help-13

[main needle]

right

[mild ore]

but where should I construct it?

[main needle]

on the base, that's the only line you have

that gives you this, and it should be clear how to finish

[mild ore]

oh

thank you so much

you're a life saver

i can't believe I didn't see that

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[crimson sedge]

<@&286206848099549185>

[main needle]

the middle 5 gets shrunk to 4

so what will (5+5+7) shrink into?

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

17

[main needle]

shrink 17 the same way 5 shrinks to 4

[crimson sedge]

What

[main needle]

to shrink you multiply by something

5 * 5/4 = 4

17 * 5/4 gives you the answer

[cedar kiln]

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[crimson sedge]

how should i approach this?

[crimson sedge]

i drew a right angle triangle and found the hypotenuse but i don’t think that’s what the question wants you to do

[main needle]

so you want to go in straight lines, and if you get to the shore you get to go more cheaply*

so the path should look like this red line

and you need to figure out where that red line hits the shore to solve the problem

say the right triangle with the red diagonal line has base x on top

then the red diagonal is sqrt(5^2+x^2) and the horizontal red line is 11-x

see if you can use that to minimize the cost

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

Yep ima try to do this on my own

Ty

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[valid bolt]

I promise it's a simple question

Why are the units for (b)'s answer rad/sec?

I understand why it's something/sec

[main needle]

radians are the angle units we choose to use

going from tan to sec^2 only works in radians

if it was degrees or something else you'd have weird fractions with pi on the outside

https://math.stackexchange.com/questions/214912/derivative-of-the-sine-function-when-the-argument-is-measured-in-degrees for details

[valid bolt]

So, basically, the unit for trig functions is radians and not degrees

[main needle]

at least in calculus yes

if you didn't do derivatives the units don't really matter

[valid bolt]

Okay

I guess it's because we calculated those derivatives we know based on the assumption that the trig functions' inputs are in radians*

[main needle]

you mean radians? yeah

[valid bolt]

Yes, radians mb

I think I got it

Thank you

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[dry plume]

im having trouble with energy

[dry plume]

im stuck on (f)

I am confused as to where to find the kenetic energy from this question

[cedar kiln]

@dry plume Has your question been resolved?

[dry plume]

<@&286206848099549185> if you could help it would be appreciated

[sweet forge]

i don't know any chemistry!

[dry plume]

this isnt a chemistry class. i dont know chemistry either

pepehands

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[dry plume]

.cose

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[lilac bloom]

so i am here now

[buoyant perch]

use the fact that the angles in a triangle add to 180 degrees

[lilac bloom]

mhm

oh

60+55+85+d=180

[buoyant perch]

no because thats not a triangle

you need to find the angle at E

[steel canopy]

Ye using 55+60

You can find angle for E

Just like what snow says

[buoyant perch]

the opposite angles at E are equal

[steel canopy]

Yup

And then you add that E angle with 85 =180

And you find last beautiful Angle

[lilac bloom]

ohhh 30

[buoyant perch]

that seems right to me

[cedar kiln]

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[steel canopy]

This isn't even math at this point

[bright surge]

1. that's not maths
2. don't ask in an occupied channel

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[slow locust]

Not allowed if I mistakenly put that here? A big problem indeed🙄

[bright surge]

yeah...

[radiant topaz]

It's okay if that's a mistake just recheck from now onwards

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[bright surge]

we have this property
$$log_a(u^n)=n.log_a(u)$$

[wraith dagger]

Adavocowana

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✅

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[steel canopy]

I believe you have to apply the exponential rule

First

But hang on

HMM

i am not sure

Lmao

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[cedar kiln]

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[verbal plover]

Inverse variation | y=k/x where k = 10 and y = 10 | how do u find the x

[verbal plover]

my teacher hasnt teach me how to find the x can someone help me

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[steady breach]

I need assistance in solving this problem

[steady breach]

I tried substituting the polar coordinate method

then completing the square but didnt know where to go from there

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[sage obsidian]

https://cdn.discordapp.com/attachments/359052604149465088/910037809753108520/unknown.png
https://cdn.discordapp.com/attachments/359052604149465088/910037872416030720/unknown.png

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— Yesterday at 10:25 PM
Campanula — Yesterday at 10:25 PM
how is this right
this does not make any sense at all
if we use this equation we would get
y=5.8x-23.2

[sage obsidian]

how is this wrong too

[cedar kiln]

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[sage obsidian]

.reopen

[cedar kiln]

✅

[sage obsidian]

can this shit stop closing

[cedar kiln]

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[tacit drift]

Can anyone please explain how the answers are correct?

<@&286206848099549185>

[vital whale]

Line 1 from A is?

[tacit drift]

What

1 -1 -4

[vital whale]

Ok

This times -2 is?

[tacit drift]

-2 2 8

[vital whale]

Great

Line 2 is?

[tacit drift]

2 -1 24

[vital whale]

If you add line 1 times -2 on line 2 we will get?

[tacit drift]

This

Plus

[tacit drift]

This?

[vital whale]

Yeah

[tacit drift]

0 1 32

[vital whale]

Yes

[vital whale]

A1 will be A but with this as the second line.

Can you write A1?

Hold on

[vital whale]

This is wrong.

[tacit drift]

Why

[vital whale]

Look at A

Line 2 is
2 -1 -24

Missed the signal

Now do line 1 times -2 and add to line 2

This is will be A1 line 2

Write it.

[cedar kiln]

@tacit drift Has your question been resolved?

[tacit drift]

ok

[vital whale]

A1 will be?

[tacit drift]

when say line 1

u mean the original one right?

[vital whale]

Yeah because we are A to A1 yet

[tacit drift]

2 -1 -24
+

1 -1 4

?

[vital whale]

No

Let's simplify notations

Let's call Line x as LX

[tacit drift]

whats line x anyways

[vital whale]

Example

Let's start from A

L1 from A is [ 1 -1 -4]

L2 from A is [ 2 -1 -24 ]

Got it?

[tacit drift]

yes

[vital whale]

So L2 from A1 is -2×L1 from A + L2 from A

We can write this as....

L2 = -2×L1 + L2

Can you write A1?

[tacit drift]

but thats what i asked u

[tacit drift]

thi

isnt it the same?

[vital whale]

This is not -2×L1

[cedar kiln]

@tacit drift Has your question been resolved?

[vital whale]

@tacit drift without your feedback it's hard to help you.

[tacit drift]

Sorry I’m but confused 2 mins

I’m getting

0 1 -16

[vital whale]

For?

[tacit drift]

A1

[vital whale]

Ok

Write to me all A1

[tacit drift]

All?

U mean do the same for all rows?

[vital whale]

Yes!

I mean

After you did the row operation you will get A1.

Write the full matrix.

[tacit drift]

ok

[vital whale]

Every time you do a row/column operation you get a equivalent matrix.

My time there is over just Google row operations and you will get what you want

[tacit drift]

done after that?

can u just tell what all to do

ill figure it out

[vital whale]

Google matrix rows operations

[cedar kiln]

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[obsidian coral]

What are you stuck on?

[brittle sandal]

every single thing

[obsidian coral]

Just convert those all to the same units and compare them

[brittle sandal]

u expect me to understand whatever u just said-

[obsidian coral]

It's not hard, convert to same units and compare them all

A hint is the common unit to convert to is kg

[brittle sandal]

so im gonan convert each of those numbers to kg-

?

gonna*

[obsidian coral]

Yes

[brittle sandal]

o

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[covert finch]

a typist types 1064 words in 28 minutes, how many words can they type per minute?
need some help with this

[fair geyser]

divide the numbers

[covert finch]

but by what?

nvm

.close

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[heady lava]

A group of 20 people greet each other with a handshake. How many handshakes will there be if every person can only greet the other person once. The memo says its (20*19)/2=190 but i dont know why they divide by 2.

[livid hound]

a shaking b and b shaking a is the same handshake

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[gray vine]

12. A tunnel has the shape of a quadratic curve, according to the figure below. The equation for this is:
    f (x) = - ((5/8) ((x-4) ^ 2) +10)

Calculate the maximum width of a truck with a height of 7.9m can have to pass through the tunnel.

[gray vine]

This is what I got:

x1 = 4+1,833030278 and x2 = 4-1,833030278.

Answer: 3.666060556m.

The image is the tunnel in question

[frozen valve]

just find x for y=7.9

and find the dist between the two opints

[gray vine]

Can you confirm you got the same solution?

,w ((-5/8) ((x-4) ^ 2) +10) = 7.9

[wraith dagger]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Theophania#1704

[gray vine]

,calc 4+1,833030278

[wraith dagger]

The following error occured while calculating:
Error: Unexpected operator , (char 4)

[gray vine]

,calc 4-1.833030278

[wraith dagger]

Result:

2.166969722

[gray vine]

,calc 4+1.833030278

[wraith dagger]

Result:

5.833030278

[gray vine]

,calc 5.83303-2.16697

[wraith dagger]

Result:

3.66606

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[lyric flint]

Can I have help finding the last one? I have all the other answers but I'm also not sure if they are correct.

[lyric flint]

Here are the double angle Identities for the problem

[graceful karma]

Remember that tan = sin/cos

That should be enough to solve

[lyric flint]

Ahh ok

could u check to see if my other problems are correct?

[graceful karma]

I can try

[lyric flint]

For (a) I got 4/5 using regular trig using a right triangle

for (b) I got 24/25 using the sin(2x) = 2sinxcosx

and for (c) I got -7/25 using cos(2x) = 1 - 2sin^2X

[graceful karma]

Looks good

[lyric flint]

Ok thank you so much, you might see me in another math help seeing if I did (d) right haha

[graceful karma]

Np

[lyric flint]

thank you for the help!

.close

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[storm thunder]

is LHS = RHS ?

[cedar kiln]

@storm thunder Has your question been resolved?

[storm thunder]

<@&286206848099549185>

[leaden otter]

Not necessarily

[storm thunder]

Ok

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[ancient gate]

what did BD : AC = 8:15 mean

I forgot that

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[stable trout]

Its trigonometry I haven’t a clue we done last year so I haven’t even tried to do it

[cedar kiln]

@stable trout Has your question been resolved?

[wet crow]

perimeter is the sum of all 3 sides

26 = 10cm + 4cm + n (side that is missing)

[stable trout]

so the other side would be? 12cm?

i didnt notice it showed permiter in question-

[wet crow]

Yes, it should be 12cm

[stable trout]

Ahh okay and how would I find Y here?

[cedar kiln]

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[crimson sedge]

how would you procede if it was just some number instead of the parameter ?

think about methods to solve a quadratic equation

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@crimson sedge Has your question been resolved?

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[olive sierra]

how you can find the solutions to a quadratic equation by looking at its graph?

[frozen lantern]

If the lenght of a rectagle is 15 meters less than double the width.If the perimiter is over 120 meters what are the values the width can be.

[olive sierra]

channel in use

[spiral orbit]

We call a solution to a quadratic equation - the values of x for which the quadratic is equal to 0.

Look at the graph of the quadratic, if you see any points where the y value is 0, then it is a solution

In other words, the intersection with the x axis are your solutions

,w plot x(x-1)

[wraith dagger]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by peaceGiant#0724

[olive sierra]

so the solution would be 0 and 1?

[spiral orbit]

In the above quadratic, the intersections between the x axis (which has equation y = 0) and the quadratic is exactly what you said, at x = 0 and x = 1

[olive sierra]

.close

[cedar kiln]

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[olive sierra]

@spiral orbit ty

[cedar kiln]

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[plain river]

how do i differentiate ln(ln x) ?

[plain river]

i keep getting x ln x / 1 but apparantly its that but reversed?

[floral thistle]

@plain river chain rule

the answer is (1/x)(1/ln(x))

or 1/(x*ln(x))

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[plain river]

ty

[cedar kiln]

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[ruby garden]

this is quite useful

[cedar kiln]

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[ruby garden]

.reopen

[cedar kiln]

✅

[cedar kiln]

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[silent lark]

Probability:
An event has a 7/11 chance of succeeding.
the probability doesn't change with iteration.
How often does the event need to happen to succeed by majority with at least 95%.

[silent lark]

How to approach this?

[bright surge]

so there's a 4/11 chance of the event not succeeding the first time

right?

[silent lark]

and every other time yes

[bright surge]

so the probability that the event happens the second time is also 4/11

[silent lark]

jup

[bright surge]

so the probability that the event succeeds the second time will by 4/11 * 7/11

right?

[silent lark]

jep

[bright surge]

so the probability that the event will succeed in the 2 times

(first time + second time)

will be 7/11 + 4/11*7/11

right?

[silent lark]

I do see where you're giong but I'm looking for suceeding by majority

[bright surge]

yeah?

after that just repeat the steps

[silent lark]

if I iterate as you did I'm getting the min chance of one success

[bright surge]

and then add that with the previous chances

you'll get the total chance

[silent lark]

with
sucess/attempts = 1/attempts
I need
P(1+(attempts)/2 /attempts) >= 0.95

[cedar kiln]

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[sage obsidian]

https://cdn.discordapp.com/attachments/359052604149465088/910037809753108520/unknown.png
https://cdn.discordapp.com/attachments/359052604149465088/910037872416030720/unknown.png

how is this right
this does not make any sense at all
if we use this equation we would get
y=5.8x-23.2
4th times the charm
5th times the charm
6th times the charm
7th times the charm !! woo!

[radiant topaz]

Do you have your working?

[cedar kiln]

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[prime garnet]

Ask I have a question about integrating partial fractions

[prime garnet]

this is the problem

trying to solve for a and b I ran into issues

posting my work

[radiant topaz]

What did you try?

[prime garnet]

I scribbled out my stuff when it didnt work so im not sure if its worth posting

[bright surge]

uh
it hink we can just expand the denominator?
x^2 - 4

[obsidian coral]

It's partial fractions

[radiant topaz]

Partial fraction is better

[obsidian coral]

So not really

[wraith dagger]

it's Sam

[prime garnet]

work 2

work 1(sorry for scribbling)

[obsidian coral]

You're doing extra nonsense that doesn't achieve much

[obsidian coral]

@prime garnet Start with this here

[prime garnet]

closest thing I got was a way of solving a, and x. I couldve solved b in terms of and x but it would be like a.

[obsidian coral]

Like I said, it's extra nonsense that isn't needed

[prime garnet]

should I have just assumed x was 0?

[obsidian coral]

No

[obsidian coral]

Step one, start with this @prime garnet

[prime garnet]

im sorry but the@wraith dagger it's Sam isnt showing up for me.

[obsidian coral]

Click on it

It'll jump to that

[prime garnet]

aaah ty.

[radiant topaz]

I just wrote what you wrote

[obsidian coral]

Next, clear fractions, by multiplying by the common denominator

[obsidian coral]

It is but theirs is a bit messy with whatever method that was done

So starting with a clean slate at that initial step works

[prime garnet]

am I solving for a?

[obsidian coral]

You'll be finding both A and B

So did you clear denominators?

[prime garnet]

so a and b have the same denominator?

[obsidian coral]

No

Clearing denominators, meaning getting rid of fractions

So multiply everything by the common denominator

[prime garnet]

I did this for a

[obsidian coral]

What happened to B?

[prime garnet]

I moved it to the other side

[obsidian coral]

Don't

There's no need to

You're making it more complex than it needs to be

If you multiplied by a common denominator, you should have something like A(x - 2) + B(x + 2) = 1

[prime garnet]

I have that but they both have (x+2)(x-2)as the denominator

[obsidian coral]

Why do you still have denominators? You multiplied everything by the common denominator to get rid of the fractions

So you shouldn't have any fractions

When you multiplied by the common denominator to all the terms, you would be left with A(x - 2) + B(x + 2) = 1

[prime garnet]

this is what I have

[radiant topaz]

What is that equal to?

[obsidian coral]

Why did you combine them?

[prime garnet]

1/(x+2)(x-2)

[obsidian coral]

You needed to get rid of fractions

[prime garnet]

how?

[obsidian coral]

Did you start with A/(x + 2) + B/(x - 2) = 1/(x + 2)(x - 2)?

[prime garnet]

yeah

oh wait

[obsidian coral]

Multiply everything by the common denominator

Which is (x + 2)(x - 2)

[prime garnet]

I see now

im multiplying both sides by (x+2)(x-2)

[obsidian coral]

So you'll be left with A(x - 2) + B(x + 2) = 1

[obsidian coral]

Yes

That's how you can clear fractions, if you multiply everything by the common denominator

[prime garnet]

thank you

[obsidian coral]

Now, you want to find A, meaning B needs to "disappear", what value of x makes that happen?

Because it's B * (x + 2), you want B * 0, right?

[prime garnet]

this is what I have

[obsidian coral]

No

Stop making it more complex

Leave it at A(x - 2) + B(x + 2) = 1

And let me explain

Whatever nonsense you're doing doesn't help

Now as I stated, you want B to be gone because you want to end up with B * 0

What value of x achieves that?

[prime garnet]

-2

[obsidian coral]

Okay, plug in that value for all the x

[prime garnet]

I get a = -1/4

[obsidian coral]

So if you did it right, you should have A(-2 - 2) + B(-2 + 2) = 1

[prime garnet]

b = 1/4

[obsidian coral]

Yes

There you go, you did partial fractions

[prime garnet]

but what if I did something that didnt make b=0?

like x=1?

[obsidian coral]

Then you still have 2 unknowns in one equation

You want 1 unknown in 1 equation

[prime garnet]

if I tried different values of x would it build a system of equations that I could do elimination on? and would a and b stay the same?

[obsidian coral]

There's no need to

You keep trying to make things more complex

[prime garnet]

thank you I think I know how to solve this now.

[prime garnet]

.close

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[acoustic sail]

is this the equation for lower left [x≠-2 X ∈ R]

[pliant shell]

$y = 1 \forall x \in (-\infty, -2)$

[wraith dagger]

jellyfish

[acoustic sail]

So this is incorrect

[pliant shell]

yes, 1 belongs to your domain

but the graph shows x < -2

[acoustic sail]

what would be the correction same format

[pliant shell]

replace x != -2 with x < -2

[acoustic sail]

how would you differentiate the graph from the others using equations

[pliant shell]

im not sure what the question is asking for

[acoustic sail]

decide which graph does not belong chose lower left comparing all equations finding reason for graph to not belong

[pliant shell]

well, bottom right is the only one with a finite domain

[acoustic sail]

The domain is different from the rest how do I prove that with equations

[cedar kiln]

@acoustic sail Has your question been resolved?

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[shadow moon]

Im not sure about this one

[tropic ingot]

For every pair of sets A, B, if their intersection is the same as one taking out all of the other, must b be empty

You could write the intersection in terms of set differences to know immediately, or think of the example when A is empty and B isn't

[shadow moon]

oh ok

Would this statement be false then?

@tropic ingot thanks i got it! 🙂

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[grand nymph]

can someone eplain why tan^-1(inf)= pi/2?

[grand nymph]

figured it out

.close

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[severe flume]

gotta find the size of these angles

[severe flume]

idk how to find out the value of the y?

[tropic oxide]

@severe flume can you say what kind of shape this is?

there's a special name for it that you would do good to recall

[severe flume]

it is a rhombus?

[tropic oxide]

no, it's not a rhombus.

[severe flume]

nvm its a parallelogram

[tropic oxide]

yes, it's a parallelogram.

what do you know about parallelograms?

[severe flume]

opposite sides are equel and the angles are too

[tropic oxide]

what else?

[severe flume]

and the opposite sides are parallel

[tropic oxide]

that's right

and what do you know about the angles that happen when two parallel lines are intersected by a third?

[severe flume]

dont know?

[tropic oxide]

so if i gave you this picture, you would not be able to tell me anything about these angles i've marked?

[severe flume]

they are both acute angles

[wise trench]

No

[severe flume]

then what?

[wise trench]

They add to 180

[severe flume]

ooh

[wise trench]

That only works if the two horizontal lines are parallel tho

[tropic oxide]

and parallel they are

as i intended them to

[wise trench]

Ye

[tropic oxide]

this is also relevant to the original problem bc the diagram had a parallelogram on it

[severe flume]

ye

[tropic oxide]

the two marked angles add to 180°, if that was not clear already

[severe flume]

yeah

<@&286206848099549185>

[tropic oxide]

the two marked angles in here add to 180°.

[severe flume]

now im starting to understand it

.close

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[warm depot]

could anyone help me.. maybe in dms, or vc with maths? i have a math test soon and i dont understand vectors.. if could someone explain it to me please, i'd be greatful.

[warm depot]

<@&286206848099549185>

[eager elk]

watch khan academy

[warm depot]

oh

ok

ty

[tidal bough]

What part of vectors do you need help with?

[warm depot]

so basically everything.. but sadly my teacher isnt the type of teacher who adds like 10 questions and half are ez

usually he adds like 3 somewhat hard questions

uhh

idk what kind of problems

[tidal bough]

About vectors?

[warm depot]

just like... 2d geometry but with vectors

[tidal bough]

OK show me one

[warm depot]

well idk if its everything, but everything we did until now

[tidal bough]

Show me the vectors problem hệ gave you.

[warm depot]

if u can read what i wrote lol

so basically problems like that yk

[tidal bough]

xD

[warm depot]

f

[tidal bough]

Let me gave it a try

[warm depot]

the pic aint loading

in theory seems ez, but i get confused and then get more confused

[tidal bough]

Oh this is easy

[warm depot]

for you maybe lol

[tidal bough]

But your handwriting is confused

Can you give me the image of the question?

[warm depot]

idk how to translate it so like decomposition of vectors after 2 other vectors that are not colinear

[warm depot]

those were problems we did in the class

i dont have the problem since he dictated it, but i can show u whats in my textbook.. but its not in english

[tidal bough]

What you need to prove here?

The second pic

Oh got it

[warm depot]

needed to find AF BE AND AB

[tidal bough]

You need to calculate those vectors?

[warm depot]

yes

or

on other problems

[tidal bough]

Give me 3 min?

[warm depot]

i need to find smth like

MN=a*AB+b * AC

and find who a and b are

(btw those are all vectors)

[tidal bough]

I will solve the second one for you

[warm depot]

no

[tidal bough]

Ik

[warm depot]

thats not what i meant

like

XD

[tidal bough]

Like what do you mean

[warm depot]

i sent u questions we did in class

as an example

to what i am doing

[tidal bough]

Oh ok

[warm depot]

but its already done

[tidal bough]

What in there you don't understand?

[warm depot]

i sometimes get confused when i need to decompose a vector into other vectors

like i know that AB+BC=AC

but

sometimes its more complicated and i just get all messed up and fuck shit up

[tidal bough]

AB+BC=AC is basic

[warm depot]

yes

[tidal bough]

Don't treat them as a line

Vector is normally use to present force in physic.

[warm depot]

yeah

[tidal bough]

So this is kinda make sense

[warm depot]

i also have vectors in physicss

but i understand a lil more in physics because well

it has more sense when u work with forces and weight and stuff

[tidal bough]

Yeah xD

[warm depot]

but in math with all those letters i get confused

[tidal bough]

You need to learn some of those formula to solve hard vectors problem.

[warm depot]

in physics yes

[tidal bough]

But i don't think you need to use those in your problems.

Yeah and in geometry

Show me one of your problem and i will solve it within 2 min.

[warm depot]

ok uh

[tidal bough]

But why you confused?

[warm depot]

i dont know bro

[tidal bough]

It is ok

[warm depot]

last year when i had like plane geometry in space

it was easier

but now those vectors mess me up real bad

[tidal bough]

I have the same problem with you when i first start learning

[warm depot]

how so?

am so bad i literally got a bet with my teacher

he told me

that

if i dont get more than 6/10 in the test

he will put me alone in the first bench XD

[tidal bough]

You will learn one by one when you solves problem.

[warm depot]

okay

back to the question

so theres

[tidal bough]

Ok

[warm depot]

ABCDEF a regular hexagon, and vectors u = AB, v= AC. Decompose after u and v the following vectors: CD; DE ; EF; FA ;AC; AD and AE

[tidal bough]

Send me those pic again pls

[warm depot]

from what

this problem is from a book

and its not in english

[tidal bough]

Just send me the hexagon images you draw from the question.

[warm depot]

none

[tidal bough]

And tell me what they give and what we need to prove

[warm depot]

thats the whole problem

uh ok

[tidal bough]

Ok

Wait a min

[warm depot]

dont know what u understand from this but.. XD

[tidal bough]

It is easy

It mean that you must calculate those vectors depend on u and v

[warm depot]

well

[tidal bough]

Yeah?

[warm depot]

can u tell me how

[tidal bough]

Can you tell me what descompuneti dupa mean in English pls

[warm depot]

decompose

after

[tidal bough]

Kk

Wait

Wait 2 min?

I will show you how to solve one as an examplr

Still there pal?

[warm depot]

yes

brb tho gotta do smth

[tidal bough]

I done two and i will guide you how to do the rest?

[warm depot]

yes pls

[tidal bough]

You see the image?

?

Dude

[warm depot]

yes

[tidal bough]

We need to calculate vectors CD right?

Bro?

[warm depot]

yes

[tidal bough]

CD=CA+AD in vectors.

Right?

[warm depot]

oooooooooo

yes

[tidal bough]

AD=?

Vectors AD=?

[warm depot]

ac+cd?

[tidal bough]

It is true but if you do it you will start from the beginning again xD

You know that vector BC can be decomposed into AB and AC right?

[warm depot]

yes

wait

no

what?

[tidal bough]

BC=BA+AC

[warm depot]

aaa yes

[tidal bough]

And since it is hexagon

Vectors AD=2 BC

[warm depot]

ye

[tidal bough]

You can totally prove it

So we got CD=CA+2BC so far

Right?

[warm depot]

ye

[tidal bough]

2BC = 2BA+2AC

[warm depot]

ye

[tidal bough]

CA=-AC

BA=-AB

[warm depot]

ye

[tidal bough]

So CD=-AC-2AB-2AC

Right?

[warm depot]

yes

[tidal bough]

So CD=-3AC-2AB

Since AB=u

AC=v

[warm depot]

yes

[tidal bough]

CD=-3v-2u

[warm depot]

ooo

ok

[tidal bough]

Problem solved

[warm depot]

tyy

[tidal bough]

And because it is hexagon

There are some vectors that equal to others

So you can just do the sảm

Same

[warm depot]

oh ok ty

[tidal bough]

OK if you got any problem, pls dm me

[warm depot]

okay

[tidal bough]

I gtg poop soon.

[warm depot]

tysmm

lol

[cedar kiln]

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[crimson sedge]

Can anyone help with 9bi

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

@crimson sedge Has your question been resolved?

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[granite raven]

How can I find the next number in this sequence?

144 73 14 8 236 ...

[tropic oxide]

you can't

there is no information beyond the first five terms

[cedar kiln]

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[native grail]

Here's the question

[native grail]

And here's what I've done

I am aware that I went wrong in trying to find the complimentary function but i could have also gone wrong previously. Plus i am unsure as to how to tackle the rest of the problem after that point.

[cedar kiln]

@native grail Has your question been resolved?

[native grail]

<@&286206848099549185> would someone be able to point me in the right direction?

[radiant topaz]

Try yp= at²+bt+c

Ah that won't work

y seems right then

[native grail]

so to find x is it 5x=dy/dt - 5y +5

well i mean x = (dy/dt)/5 - y + 5

[radiant topaz]

This will work

And then by given x(0) and y(0) you can get value of A and B

[native grail]

cool, nice to know i hadnt gone wrong

[radiant topaz]

I haven't checked thoroughly but it seems right so far

,w x²+6x+10=0

[wraith dagger]

Results provided by WolframAlpha

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Requested by it's Sam#5028

[native grail]

This is my x, finding a and b will be a pain

[radiant topaz]

,rotate

[native grail]

Oh sorry

Actually I'll just add the 1/5 in

[radiant topaz]

Find A and B so most of the things will cancel out

[native grail]

On it

@radiant topaz i believe that's finished

[radiant topaz]

You gotta input this in x(t) and y(t) then simplify it

[native grail]

Ah okay

[native grail]

Sorry that took so long

[radiant topaz]

Not your fault

I'm not sure how much the calculation is right its hard to check so I hope it's right

The answer looks similar to form I thought would be

[cedar kiln]

@native grail Has your question been resolved?

[native grail]

Yes

[cedar kiln]

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[floral geyser]

can somebody help me in algebra

[obsidian coral]

Don't ask people to help you. Just post your question and someone will come along to assist

[floral geyser]

ok

[silent pawn]

Do you know that you have to post the question if you want to get an answer?

[cedar kiln]

@floral geyser Has your question been resolved?

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[crimson sedge]

what does these two represent in formula ? n = no. of years/period? no idea about t = 1

[crimson sedge]

What is the formula for?

[crimson sedge]

Value of a bond or debenture that is amortised every years as follows:-

[wraith dagger]

it's Sam

[crimson sedge]

Then t would be the first period of time being amortised, and n is the total number of years

ok thanks

You're welcome

[radiant topaz]

You're welcome

[crimson sedge]

.close

[cedar kiln]

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[crimson sedge]

I forgot how to do this could someone quickly help me ?

[crimson sedge]

Seven

1/2*8 is 8/2 resulting in f(8) = 4+3

[glad kestrel]

don't do people's work for them

[crimson sedge]

Ok im right then

[crimson sedge]

alright sorry lol

[crimson sedge]

Nah bro you're good I already knew how to do it I just needed a refresher

You actually did exactly what I needed lmao

[glad kestrel]

yeah he gave you the answer lol

[cedar kiln]

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[forest steeple]

i need help with unit fractions

i want the answer 2/7 but can only use addition and have to use 1/4

• something else

[cedar kiln]

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[crimson sedge]

$\int_{-1}^1 \frac{\sqrt{x^2+1} + x - 1}{\sqrt{x^2 + 1} + x + 1}dx\\
\int_{-1}^1 \frac{\sqrt{x^2+1} + x + 1 - 2}{\sqrt{x^2 + 1} + x + 1}dx\\
\int_{-1}^1 \frac{\sqrt{x^2+1} + x + 1}{\sqrt{x^2 + 1} + x + 1}dx + $\int_{-1}^1 \frac{-2}{\sqrt{x^2 + 1} + x + 1}dx\\
\int_{-1}^1 1 dx - 2$\int_{-1}^1 \frac{1}{\sqrt{x^2 + 1} + x + 1}dx\\
x = \tan \theta \
dx = \sec^2 \theta d\theta \\
x\Big|{-1}^1 - 2\int{-\frac{\pi}{4}}^\frac{\pi}{4} \frac{\sec^2 \theta}{\sec \theta + \tan \theta + 1} d\theta \\
1 - (-1) - 2\int_{-\frac{\pi}{4}}^\frac{\pi}{4}\frac{\sec \theta}{1 + \sin \theta + \cos \theta} d\theta \\
2 - 2\int_{-\frac{\pi}{4}}^\frac{\pi}{4} \frac{\sec \theta}{1 + \sqrt{2}cos\left( \theta - \frac{\pi}{4}\right)} d\theta \\
2 - 2\int_{-\frac{\pi}{4}}^\frac{\pi}{4} \frac{1}{\cos \theta + \sqrt{2}\cos \theta cos\left( \theta - \frac{\pi}{4}\right)} d\theta \\
2 - 2\int_{-\frac{\pi}{4}}^\frac{\pi}{4} \frac{1}{\cos \theta + \frac{\sqrt{2}}{2}\left(\cos\left(\theta + \theta - \frac{\pi}{4}\right) + \cos\left(\theta - \theta + \frac{\pi}{4}\right) \right)} d\theta \\
2 - 2\int_{-\frac{\pi}{4}}^\frac{\pi}{4} \frac{1}{\cos \theta + \frac{\sqrt{2}}{2}\cos\left( 2\theta - \frac{\pi}{4}\right) + \frac{1}{2}} d\theta \\
2 - \int_{-\frac{\pi}{4}}^\frac{\pi}{4} \frac{1}{2\cos \theta + \sqrt{2}\cos\left( 2\theta - \frac{\pi}{4}\right) + 1} d\theta $

[crimson sedge]

what am i missing here on this integral? i've reached this far but it seems like a dead end to me

[wraith dagger]

Mofumofu
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

<@&286206848099549185>

[floral thistle]

wow

@crimson sedge i have an idea, on the line before you did a trig sub, multiply the top and bottom by sqrt(x^2+1)-x-1

then complete the square of the denominator

[crimson sedge]

but the denominator is just -2x though

wait hold on a second

[floral thistle]

it’s a conjugate

(a+b)(a-b) = a^2-b^2 with a = sqrt(x^2+1) and b = x+1

[crimson sedge]

i've been trying so hard to get rid of the denominator. god

yeah i forgot about that

[floral thistle]

yeah that took me a minute too

[crimson sedge]

it was 3 terms so i thought "well that wouldn't work" but then you can just use (x+1) as one term

[floral thistle]

you might have to do integration by parts but i’m not sure

[crimson sedge]

now you can split it into 3 parts

sqrt(x²+1)/x
-x/x
-1/x

[floral thistle]

wait what?

[crimson sedge]

lemme write this better

$2 - 2\int_{-1}^1 \frac{1}{\sqrt{x^2+1}+x+1}dx \\
2 - 2\int_{-1}^1 \frac{\sqrt{x^2+1} - (x + 1)}{(\sqrt{x^2+1}+x+1)(\sqrt{x^2+1} - (x+1))}dx \\
2 - 2\int_{-1}^1 \frac{\sqrt{x^2+1} - x - 1}{(\sqrt{x^2+1})^2 - (x+1)^2}dx \\
2 - 2\int_{-1}^1 \frac{\sqrt{x^2+1} - x - 1}{x^2 + 1 - x^2 - 2x - 1}dx \\
2 - 2\int_{-1}^1 \frac{\sqrt{x^2+1} - x - 1}{-2x}dx \\
2 + \int_{-1}^1 \frac{\sqrt{x^2+1} - x - 1}{x}dx \\
2 + \int_{-1}^1 \frac{\sqrt{x^2+1}}{x}dx - \int_{-1}^1 \frac{x}{x}dx - \int_{-1}^1 \frac{\sqrt{1}{x}dx \\$

[wraith dagger]

Mofumofu

$2 - 2\int_{-1}^1 \frac{1}{\sqrt{x^2+1}+x+1}dx \\\\
2 - 2\int_{-1}^1 \frac{\sqrt{x^2+1} - (x + 1)}{(\sqrt{x^2+1}+x+1)(\sqrt{x^2+1} - (x+1))}dx \\\\
2 - 2\int_{-1}^1 \frac{\sqrt{x^2+1} - x - 1}{(\sqrt{x^2+1})^2 - (x+1)^2}dx \\\\
2 - 2\int_{-1}^1 \frac{\sqrt{x^2+1} - x - 1}{x^2 + 1 - x^2 - 2x - 1}dx \\\\
2 - 2\int_{-1}^1 \frac{\sqrt{x^2+1} - x - 1}{-2x}dx \\\\
2 + \int_{-1}^1 \frac{\sqrt{x^2+1} - x - 1}{x}dx \\\\
2 + \int_{-1}^1 \frac{\sqrt{x^2+1}}{x}dx - \int_{-1}^1 \frac{x}{x}dx - \int_{-1}^1 \frac{\sqrt{1}{x}dx \\\\$
```Compilation error:```! File ended while scanning use of \frac .
<inserted text> 
                \par 
<*> 236662055615070208.tex
                          
I suspect you have forgotten a `}', causing me
to read past where you wanted me to stop.
I'll try to recover; but if the error is serious,
you'd better type `E' or `X' now and fix your file.```

[crimson sedge]

$2 - 2\int_{-1}^1 \frac{1}{\sqrt{x^2+1}+x+1}dx \\
2 - 2\int_{-1}^1 \frac{\sqrt{x^2+1} - (x + 1)}{(\sqrt{x^2+1}+x+1)(\sqrt{x^2+1} - (x+1))}dx \\
2 - 2\int_{-1}^1 \frac{\sqrt{x^2+1} - x - 1}{(\sqrt{x^2+1})^2 - (x+1)^2}dx \\
2 - 2\int_{-1}^1 \frac{\sqrt{x^2+1} - x - 1}{x^2 + 1 - x^2 - 2x - 1}dx \\
2 - 2\int_{-1}^1 \frac{\sqrt{x^2+1} - x - 1}{-2x}dx \\
2 + \int_{-1}^1 \frac{\sqrt{x^2+1} - x - 1}{x}dx \\$

[wraith dagger]

Mofumofu

[floral thistle]

oh my gosh

that’s a lot of stuff

[crimson sedge]

now you just split the integral

into 3 integrals

sqrt(x²+1)/x
-x/x
-1/x

[floral thistle]

oh ok

yeah i see

this actually worked very well

i didn’t expect it to work this well

[crimson sedge]

i think the integral of sqrt(x²+1)/x is a special value that you should memorize but i haven't so i'm gonna figure it out myself

[floral thistle]

i thought you would still have at least a binomial in the denominator

[crimson sedge]

nope

hold up-

you can't split the -1

because if you do

you'll have this

$- \int_{-1}^1 \frac{1}{x} dx$

[wraith dagger]

Mofumofu

[crimson sedge]

and that doesn't converge

[floral thistle]

this function isn’t even continuous in the interval lol

[crimson sedge]

this whole integral doesn't converge?

[floral thistle]

probably not

but maybe check?

[crimson sedge]

maybe it works if you don't split it up

[floral thistle]

yeah

maybe

[crimson sedge]

it does converge

i just don't know how to integrate this now lmao

[floral thistle]

bruh i just used a calculator on the square root divided by x

it’s a lot

like

a lot

[crimson sedge]

$\int_{-\frac{\pi}{4}}^\frac{\pi}{4} \csc\theta \sec \theta (\sec \theta - 1) d\theta$

[wraith dagger]

Mofumofu

[crimson sedge]

lemme try integration by parts

[floral thistle]

you could still do trig sub if you forgot

[crimson sedge]

i used x = tan theta and i got this

[floral thistle]

did you remember the absolute value part

or did you already do the limits

[crimson sedge]

already did

[floral thistle]

ok

[crimson sedge]

you just can't do the definite integral of 1/x from -1 to 1 because of the 0 in between

[floral thistle]

yeah this needs integration by parts if you didn’t see

[crimson sedge]

i did

problem is finding values for D and I, but i'm trying some

[floral thistle]

ok i can try to think of some

maybe try representing them as 1/sin or 1/cos

wait that won’t help

[crimson sedge]

[floral thistle]

yeah i have no clue what to do

you might have to do a method where you do integration by parts until you get the integral you’re trying to integrate in the answer

[crimson sedge]

if you can show these two things

1. f(x) = y and f(-x) = -y
2. all points are defined in the interval
   then doesn't that mean that the integral = 0? since the range is from -a to a

[floral thistle]

maybe

i don’t want to say definitely yes because i’m not 100% sure

but that does seem right

oh wait yeah i think that’s right

bruh i used a calculator again and this integral, or at least the integral of cscxsecx isn’t even that hard

trig integrals are annoying

[crimson sedge]

yeahh

i'll go to school now and see if i can do it there

so i'll close this

!close

if i knew how to

[floral thistle]

do .close

[crimson sedge]

thanks

.close

[cedar kiln]

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[floral thistle]

no problem

[cedar kiln]

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[crimson sedge]

I don’t understand this step, how is it rationalizing it?

ah nevermind I got it

.close

[cedar kiln]

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[night moss]

this is a question about pseudoconvexity of functions (defined below)

[night moss]

if a differentiable function has no zero-gradient points over a set, can you conclude that it is pseudoconvex over that set?

my understanding is just that the above statement only ever has contradictions when you have a grad(f)=0 at some point since then it could decrease while still satisfying the left statement

so if you never have a point where grad(f)=0 over a set, does that mean the statement always holds true if the function is smooth?

<@&286206848099549185>

[cedar kiln]

@night moss Has your question been resolved?

[cedar kiln]

@night moss Has your question been resolved?

[night moss]

.close

[cedar kiln]

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[crimson sedge]

exercise 13

I have to use lim h -> 0 (f(x + h) - f(x))/h
to solve this, if sqrt(|x|) has a tangent line.

m = slope

what do I do next?
do I need continuity and discontinuity?
lim x -> c f(x) = f(c)

this is the answer of exercise 13, but I just don't why the tangent line at x = 0 does not exist?

[glad kestrel]

function needs to be differentiable to have a tangent line

[crimson sedge]

What do you mean with differentiable?
I haven't got differentiation yet in class.

I have to use Newton quotient or difference quotient.

[glad kestrel]

your right hand and left hand limits are not written properly

you got rid of the absolute value sign

also your question just seems to say "look at the graph"

[crimson sedge]

I did that, because h comes near 0 from the left side, so |h| becomes -h.

and h comes near 0 from the right side, so |h| becomes h.

or maybe I am wrong on that one, I don't know why 😰

[glad kestrel]

you're supposed to just graph the original function

[crimson sedge]

sorry

[glad kestrel]

then refer to what your text says here

[crimson sedge]

there is a cusp at the origin (0,0)
but I don't know how to proof it.
I got this:

A cusp, is when the right limit is infinity and the left limit is minus infinity.
but how is lim h -> 0- (sqrt(-h)/h) = minus infinity?

[glad kestrel]

you took the absolute value sign out

you also didn't make both h's negative when you approach 0 from the left

[crimson sedge]

$\lim_{h \to 0-} \frac{\sqrt{-h}}{-h}$

[wraith dagger]

Bleidorb