorem on right triangles

[stark coral]

oh

so use trig to find the side?

lemme try

i have angle KMN already

so it should be possible

since the angle is pointing to LN

we have to use tanjent right?

tan 22 = 7.61/LM

lemme try that

bro i still didnt get it

i got 18.8

[thin epoch]

so start over

[stark coral]

WHEN WILL THIS END

[stark coral]

ok

[thin epoch]

not entirely, but clean slate

[stark coral]

ok

[thin epoch]

that's what you've calculated so far

[stark coral]

btw the angle is 22.6

anyways

[thin epoch]

which one?

KMN?

[stark coral]

KMN

yeah

so im gonna calculate LN

[thin epoch]

oh right

[stark coral]

got 5.99 for LN

[stark coral]

completely different answer right off the bat 💀

im working with triangle NLM instead of NKL

to find NL

we gotta use sine ratio

so sin 22.6 = LN/15.6

[thin epoch]

if you use SOHCAHTOA with angle KLN

[stark coral]

15.6(sin 22.6) = LN

[thin epoch]

you want to find KL to subtract it from KM

[stark coral]

i meant LN in the working

not LM

mb

[stark coral]

LN = 5.99

[thin epoch]

why do you want LN?

[stark coral]

oh

crap i was gonna use pytha theorem again...

i should breka out of that habit

i did it unconsciously

lemme start again

[thin epoch]

you have the Opposite, you want the Adjacent

[stark coral]

wait

but

isnt the opposite LN?

and we dont have LN

[thin epoch]

no

LN is the hypotenuse

[stark coral]

so i thought we have to do cos

[stark coral]

wait

[thin epoch]

if you're working with angle KLN (which is given at 52 degrees)

[stark coral]

but isnt NM the hyp

[stark coral]

oh

[thin epoch]

you have two right triangles

[stark coral]

i wasnt working with KLN

[thin epoch]

KLN and KMN

[stark coral]

i was working with NLM

if i work with NLM, will i still get the answer?

[thin epoch]

you can't use pythagorean with NLM

[stark coral]

yeah i realized

so i was gonna not use it

we have an angle and a side

so trig can be used to find LM

maybe thats what the textbook is expecting me to do

since they asked for KMN's angle

my pomodoro break started so

im gonna start again in 1 min 40 secs

ok back again

time to solve

[thin epoch]

forget NLM

[stark coral]

ok

[thin epoch]

https://cdn.discordapp.com/attachments/903486480075354182/1051249521163780116/image.png

you have KM and you need LM

so you need to find KL

[stark coral]

aight bet

[thin epoch]

if you have an angle (KLN) and the opposite side (KN), how do you find the adjacent side (KL)?

[stark coral]

i got 4.68

should i round it up

to 4.7

[thin epoch]

sure

[stark coral]

ok

so 4.7cm for KL

then pytha theorem

[thin epoch]

you don't need pythagoras

you don't need LN

the question is asking for LM

you just have to subtract KL from KM

[stark coral]

KL = 4.7cm

NM = 15.6

NK = 6.0

since KL = 4.7

and KM = 14.4

just do KM-KL?

[thin epoch]

yeah

[stark coral]

but that gives me 9.7

textbook says 6.8

maybe the textbook is rigged

[thin epoch]

their answer doesn't make sense to me

[stark coral]

yeah

im moving on

to the next one

it is what it is

maybe its an error in the textbook

[thin epoch]

yeah i'm 99% sure it's supposed to be ~9.7

[devout snow]

@stark coral Has your question been resolved?

[stark coral]

dont close it yet bot

ok cool its open still

[stark coral]

actually um

i did the same method for another question

and the answer is different

lemme show the other question

[thin epoch]

shoot

[stark coral]

my phones charging rn so

i had to draw the diagram

sorry ab the bad drawing lol but

i) find the lenght of LO

i was like, this is a piece of cake

since LO is the hyp

[thin epoch]

i'm assuming LMO and LMN are right angles?

[stark coral]

yeah

2 right angles

[thin epoch]

so you can find MN

[stark coral]

sin 35 = 10/LO

LO(sin35) = 10

LO(sin35)/sin35 = 10/sin35

lo = 10/sin 35

= 17.4

[thin epoch]

and the textbook says otherwise?

[stark coral]

textbook says 24cm for LO

but for the length of MN

the textbook says the answer is 17.4

which is the same answer i got

this is weird

[thin epoch]

yeah so based on your drawing, MN should be 24cm

[stark coral]

lemme solve that just in case

for practice purposes

[stark coral]

thats not what i got

i got 27.7

since its a right angle triangle, we can do trig

10^2 + x^2 = 26^2

100 + x^2 = 676

x^2 = sq root 767

= 27.7

[thin epoch]

576 not 767

[stark coral]

oh

i added

instead of subtracting

i really need to stop rushing lol

24

correct

thanks for the help mane

appreciated

im gonna have to go for a run

a 1 hour run

then ill shower and work again

have a nice day/night

[thin epoch]

you too

[stark coral]

🫡

[devout snow]

@stark coral Has your question been resolved?

[devout snow]

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[devout snow]

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[frozen sonnet]

i'm not sure how to do this, I have tried splitting it in half and finding the length of the middle

[vital patrol]

u can use law of sines

[frozen sonnet]

ohhh

ok

thank you

.close

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[restive river]

Let G be a connected graph on n ≥ 4 vertices. If the degree-sequence of G is 4, 3, 2, 2, 1, 1, 1, 1, 1, is it possible that G contains a cycle? (Hint: recall
that G is connected).

[restive river]

Topic: Discrete Mathematics - Graph Theory

.close

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[devout snow]

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[upper crescent]

I am doing part d of the question

I have looked up the answer, but I don't know how the part circled in blue was derived

Since as I rmb, it should be $QA^Q^$ instead

[woven radish]

Trenton

[upper schooner]

That star is conjugate transpose yea?

In which case, they're just working out $(Q*AQ)*$ using the line above it

[upper crescent]

yes

[upper schooner]

Don't forget that when you consider the transpose, you have $(AB)^{T} = B^{T} A^{T}$...

[upper crescent]

But as I rmb, $(AB)^T=B^TA^T$

[woven radish]

chartbit

Trenton

[upper crescent]

oh nth

[upper schooner]

Yea, work that out in steps, like ((Q*A)Q)* and you should get there 😉

[upper crescent]

But then how to get $Q^*A^*Q=[IUI]_{\beta}$?

[woven radish]

Trenton

[upper crescent]

I guess $A^*=[U]_{\beta}$

[woven radish]

Trenton

[upper crescent]

and $Q^*=[I]{\beta '}^{\beta}$ and $Q=[I]{\beta}^{\beta '}$

[woven radish]

Trenton

[upper schooner]

Actually wait ignore me

Hmm 🤔

[upper crescent]

??

[devout snow]

@upper crescent Has your question been resolved?

[upper schooner]

Deleted something I said which was wrong 😂

[devout snow]

Channel closed

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[upper crescent]

.reopen

[devout snow]

✅

[upper crescent]

ah I see

[upper schooner]

Hmm, the way I see it, isn't it then just like rewriting this but for U instead of T?

At least that's what I'm gonna take it as what they're trying to say 😂

[upper schooner]

But then, shouldn't this have a $\beta'$ or not?

[woven radish]

chartbit

[upper schooner]

(That was what I said but then deleted, I wasn't sure so erased it from history 😂 )

[upper crescent]

lol

[upper crescent]

oh yes

It makes sense

[upper schooner]

Ahh, so I wasn't being slow then 😂

[upper crescent]

lol I just forget we can apply similar argument

yeah thanks a lot

.close

[devout snow]

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[upper crescent]

I am doing part g of the question

[upper crescent]

So I am going apply GS process

But the definition of the inner product is not stated

is there any definition of the inner product of 2 square matrices?

[upper schooner]

Yessss

[upper crescent]

oh really

[upper schooner]

Are you linear in your first argument, or your second argument? (Ignore me, these are real matrices)

Try (A,B) = tr(AB^{T}) (in this case, as we're real - if complex, try (A,B) = tr(AB*) )

[upper crescent]

umm ok

[upper schooner]

But the first one does form a real inner product (exercise left to the reader 😂 )

[upper crescent]

Alright let me try to verify this (the answer is really unclear)

[upper crescent]

lmao

[upper schooner]

Ah, yea, it looks like they chose that as their inner product 😂

Norm of the first matrix becomes 6, which is what they normalise with

[upper schooner]

Haha I've been dealing with a lot of complex inner products, which is why that question was in my mind 😂

[upper crescent]

cool

Right, there is a typo in that answer

They miss the division sign

[upper schooner]

Yea I was thinking the same thing!

[upper crescent]

lol

[upper schooner]

Look at me, actually trying it out 😂

What has this group done to me haha

I've even turned blue!

[restive river]

green next!

[upper crescent]

lol I also turned into blue in May 2022

but I am too inactive in the summer so I turned into white again

lol

[upper schooner]

Could you imagine 😂 I would be shocked!

Too many interesting maths problems lmao

[upper schooner]

Mind you, I joined this group some time ago but barely ever used Discord

[restive river]

Oh definitely. Good part of this server is that you also get to learn more as you help as well

[upper schooner]

But then cause I have meetings with people on here, ended up browsing and well, here we are!

[restive river]

Meetings? 😮

[upper schooner]

Yea definitely, and as they always say, one of the best ways to learn is to teach others 😂

[restive river]

Yeah definitely

[upper schooner]

Yea for work, I work from home doing web development

[restive river]

Oh wow, that's cool that you organise meetings through this server as well

[upper schooner]

On that note, fuck React Native 😒

[restive river]

Isn't that some UI software if my memory serves me correctly haha

[upper schooner]

Well, not this one, another one that they set up, but yea

[upper schooner]

Yea for mobile apps

'Tis giving me so much pain

Don't think I'm cut out for webdev tbh

[restive river]

Ohh okay this makes more sense

[restive river]

It'd funny how I know some of my cs friends say the complete opposite lmaoo

They wanna become webdevs because they hate the normal background coding

[upper schooner]

Hmm, I'm not too great with coding either, above relatively simple tasks I think 🤔

Like if my life depended on coding or development, might as well start preparing for my funeral now 😂

[restive river]

point taken: dont code

[arctic field]

lmao green next

dont forget light bloo

[upper schooner]

Veri veri activ 😂

[restive river]

i wish i had that

i got both green and light blue at the same time

[upper schooner]

Speaking of, it's snowed outside 🥶

[restive river]

[arctic field]

[restive river]

im dying to 100% humidity

[arctic field]

i joined when helpful wasnt a thing

got light bloo within a month

[upper schooner]

Ouch! 😭 Whereabouts are you in the world?

[flint pivot]

whats the problem here?

[restive river]

Somewhere in the eastern barren lands of Europe

[upper crescent]

Ahhhhhh

[restive river]

okay not barren but

[upper crescent]

Computational error

[upper schooner]

Ahhh I see, fair fair

[upper schooner]

Wait in the work you just did?

[upper crescent]

yup.....

alright forget it

I think I understand the way to compute

[upper schooner]

Do Gram-Schmidt on g wrt to the inner product (A,B) = tr(AB^{T})

[upper schooner]

Haha, that was literally me after computing the norm of the first matrix

Enough hard work for a day 😂

[upper crescent]

yes.......

alright thank you

you really saved me

.close

[devout snow]

Channel closed

Closed by @upper crescent

Use .reopen if this was a mistake.

[upper crescent]

.reopen

[devout snow]

✅

[upper crescent]

ah I forget you are asking rpyurhbu

[upper schooner]

Ah, no no, I was just telling them what the problem was 😂

They're free to try it and show us their work 😂

[arctic field]

tr(AB^T) really is just treating the entries of the matrix as a vector and taking the pairwise product and summing

[upper crescent]

oic

Noice

thanks

.close

[devout snow]

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[kindred blade]

How can I prove that this function is convex or not (it is a problem of continuous optimization):

[devout snow]

@kindred blade Has your question been resolved?

[devout snow]

@kindred blade Has your question been resolved?

[stone stump]

calculate derivative?

what characterizations of convex functions do you have

[kindred blade]

even if I calculate the gradient, it doesn't tell me too much about it

$f(y) - f(x) >= \nabla f(x)^T * (x-y)$

[woven radish]

Razvanip

[kindred blade]

at best I can only find whose the minimum in this function

[stone stump]

second derivative?

[kindred blade]

does it have the same interpretation like in R^1?

[stone stump]

what do you mean with interpretation here

second derivative here is of course a matrix and you need positive definiteness

[kindred blade]

I'm thinking about critical points

[stone stump]

critical points are still those with gradient = 0

[kindred blade]

oh yeah

if it is convex, then the second derivative is positive semi-definite

[kindred blade]

and those whose second derivative in that point is positive semi-definite*

[stone stump]

well second derivative for those tells us whether its min, max or saddle

but they are all critical points

critical points by definition are those where the gradient is 0 (or sometimes also those where the gradient doesn't exist)

[kindred blade]

yeah, but in the context of optimization, we don't want saddle points, so we add that constraint as well

[stone stump]

ok fair enough if your course includes that

[kindred blade]

Thanks

[devout snow]

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[devout snow]

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[hearty summit]

Hi I have this set and I have to see if it is a basis

[hearty summit]

I have this set that is indeed a basis

so I can analyze from the second set if the first one is also a basis

https://cdn.discordapp.com/attachments/1051492410250768495/1051500329168031754/rn_image_picker_lib_temp_3374e75f-8046-4898-95a5-ac9912a80028.jpg

I tried doing this

is it ok?

[restive river]

yup looks fine

[hearty summit]

nice, ty 😄

.close

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[restive river]

need help

[restive river]

function helps

i have replaced all x with the bracket

<@&286206848099549185>

[pseudo basin]

show us what you get after said replacement.

[restive river]

-4p^2-10p-7

but that isnt right

[versed slate]

Could you show how you square -2p?

[pseudo basin]

yeah because you screwed up.

[restive river]

shouldnt i squaret the numbers

[pseudo basin]

(-2p)^2 isn't -2(p^2)

[restive river]

i missed that step

[restive river]

ye bc u replace that w X

okay ty yall

ann and helmfirth ur help is greatly aprrectiated

[versed slate]

No problem, it’s what we’re here to do

[devout snow]

@restive river Has your question been resolved?

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[restive river]

When i divided 153 over 1703 the result will be like 000.0898 then i Multiply it by 100% so the Point change to be 8.98
So when i divided another number: 1550 over 1703 the result will be 0.9101 * 100% result will be 91.02 the question is now why there is a zero (between the bracket ) in 0.91[0]1

[tender spade]

wdym why is there a zero
it comes during division man

[merry seal]

you move the decimal right by two places

0.9101 --> 9.101 --> 91.01

why should it not be there?

[restive river]

The question in the divison

I wan

[winter patrol]

The question in the divison
I wan
?

[restive river]

Want to know why there will be zero

In the divison process

@winter patrol @merry seal @tender spade

[winter patrol]

have you every done long division before?

[restive river]

Yes

[winter patrol]

that's why

these are just numbers

if a 0 comes up in the division process, there's a 0

[restive river]

There is no zero to comes up

[winter patrol]

1550/1703 is approximately 0.9102
there's a 0 right there in bold

[merry seal]

I don't think we can really explain how to do long division it's not something that can easily be demonstrated through text even if I use pictures

I mean on one image I send there will not be any 0, on the next there will be

[winter patrol]

there's no reason why that 0 shouldn't be there

lets even use an integer example like
735/7 = 105
are you saying that 0 shouldn't be there either?

[restive river]

In long divison and in calculator

My brain will dead??!

[winter patrol]

you didn't put the correct values in the second calc and it isn't giving enough precision

[merry seal]

well it should be giving enough precision if he puts the correct values

[restive river]

The values are correct

[winter patrol]

it won't give the 4th dp

since that dodgy calc only goes up to 3

[merry seal]

oh wait it's 3 digits yeah

[winter patrol]

last i checked 155 isnt 1550

[restive river]

Okay there is an error

Why it should be 0.0898 in first number and second is 0.91

Why

[winter patrol]

what answer are you expecting...

[restive river]

I found the answer the Divisor are bigger than the other

So there will be a zero

Okay thanks

[winter patrol]

153/1703 is approximately 0.08984
converting that to a % by multiplying by 100% gives approximately 8.98%
1550/1703 is approximately 0.9102
converting that to a % by multiplying by 100% gives approximately 91.02%
these are the mathematical results of multiplication and division

too general

[devout snow]

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[dim halo]

does someone understand what's going on in this image? i'm having trouble

it's supposed to be an informal derivation showing the following equation:

Pr(E|T) = [Pr(A|T) x Pr(E|T&A)] + [Pr(¬A|T) x Pr(E|T&¬A)]

[devout snow]

@dim halo Has your question been resolved?

[dim halo]

specifically,

Pr(E & T) = Pr(E & A & T) + Pr(E & ¬A & T) because those two propositions are mutually exclusive and exhaust all ways that E & T can be true.

does anyone understand this justification?

[devout snow]

@dim halo Has your question been resolved?

[devout snow]

@dim halo Has your question been resolved?

[devout snow]

@dim halo Has your question been resolved?

[lament turtle]

You can try drawing a venn diagram or a truth table to make sense of it

[dim halo]

wdym

[lament turtle]

Red: E & A & T
Yellow: E & ¬A & T

Red + Yellow: E & T

[dim halo]

what does this tell me about the probabilities?

[lament turtle]

Well if you wanted to find the probability, you would take everything in the red area + everything in the yellow area, and then divide it by the total number of outcomes

Basically tells you that you can separate it

Kind of how like Pr(A) + Pr(¬A) = 1, either A is true or false.

When E & T is true, either E & A & T or E & ¬A & T is true and the other is false. So Pr(E & T) = Pr(E & A & T) + Pr(E & ¬A & T)

[dim halo]

so ur saying Pr(E & T) = Pr(E & A & T) + Pr(E & ¬A & T) is true?

[lament turtle]

Yeah?

[dim halo]

guess im not sure how ur getting there

[lament turtle]

That's okay

[dim halo]

is there a non-ven diagram way

to explain

[lament turtle]

Yep, truth tables work too, eventually you get the conclusion that E & T means either 1. E & A & T is true, or 2. E & ¬A & T is true.

Let me try to think of a better example

[dim halo]

oh i see

makes sense

if Pr(E&T&A)=1, Pr(E&~A&T)=0

and vice versa

[lament turtle]

mm yes, but also not exactly 1 you know

More like
if Pr(E&T&A)=x, Pr(E&~A&T)= Pr(E & T) - x

[dim halo]

so:

Pr(E&~A&T)= Pr(E&T) - Pr(E&T&A)

[lament turtle]

Yeap, and...

[dim halo]

and?