help-27

[devout snow]

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[turbid hearth]

i think i got it but can some let me know if its right

[devout snow]

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[restive river]

I need help please

[spice pike]

ok so first find the area of the triangle which is (base x height)/2

then we find the area of the trapezoid which is (a+b)/2 x height

does this make sense?

[devout snow]

@restive river Has your question been resolved?

[restive river]

wait

a+ b

10 + 26

?

i thought u multiply those both and get the square root

[spice pike]

a+b are both bases

a can be 22m

b is the bottom base

find b first

[restive river]

The base is 10 @spice pike

[spice pike]

its 10 for the triangle yes

but not for the trapezoid

[restive river]

The trapezoid is on the right side right

[spice pike]

yup

[restive river]

So we have to solve for the base?

Or is it 22

[spice pike]

22 is the top base

[restive river]

And we need the buttom base

[spice pike]

you need to do 22-10

[restive river]

That’s 12

[spice pike]

to find the remainder of thee bottom

[restive river]

So that’s the buttom vase

[spice pike]

yes

[restive river]

Ohh

Ok

[spice pike]

its like sloving the missing side

[restive river]

So now that we know 12 is the base for the right trapezoid

Do we square it

[spice pike]

we dont square anything

[restive river]

I thought u square with area

[spice pike]

do dont square for area

[restive river]

Idk what’s next

[spice pike]

did you find the area for both shapes

[restive river]

Wait since 10 is the base what’s 26?

For the left triangle

[spice pike]

whats the formula for a triangle

[restive river]

A = 1/2 base times height?

[spice pike]

yes

whats the base and height?

[restive river]

for the right trapezoid?

12 and 22?

[spice pike]

no

the triangle

[restive river]

10 and 22?

26*

[devout snow]

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[restive river]

.reopen

[devout snow]

✅

[spice pike]

not 10 and 22

thats for the trapezoid

[restive river]

10 and 26

The base for the left triegbel is 10

[spice pike]

yes!

[restive river]

and height 26*

[spice pike]

now plug that in the formula and it gives you the area of the triangle

yup

[restive river]

130?

[spice pike]

nice you got it

[restive river]

LETS GO

[spice pike]

now we need to find the area of the trapezoid

[restive river]

How

[spice pike]

(a+b)/2 times height

[restive river]

A + b from the left triangle or

Trapezoid

[spice pike]

the trapezoid

[restive river]

uhh

12 and 22 right

[spice pike]

we found the red

[restive river]

12 is the base

So

12 and 22

[spice pike]

we need to find thee right one

[restive river]

For the trapezoid

[spice pike]

yup

[restive river]

But what’s the height

[spice pike]

plug everything in the formula

[restive river]

Idk the height tho I only know a + b

[spice pike]

whats the height of the triangle

[restive river]

Ooo

13

130

[spice pike]

no thats the area of the triangle

the height is given

[restive river]

26?

[spice pike]

yup

[restive river]

When u mean

2 times

Do u just mean

12 + 22 * 26

[spice pike]

(12+22)/2 * 26

my bad

[restive river]

442

Is the answer?

Bro I have 30 more of these to solve how Df am I gonna do that in 2 hours

Oh my god

[spice pike]

yup

now add the two areas up

[restive river]

4420 + 130

440

442

[spice pike]

what do you get

[restive river]

572

[spice pike]

up

this is the long way to do this question

i can show you the easier way

[restive river]

YES

SHOW ME

Please

[spice pike]

you need to imagine you moving the triangle to the other side

like this

it makes a square

then you use the formula for the area of a square

area= L*W

22*26

=572

[restive river]

Wtf

U just solved it in 30 seconds

and this

could i make this a aquare as well

@spice pike

[devout snow]

@restive river Has your question been resolved?

[tough forge]

yes

mive it from the right one to left or the left one to right

[devout snow]

@restive river Has your question been resolved?

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[solar yoke]

When factoring polynomials what do you do when your c is a variable

[solar yoke]

Example -6t - 16 + t^2

[red sierra]

Send the question you're having trouble with

[main gull]

Reorganize that

From highest power to lowest

That's the first step you should do

[red sierra]

You should do this first

[main gull]

Reorganize that
From highest power to lowest
That's the first step you should do

[red sierra]

Find what term has the highest power

[solar yoke]

What do you do after

[red sierra]

Also do not write your t like you write +

[red sierra]

Well did you do it?

[solar yoke]

I do it a little diff I was probably just lazy

Yes

[red sierra]

What did you get?

[solar yoke]

t^2 -6t -16

[main gull]

Now factor that

Using whatever method you know

[solar yoke]

-8 x 2 right

Yo?

btw this is Factoring x^2 + bx + c

.close

[devout snow]

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[devout snow]

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[willow osprey]

which term must be subtracted from 5a+3b to get -5a+9b?

[winter patrol]

what have you tried?

[willow osprey]

I don't know if I have to add it or subtract it

[winter patrol]

would you understand what
5 subtracted from 9 means?

[willow osprey]

what do you mean?

[winter patrol]

do you understand
5 subtracted from 9
yes/no

[willow osprey]

yes

5-9 right?

no 9-5

[willow osprey]

9-5

[winter patrol]

yes

the same definition can be applied to your question

you can first introduce another variable like x to represent what you want to determine

can you write an equation that represents
x subtracted from 5a+3b gets -5a+9b

[willow osprey]

okay so 5a+3b - x = -5a+9b

[willow osprey]

is that right?

[winter patrol]

yes

and then solve for x

[willow osprey]

-10a+6b

[winter patrol]

not quite

[willow osprey]

what is wrong?

@winter patrol

[limber wagon]

Almost there

[winter patrol]

-10a+6b
i'm assuming that by typing the expression by itself, that was your result for x,
how are you getting that from:
5a+3b - x = -5a+9b

[willow osprey]

5a+3b - -10a+6a = -5a+9b

[winter patrol]

isn't right

what exactly are you doing to get that bolded part:
5a+3b - -10a+6a = -5a+9b

that's not solving for x

doing the manipulation and solving for/isolating x in the equation
gives you something that tells you exactly what x is equal to

[willow osprey]

5a+3b + -5a+9b is that right?

5a+3b + -5a+9b = x

[winter patrol]

no

[willow osprey]

i don t understand what you mean

[winter patrol]

you're not doing valid manipulation

$1 - x = -2 \red{\not\implies} 1 - 2 = x$

[woven radish]

ℝamonov

[willow osprey]

what do i wrong?

[winter patrol]

you essentially did the above
it should be clear with integers that its wrong

[willow osprey]

2-1= x

[winter patrol]

still wrong

note that one of the basic principles of manipulation is applying the same operation to both sides of the equation

[willow osprey]

what is manipulation ?

[winter patrol]

rearranging using valid operations to get something in your desired form

to solve 1 - x = -2 properly, you can:
first add x to both sides of the equation to get
1 - x + x = - 2 + x
1 = -2 + x
then add 2 to both sides of the equation to get
1 + 2 = -2 + x + 2
3 = x
which tells you that x is 3

[willow osprey]

yes

[winter patrol]

did you understand that?

[willow osprey]

yes

[winter patrol]

now apply the same idea and structure to solve

5a+3b - x = -5a+9b

[willow osprey]

ok

wait

5a+3b-x = -5a+9b

+x
5a+3b = -5a+9b+x

+5a
10a+3b=9b+x

-9b
10a-6a=x

@winter patrol is that right?

[winter patrol]

did you mean
10a-6b=x

[willow osprey]

yes

is that correct?

[winter patrol]

then yes

[willow osprey]

Do i have to do brackets?

[winter patrol]

what brackets

depends on how you're writing your work

[willow osprey]

okay

[winter patrol]

if you wanted to be efficient and do something like
$$5a+3b -\red{(}-5a+9b\red{)} = x$$
those red parentheses would be needed

[woven radish]

ℝamonov

[willow osprey]

why need i these red parentheses?

[winter patrol]

to indicate subtraction of the entire expression that was originally on the right

[willow osprey]

aha

ok thank you for your help

[devout snow]

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[regal hemlock]

Hello, can someone please explain this to me:

[regal hemlock]

Starting from how the third line was simplified, specifically the (1/2^logn ) part

I mostly don't get line 3 and 4

[violet wind]

seems like they applied geometric series formula

but it's not an infinite series

so 2 is actually an overestimate for it

it should be < cn(2)

[regal hemlock]

Oh so it's supposed to be 1+1 where the second 1 is 1/2+1/4+ 1/2^logn?

[regal hemlock]

Can you please explain why the last one is 1/2^logn?

[violet wind]

n is 2^log(n), right

so 1/2^logn is 1/n

well, since they factored n out

but really it should be 1/(cn) since they factored cn out

or more likely the original sum should be cn + cn/2 + ... + c

so that's another mistake in the slide

[regal hemlock]

Wait, why is 1/2^logn = 1/n?

[violet wind]

because 2^log(n) is n

base 2 is implied for the logs here

[regal hemlock]

Assuming is base 2?

Oh

Why do we not assume natural log or base 10?

[violet wind]

in CS all logs are assumed to be base 2 unless stated otherwise

since those are by far the ones that occur the most

[regal hemlock]

Thank you!

[devout snow]

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[open apex]

n is a natural number and its bigger or equal than 1, demonstrate that for each n there are x, y, z that follow the condition

[devout snow]

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[open apex]

<@&286206848099549185>

[devout snow]

@open apex Has your question been resolved?

[devout snow]

@open apex Has your question been resolved?

[stone stump]

induction

try it for small values of n first

see how you can use the previous solutions you found

[devout snow]

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[ionic parrot]

A cyclist reaches an average of 24 km/h on the flat road sections between A and B. In the uphill sections, his speed drops to an average of 12 km/h. In the downhill sections, the average speed rises to 30 km/h. The road from A to B is covered in 4 hours 33 minutes. The way back takes 5 hours. The total distance traveled is 198 km. Determine the number of flat kilometers, the number of kilometers uphill and the number of kilometers downhill from A to B.

[devout snow]

@ionic parrot Has your question been resolved?

[ionic parrot]

<@&286206848099549185>

pls someone

<@&286206848099549185> someone

[devout snow]

@ionic parrot Has your question been resolved?

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[ionic parrot]

.close

[devout snow]

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[devout snow]

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[worthy vault]

Why is c wrong?

[worthy vault]

i know the answer is b,d

but why is c not an answer

[signal crag]

Domains

And general solutions

Put it into Desmos

[worthy vault]

ok i see now

[signal crag]

You will see something cool

[worthy vault]

is this c?

cuz 45 degree diagnol flip of f(x)

[signal crag]

You can also test points

The inverse is just the swap of the x and y coordinates

But yeah you’re right

C is the answer

[worthy vault]

is this also c because of IVT therom

[signal crag]

I don’t know about ivt

But yeah from common sense I can say that c is correct

[lone ravine]

Intermediate value theorem

[worthy vault]

would this be c and e
because of cast rule?

[signal crag]

Interesting

[worthy vault]

@lone ravine

[lone ravine]

Yes, should be correct

[worthy vault]

is this exponential or quadratic

cuz the second difference is 0.14, 0.15

and the difference in ln Q is linear of 0.17

@lone ravine

would this be exponential then

@zenith jacinth

.close

[devout snow]

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[torpid ocean]

how would we use set operations to solve this

[cursive totem]

$$|A\cup B\cup C| = |A| + |B| + |C| - (|A\cap B| + |A\cap C| + |B\cap C|) + |A\cap B\cap C|$$

[woven radish]

please request a new nickname

[torpid ocean]

yes

[torpid ocean]

what can we do with this?

[cursive totem]

this is the formula for the order of a union of sets

do you know what this means?

[torpid ocean]

yeah, but what about the none?

when no one picks

[cursive totem]

just subtract from 40

[torpid ocean]

okay

This is an easier example

I did 42 = 38 + A - (8+7+10) + 3

but got 26

and it wasn't rtiht

[cursive totem]

did you subtract 7

[torpid ocean]

no

[cursive totem]

why not

[torpid ocean]

Also how would we take account of something like this

[cursive totem]

simplify first whats in the parentheses

M U B^c, so math and not biology

[torpid ocean]

yeah

[cursive totem]

intersect with art

[torpid ocean]

seems a bit difficult to do without a venn diagarem

[torpid ocean]

any ideas

[devout snow]

@torpid ocean Has your question been resolved?

[torpid ocean]

<@&286206848099549185>

[devout snow]

@torpid ocean Has your question been resolved?

[torpid ocean]

<@&286206848099549185>

[devout snow]

@torpid ocean Has your question been resolved?

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[devout snow]

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[serene tundra]

Yo what does this converge to?

[serene tundra]

$\sum_{n=2}^{∞} (-1)^n /ln(n)$

[woven radish]

Alexandria!

[warped relic]

$\sum_{n=2}^{\infty} \frac{(-1)^n}{\ln(n)}$

[woven radish]

VulcanOne

[warped relic]

So like 1/ln2 -1/ln3 + 1/ln4 -1/ln5 +...

[serene tundra]

yup

[warped relic]

So if 1/1 -1/2 + 1/3 -1/4 +... = ln(2), we can multiply both sides by -1 and add 1 to the other side

It will be 1/2-1/3+1/4-1/5+...=1-ln(2)

I think we can do something similar with our friend here

Wait

Hmm

It's like

Less than 1

[serene tundra]

yeah

.close

[devout snow]

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[frail lake]

Im trying to prove this via strong induciton, here's what i have so far:

Base case: If there is only 1 container, it must contain all m items, and m > 1.

Assumption: Assume that if m items are put into k containers with m > k, k = 1,2,3,...n, at least one container must contain more than one item.

But I don't really know how to do the induction step.

because i know that m > n, but that doesn't ensure that m > n+1...like can't it be m >= n+1?

<@&286206848099549185>

[devout snow]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[frail lake]

.close

[devout snow]

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[worldly juniper]

.close

[devout snow]

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[acoustic vault]

oh nvm

lol

[worldly juniper]

dang 💀 i was just about to put it in the discrete channel

[acoustic vault]

you did it completely right

[worldly juniper]

wait hold on let me just

.reopen

[acoustic vault]

just at the end you have k^3 and you subtract a positive number

k^3-2k^2+5k

[devout snow]

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[acoustic vault]

you can just say k^3-2k^2+5k<k^3-2*5k+5k

[worldly juniper]

oop

bot did a funny

[acoustic vault]

and thenyou have k^3-5k < k^3

which obviously holds

[worldly juniper]

if you don't mind me asking, where did you get this part from?

[acoustic vault]

-k^2< -5*k

becuase of our restriction on k

[worldly juniper]

i'm sorry, i don't think i follow

[acoustic vault]

we have $k\geq 5$ i think?

[worldly juniper]

where did we obtain the values on the right of k^3? in the part where i stopped, that's it

[woven radish]

Køter

[worldly juniper]

oop

oh so then you'd multiply both sides of that inequality by -k?

[acoustic vault]

ye so you have -k*k < 5k

and you end up with k^3-5k right?

[worldly juniper]

unfortunately i think i'm still not following

we end up with -k*k < 5k but how do we get to k^3-5k

[acoustic vault]

oh ok you ended up with k^3-2k^2+5k

and you want to show this is < k^3

first what you do is use the fact that -k*k<5k so you have

$k^3-2k * k+5k < k^3-2 * 5k +5k$

[woven radish]

Køter

[acoustic vault]

right? we used -k*k < 5k. is that clear so far?

no wait im so sorry im the one fucking up lmao

wait let me think sec

i just flipped it sorry

[worldly juniper]

k i was like 😅 how'd we go from modifying an inequality to strictly modifying one side of an inequality

oop

[acoustic vault]

i was right before i just gaslit myself

if you have -2k^2 it will be less than -2 * 5k because k>5

this makes sense no?

[worldly juniper]

not at all, my apologies

ok let me just

reframe what i have

strictly speaking, this is all that i've done up to

[acoustic vault]

yea then you multiplied it out right?

[worldly juniper]

i don't know if this is the correct next step, all i'm trying to prove is that the summation above on the left of the inequality is less than ((k+1)-1)^3

yep yep

[acoustic vault]

this is 100% correct

but you have to look at only $k^3-2k^2+5k$

[worldly juniper]

i multiplied it out but im not sure if i should be using whats on the right to prove that since its an intermediary step ^^;

oop

[woven radish]

Køter

[acoustic vault]

and you rewrite it

show that its less than other stuff

which in turn makes it easier to say that its less than k^3

[worldly juniper]

yeah, its the rewriting part thats getting me

ough

[acoustic vault]

because just looking at it. its not totally obvious that its less than k^3 right?

[worldly juniper]

nope

i was thinking, the -2k^2+5k obstructs that

[acoustic vault]

so we use our initial condition that $k\geq 5$

[woven radish]

Køter

[acoustic vault]

if we multiply with -1 we have $-k < -5$

[woven radish]

Køter

[acoustic vault]

then multiply with k

$-k * k < -5k$

[woven radish]

Køter

[acoustic vault]

so now we go back to our original $k^3-2k * k +5k$

[woven radish]

Køter

[acoustic vault]

we rewrite k * k to -5k

but obviously then its not equal anymore it will be less than

so $k^3-2k * k + 5k < k^3-2 * 5k+5k$

[woven radish]

Køter

[acoustic vault]

only the 2nd term is touched and we made it into something bigger

all good ?

[worldly juniper]

i'm missing something between the second to last step and last step

like i get how k squared and flipped is less than -5k but how does that translate to the last bit

just because we supplant the second k with 5k because of the inequality?

not sure how that applies to k^3 itself on its lonesome

[acoustic vault]

but we didnt touch k^3

or the last 5k

we only changed -k*k to -5k

[worldly juniper]

yeah

but how would that translate to k^3?

[worldly juniper]

since k^3 is the simplified version of the right side of the induction hypothesis applied for k+1

[acoustic vault]

ye ik we can get to that

i just want to make clear you understand this step first

[worldly juniper]

i do now, though i'm not sure this'd be admissible to my prof haha

[acoustic vault]

$k^3-2k * k + 5k < k^3-2 * 5k+5k=k^3-5k$

[woven radish]

Køter

[acoustic vault]

-10k+5k=-5k right

[worldly juniper]

ja

[acoustic vault]

now k^3 -5k < k^3

because k is positive

so k^3 subtracted something positive is obviously always less than k^3 itself

[worldly juniper]

holy sh--

dude, thank you

i've been staring at this since yesterday

[acoustic vault]

yea np

[worldly juniper]

also

i think you might have to close this channel since the bot glitched

[acoustic vault]

we basically shown that something even bigger than $k^3-2k^2+5k$ is still less than k^3

[woven radish]

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

[acoustic vault]

ye sure

.close

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[wooden bridge]

hi can i get help seeing where i went wrong?

apparently this is the correct answer

original problem:

[restive river]

its x^4 on denom

I spent like 10 minutes trying to figure out where u went wrong lol

[wooden bridge]

wait omg

[restive river]

ur working out is correct if it was x^2 though 🙂

[wooden bridge]

it's always the little errors .-.

[wooden bridge]

ty for the catch!

.close

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[mild basin]

how is this being simplified?

I tried but to no avail

[pseudo basin]

no, you stopped midway.

now add the fractions and factor out (2x-3)^3 from the numerator.

what's that stray 40x doing there?

[mild basin]

I need to keep 40x+32 in brackets?

[pseudo basin]

well of course you do??

why would you rip them off

(a+b)c isn't the same as a+bc

[mild basin]

my bad.. associativity is still something I need to practice on. can only remove brackets when addition is the only operator left

I think I factored out the (2x-3)^2 but I'm not sure about the last term(s)
20(20x-3)

it's weird that I still don't see how I can get to this

maybe I need brackets around - [20(20x-3)]? Otherwise I will be distributing onto -20 and (20x-3) individually?

wait no, maybe that's fine

brackets are only needed to make multiple terms into one

20(20x-3) is still one term, I think

like 20x

it's the first terms I can't multiply.. powers are different

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[restive river]

Hey

[restive river]

How do you do reciprocals in transformations

[warped relic]

Hmm

Could you show an image of what you mean?

[devout snow]

@restive river Has your question been resolved?

[restive river]

1 over m(x-3) @warped relic

You there @warped relic

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@restive river Has your question been resolved?

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[brazen star]

How to explain why the product of the slope of 2 perpendicular lines is -1

[brazen star]

To a 6th grader

[bleak rain]

Think of x and -x

Theyre the most obvious perpendiculare lines other than the axis

[bleak rain]

From those you can get all other perpendicular lines by rotating and translating them around the plane

And rotating /translation dont change the product of the slope cause when you rotate they both change together in a way that conserves the product

[brazen star]

They havent learned translation and rotation

[warm lantern]

If they want to see something more concrete, you can draw a diagram of two perpendicular lines L1 and L2 (passing through the origin O for simplicity)
Pick one point P1 from L1
Pick one point P2 from L2
If they know the pythagorean theorem then the distance from O to P1 squared plus the distance from O to P2 squared = distance between P1 and P2 squared

make sure to use a diagram

a big one

I think from there you can show that the product of the gradients are -1

I'm going to check this myself now

[warm lantern]

or you can do this ⤴️

[arctic field]

image that isn't chopped

[warm lantern]

I suppose you could start off with snow's explanation

if that's not enough then dabble's explanation

and then mine

[restive river]

You are so based big bro

[arctic field]

from when i had plans for writing a textbook

[restive river]

[warm lantern]

O7

[restive river]

Now you camp #latex-testing on a discord server

Gg

[arctic field]

[brazen star]

Ima try

[arctic field]

does the graphic make sense to you?

[warm lantern]

and by the way, I just checked this. It works

lemme get an image of it

[brazen star]

Ok now it doesnt

[arctic field]

where doesn't it

[brazen star]

Nvm it made sense now

[arctic field]

the rise and run ive labelled correspond to m1

then m2 reuses those in its calculation

the negative sign is introduced because in the diagram the rise of m2 is the run of m1 but going downwards

[warm lantern]

done

[devout snow]

@brazen star Has your question been resolved?

[brazen star]

.close

[devout snow]

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[pulsar ermine]

I’m not quite sure whether this is true

[sick fulcrum]

try to prove it then

or find a counterexample

[pulsar ermine]

I don’t think it’s true, since the identity function isn’t in H

[sick fulcrum]

well you answered your own question didn't you?

be more confident in your logical reasoning

[pulsar ermine]

Suppose a and a-1 was in H, the composition would be the identity function which doesn’t belong in H hence it’s not closed under the binary operation

[sick fulcrum]

Eh you don't need to do it that way

let eps denote the identity function

then you know eps(i) = i for all i

but then by definition of H...

[devout snow]

@pulsar ermine Has your question been resolved?

[devout snow]

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[slim radish]

What did I do wrong

[stark anvil]

It's kind of difficult to understand what you did

But since we have been given the value of $cos\beta =- \frac{2}{5}$ you can find the value of $sin\beta$ and substitute it in $cos(\alpha - \beta)$ for your answer

[woven radish]

ColdTee

[stark anvil]

You have to do the same with $sin\alpha$ to find $cos\alpha$

[woven radish]

ColdTee

[devout snow]

@slim radish Has your question been resolved?

[slim radish]

no i already know that

its kind of hard to see

but i already made two triangles and labeled them

and its csc

so i know that i need to find sin first

then reverse it

[devout snow]

@slim radish Has your question been resolved?

[devout snow]

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[warm turtle]

Hi! Is there a way for me to draw 2 circles in a graphing calculator? I need to determine where two circles lie in order to see whether they intersect or not. This is the math problem in the first attachment translated:

Draw the circles c1 and c2 and calculate exactly d(c1,c2).

In the second attachment there is more explanation, which I'll provide the translation for here:

To calculate the distance between two circles, use the line through the centres of the cirels. If the circles intersect or touch, the distance is 0. If the circles do not intersect, they lie completely outside each other or one circle lies inside the other. In the figure you can see that in the first case the distance is equal to d(c1,c2) = MN - r1 - r2 and in the second case to d(c1,c2) = r1 - MN - r2.

Thanks in advance!

[main python]

I think you have to draw without a graphing calculator

[warm turtle]

second attachment:

But is there a way to find out without having to draw?

because drawing takes more time

[main python]

Your book explains it

if the distance between the centers is equal than the sum of the radius

what do you think happens?

[warm turtle]

i'm not sure what you mean, sorry

than the sum of the radius is what?

im confused by your sentence a little bit

[main python]

Look at this drawing

You can find easily the distance between M and N

call that d

In the first example, d is greater than the sum of the radius (r_1+r_2)

look at what happens

[warm turtle]

i'm not sure what happens

like

if u have MN

and u remove r1 and r2 u get d

but like

[main python]

d is everything

[warm turtle]

I don't really understand what is supposed to happen

[main python]

wait

[warm turtle]

d is the distance between the two circles

[main python]

call d = d(A,B)

[warm turtle]

but d doesn't go to the middle of the circle i thought

[main python]

what?

[warm turtle]

d is the distance between the two circles, like where they touch

or at least thats what the image shows

[main python]

d is the distance between the 2 centers

[warm turtle]

it says if u want d u have to do MN - r1 - r2

[main python]

NO

d is the distance between the centers

[warm turtle]

im confused D;

[main python]

d is the distance between the centers

[warm turtle]

okay

[main python]

the blue line is what you are thinking it is

[warm turtle]

yeah

[main python]

So we have d = a+b+c

[warm turtle]

okay

[main python]

In this case obviously d> a+b

correct?

[warm turtle]

yes

because d also has c

[main python]

What about this one

should be B on the right

[warm turtle]

well

they're just touching

so

wouldn't d be like 0

or no

wait

[main python]

d is the distance between A and B

[warm turtle]

well d is the same as r1+r2 then

why would it say d(c1,c2) = MN - r1 - r2

[main python]

ok

lets use that

then d is zero

and MN is r_1+r_2

and this one

[warm turtle]

i'm sorry i don't really understand

what i'm supposed to do with this

[main python]

find if d>0, d=0 or d<0

Basically you know that if circle A and circle B don't intersect -> d>0
If they are tangent -> d = 0

if they intersect twice -> ?

[warm turtle]

tangent?

[main python]

intersect once

[warm turtle]

like just touching?

)( like this?

[warm turtle]

d<0?

[main python]

yes

[main python]

yes

what about this one

[warm turtle]

what am i supposed to determine? d?

[main python]

Yes

[warm turtle]

r1 - AB - r2?

i dont really get it, d really seems to be the distance between the circles and not middlepoints

because they literally have this example in the book

to calculate d

[warm turtle]

and it says this

and then you'd get this

it does say though that to determine the distance u need the line in between the two middlepoints

so that's a bit confusing to me

c1 and c2 are the middle points and the line between them would be d according to you

but then why do they draw it like that in the example it's so confusing

but yes i understand how it would be the distance between the two centres

because you're excluding the radius of the circles

and what's left is the distance between the middle points

but then again i'm not entirely sure

[main python]

Let's recap

If A and B don't intersect -> d > 0

if A and B intersect once -> d = 0

if A and B intersect twice -> d<0

In this case

d<0

but they don't intersect

[warm turtle]

because it's inside the other circle

[main python]

Then you would have to check that

[warm turtle]

okay well i think i understand now. sorry for being so stubborn, i just didn't realize that i had to take the other part of the circle into account

i think i can continue now

thank you for all your help!

i'm gonna close the chat

.close

[devout snow]

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[devout snow]

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[cyan rivet]

Quick question: how did my teacher go from this step to the next?

[cyan rivet]

Pre Calc, where did the 1 come from?

[cyan rivet]

@dry robin

<@&286206848099549185>

[devout snow]

@cyan rivet Has your question been resolved?

[cyan rivet]

I need help ;((

Test in 20 mins <@&286206848099549185>

[radiant dune]

The numerator is 2sec²x+2secx. Take 2secx common, factor it out, and you get,
2secx(secx+1)

You can test that this is correct by distributing the 2secx

[cyan rivet]

Ok I get it now, thank you 🙂

[devout snow]

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[devout snow]

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[stable merlin]

only number 8

i did (x+7)square + x square = (x+8) square

[pseudo basin]

use the symbol ^ for exponents

(x+7)^2 + x^2 = (x+8)^2

ok, so did you do anything after this

[stable merlin]

then im stuck on 3x^2 + 30x - 113 = 0

[pseudo basin]

how did you get to that?

show your work.

i think you screwed that one up in at least two places, but i'd like to see it.

[pseudo basin]

,rccw

[pseudo basin]

let's see here.

2x^2 + 14x + 49 = x^2 + 16x + 64 is correct

can you explain what you did afterwards?

in words, if possible.

[stable merlin]

I tried to solve it

[pseudo basin]

i want to know your actions, not your goal or intentions.

[stable merlin]

i made it in a quatratic

[pseudo basin]

you "tried to solve it" sure but that describes your entire sheet of work. im focusing on the one line that i know you screwed up in.

"making it a quadratic" is not a description of an algebraic operation.

do you want me to interrogate you on how you "made" "it" "in a quadratic"?

surely not.

we would both be unhappy if it came to that.

so let me instead say this:

you wanted to move all the terms to one side, but in doing so, you subtracted x^2 + 16x + 64 from the right but then added the same to the left (and something very strange went down with the constant, but that's a different matter)

[stable merlin]

yes

[pseudo basin]

do you understand why this is erroneous?

[stable merlin]

no

[pseudo basin]

when you manipulate an equation in algebra, you do the same thing to both sides

you cannot add to one side and subtract from the other -- doing so would throw the equation off balance.

[stable merlin]

its Quatratics tho

[pseudo basin]

do you mean "This is a quadratic equation, therefore I proclaim the thing you just stated flat-out doesn't apply"?

[stable merlin]

Yes

[pseudo basin]

you're wrong

the laws of algebra apply no matter if you're dealing with a linear or quadratic or any other type of equation

you still cannot add to one side but subtract from the other

[stable merlin]

Ok what do i have to do now

[pseudo basin]

well, in keeping with the "do the same thing to both sides" principle, i would have you instead subtract (x^2+16x+64) from both sides.

or, if you're squeamish about it, or if you're not 100% certain you can do it without fucking up, make it three steps:

1. subtract x^2 from both sides
2. subtract 16x from both sides
3. subtract 64 from both sides
   (in no particular order)

[stable merlin]

Ok wait

[pseudo basin]

would you like me to check your work once you do this, or do you think you can continue on your own?

[stable merlin]

Check

[placid rover]

Quadratic or no quadratic, its an equation

x, the variable, merely stands for a number 'which can change'

To solve the equation, you are finding the x's which satisfy it (make the equals sign true).

[pseudo basin]

did you do it in one step somehow?

or did you do as i told you to in three steps

[stable merlin]

three steps

[pseudo basin]

and why didn't you show me those three steps?

[stable merlin]

i did 2x^2 - x^2 then 14x - 16x then 64-49

[pseudo basin]

you say you want me to check your work, but then you don't show the work that needs to be checked.

[stable merlin]

i do it in the head

[pseudo basin]

bad

don't do things in your head, that's how you commit ten different fuckups all at once

and for consistency's sake the constant would've been 49 - 64, not 64 - 49. though you did put it down correctly as -15, so whatever.

[stable merlin]

I wrote wrong in chat

i got answer 3

x = (x+3)?

[pseudo basin]

incorrect

no, x is not equal to (x+3) [and in fact, those two are NEVER equal]

x^2 - 2x - 15 does not factorize as (x+5)(x-3). expanding (x+5)(x-3) would have given x^2 + 2x - 15. check your factorization.

[stable merlin]

btw this is the triangle

[stable merlin]

my bad i forgot to put (x-5)

[pseudo basin]

ok, great, now that's in order.

now only one step remains: to tell which of the two values you got for x is the one you care about and which needs to be discarded

[stable merlin]

the correct one must be 5 since -3 is rejected

and it said positive interger

so 5

what i have to do now

[pseudo basin]

you got the value of x, and you know the sides of your triangle in terms of x...

[stable merlin]

so i just do 5+7 and 5+8?

.close

[devout snow]

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[craggy plaza]

hi i would like to ask how can i find 0.236 here?

But Idk how find p(convict n guilty)

[devout snow]

@craggy plaza Has your question been resolved?

[craggy plaza]

<@&286206848099549185>

[devout snow]

@craggy plaza Has your question been resolved?

[craggy plaza]

<@&286206848099549185>

[tepid smelt]

what happen

[craggy plaza]

the question has not been solved yet

[devout snow]

@craggy plaza Has your question been resolved?

[devout snow]

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[restive river]

Can someone help me with turning this into a reduced row matrix

[restive river]

okay

set up the matrix

So far this is what I have

I’ve already done 1/4R1 —> R1 and then 2R1–R2

[pseudo basin]

you really should not write minus one-half as -.5

that decimal point will be lost when you least expect it and you will be stuck writing minus five instead.

the same goes for any decimal between 0 and 1 -- NEVER omit the leading zero on those, EVER!!!

[restive river]

Oh ok

[pseudo basin]

anyway, it sounds like you tried to kill two birds with one stone and they both got away,

or you've written down some intermediate steps and chose not to show us

[restive river]

I forgot 1 picture my bad

I’ll show it

This is all I’ve done so far

[grave nebula]

I would times 2 the bottom row

-4x -4y = 12

[pseudo basin]

ok, hold on...

[grave nebula]

That would then nullify x

-6y = 18

[pseudo basin]

that... okay, yeah, this tracks.

[grave nebula]

y = 18/6

[pseudo basin]

@restive river the thing to do here would be to just keep going

[grave nebula]

y=3

[restive river]

@pseudo basin I just done know what step to do next

[pseudo basin]

well, do you recognize that your matrix as it currently is, i.e. $$\left[\begin{array}{cc|c} 1 & -\tfrac12 & \tfrac32 \ 0 & -3 & 9 \end{array}\right]$$ is already in row-echelon form?

[woven radish]

Ann

[restive river]

Yes but I still need to reduce it

[pseudo basin]

well,

can you tell me at least one way in which it is not yet reduced?

do not yet suggest a course of action.

[restive river]

Well the -1/2 needs to be a 0 and the -3 needs to be a 1

@pseudo basin

[pseudo basin]

how can you make the -3 into a 1?

(now suggest a course of action,)

[restive river]

Divide by -3 right?

So -1/3R2–>R2

Then I could do 1/2R2–>R1

And that would be the final answer right @pseudo basin

[pseudo basin]

yeah, except don't forget the closing ] on your matrix...

[restive river]

Ya

Thanks

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[restive river]

Assuming x is a positive real number, find the minimum value of:

[restive river]

$$x^3+48/x$$

[woven radish]

Rijad

[restive river]

How do I even start

[supple knot]

do you know derivatives

[restive river]

No

I wanna solve it without any derivatives

[supple knot]

,w plot x^3 + 48/x

[woven radish]

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[supple knot]

is x > 0?

[restive river]

Yes

[supple knot]

,w min x^3 + 48/x for x > 0

[woven radish]

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[restive river]

I just plugged in whole numbers and got 32

[supple knot]

[restive river]

But I need a concrete way of finding out how to solve these kinds of problems

Hello?

[supple knot]

no idea

derivatives are concrete

[restive river]

Hmm

I guess I will find out myself

Nonetheless, thanks

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[valid axle]

Is it convergent or divergent, if it is convergent then find the value it converges to $\sum_{k=0}^{\infty} \frac1{2^k}$

[sonic smelt]

\sum

This is an infinite geometric sum, do you know the general formula of it?

[woven radish]

afeAlway

[valid axle]

So I came to the conclusion that it is convergent using the ratio test

But I'm not sure how I am supposed to find the number it converges to

[woven radish]

hibyehibye

[valid axle]

in any case the answer is two

[restive river]

@valid axle

[valid axle]

yeah, shit that is what they meant (they said tip: geometric sum) I assumed they mean that I was supposed to check the value of r.

let me try it and get back to yall

Thanks guys

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[dry sky]

any help on why the top integral works but the bottom doesnt?

[devout snow]

@dry sky Has your question been resolved?

[devout snow]

@dry sky Has your question been resolved?

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[uneven bloom]

could someone help me with proving this trigonometry identity

[uneven bloom]

$\frac{cos(A+B)}{cos(A-B)}=\frac{1-tan(A)tan(B)}{1+tan(A)tan(B)}$

[woven radish]

Astassa

[uneven bloom]

<@&286206848099549185>

[bronze oak]

you can take it from here i belive

[uneven bloom]

why do you divide the top and bottom by cos(A)cos(B)?

[devout snow]

@uneven bloom Has your question been resolved?

[uneven bloom]

ok cool, got it

thank you

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[jaunty gazelle]

Can someone explain how the circled part factored out to that?

[unreal thunder]

@jaunty gazelle

That is the identity (a+b)^2

notice the middle term is $\frac{1}{2}= 2 * x^2 \frac{1}{4x^2}$

[woven radish]

Dubleyou

[jaunty gazelle]

How does 1/2 equal that..?

[cloud remnant]

Why not?

[unreal thunder]

cancels the x^2 and the gcf 2

[jaunty gazelle]

Oh wait I see

Hold on

[jaunty gazelle]

I’m still confisedzzz hpw@did we get 1+x^4-1/2+1/16x^4…

confused*

How did we get*

[unreal thunder]

can you rewrite $x^4+\frac{1}{2} + \frac{1}{16x^4}$ as some a^2 + 2ab+ b^2 ?

[woven radish]

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

[jaunty gazelle]

At least, how did we get from 1+(x^2-1/4x^2)^2...

[unreal thunder]

well 1- 1/2 is 1/2

[unreal thunder]

so the rest of them is arranged to this

[jaunty gazelle]

Not sure if I'm overcomplicating this or not...were not actually adding the one to anything right...were just distributing it. All were doing is foiling (x^2-1/4x^2)^2...

[unreal thunder]

It seems like you need help from step 1

[jaunty gazelle]

I might need help from step 1 yes..

[unreal thunder]

I don't know what exactly are you doing, i guess it's arclength

i'll help with you with algebra for now

We know $y'=x^2-\frac{1}{4x^2}$ right?

[jaunty gazelle]

I'm doing arc length but the biggest trouble right now is algebra..

[woven radish]

Dubleyou

[jaunty gazelle]

Yes I was able to get that.

[unreal thunder]

now we have to "square this term and add one" so that it fits with arclength equation