help-26

[toxic marsh]

power rule

[empty sail]

My suggestion, make all the x's have powers so you can combine it all. Like $\sqrt{x} = x^{\frac{1}{2}}$ so then something like $x \sqrt{x} = x^{\frac{3}{2}}$

[thorny flame]

dldh06

[toxic marsh]

yea I did that in the scratch work 😛

[empty sail]

Doesn't look like it

Because if you did $$x^5 \sqrt{x} = x^{\frac{11}{2}}$$

[thorny flame]

dldh06

[empty sail]

And I don't see a 11/2

[topaz sinew]

@toxic marsh Has your question been resolved?

[toxic marsh]

oh

[topaz sinew]

Channel closed

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[toxic marsh]

ty

[topaz sinew]

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[neon iron]

I need help

[neon iron]

,rotate

[neon iron]

So I know I should split this into cases

But from there…

[wooden matrix]

from there..... what?

[neon iron]

Well

My cases were I think x>4 and x<4

[wooden matrix]

no

there are 3 cases.

[neon iron]

x=0?

[wooden matrix]

Use the hint you've cropped out

How do you usually handle abs value equations

[neon iron]

Positive or negative

I think

Right?

[topaz sinew]

@neon iron Has your question been resolved?

[neon iron]

<@&286206848099549185>

Please

[vital relic]

ooh this is fun

@neon iron hi

If there was just one absolute value, do you know how to approach?

^ Are you familiar with this

[drifting swift]

i think this might be technically correct but misleading

[vital relic]

Was gonna generalise to 3

misleading how?

[drifting swift]

someone who's not fluent in algebra can easily get confused by this

i mean, even i took a second to parse it

anyway, ||the most sensible case breakdown is into three cases: x ≥ 4; -4 ≤ x < 4; x < -4. the rest is algebra & linear inequalities||

[drifting swift]

maybe misleading isnt the right word. 'unnecessarily confusing' would be better

[neon iron]

Yes

[vital relic]

Yh maybe. I was taught to solve equations/inequalities by making sure you had both forwards and backwards directions.

Anyways --- take Ann's hint. You use the same idea

except now you need to split into 3 cases

[neon iron]

How about x>4

[vital relic]

(Case 1) OR (Case 2) OR (Case 3)

That was what I was trying to get at

[neon iron]

Wait oh

[drifting swift]

might be better to phrase it in terms of unions and intersections of per-case solution sets

also @neon iron if you want you can put the point 4 into the middle case

[neon iron]

I see

[drifting swift]

it doesn't really matter exactly which cases the boundary points go in

they can even be their own cases

[vital relic]

All of this thinking comes from the definition

Notice that x = 0

Can be split into the top or bottom (it doesn't matter because 0 = -0)

[neon iron]

So going one case at a time, if x>=4 then

What can I do

[drifting swift]

if x ≥ 4 then what is |x-4|?

[neon iron]

x-4

[drifting swift]

and what is |x+4|?

[neon iron]

x+4

[drifting swift]

so what does your inequality become?

[neon iron]

x-4+x+4<=10

2x<=10

[drifting swift]

exactly

do the same rewrite procedure in each case to reduce the inequality to linear

[neon iron]

x>=5

[drifting swift]

...what

[neon iron]

Flip the sign

[drifting swift]

why flip the sign?

[neon iron]

When dividing in inequalities, right?

[drifting swift]

no

just because you're dividing by something in an inequality does not by itself mean the sign must be flipped

otherwise, according to your own logic, 10 < 30 implies 5 > 15

because "fLiP tHe SiGn"

you flip the sign when you multiply or divide by a negative number

[sweet shard]

also, always check convenient numbers when doing inequality problems. x=1: 2 * 1 <= 10 is true but 1 >= 5 is false. it takes a second of more work but saves lots of bad mistakes

"sanity check"

[neon iron]

Ah

So then x <= 5

That is for the first case

Now for the next

The next case is a bit different

I would probably want to split it

[drifting swift]

do you mean the middle case or the left case

[neon iron]

Middle case

When x>=-4

x+4+4-x

8<=10

Hmm

[drifting swift]

x ≥ -4 is not the condition of the middle case

the condition of the middle case is -4 ≤ x < 4

[neon iron]

I split it

[drifting swift]

??

[neon iron]

Since I don’t know how to use the whole inequality

[drifting swift]

without knowing x < 4, you cannot conclude |x-4| = 4-x

[neon iron]

Right

[drifting swift]

but anyway, what you should conclude here is that ||the entire interval covered by the middle case, in this case [-4, 4), is part of the solution.||

[neon iron]

I don’t understand

[drifting swift]

what do you not understand?

[neon iron]

This

[drifting swift]

very helpful /s

which part of that message do you not understand?

[neon iron]

The hidden part

[drifting swift]

...

ok

fine

i thought i was going to get some more specifics out of you

but i guess not

are you willing to take a step back and having me try to appeal to some general knowledge that i imagine you should have?

[neon iron]

Idk Ig you are right

I mean you are

[drifting swift]

oh

[neon iron]

But I don’t see how you got there

[drifting swift]

how could i forget

your refusal to answer yes-no questions

[neon iron]

You didn’t ask one

[neon iron]

Yes

[drifting swift]

"You didn't ask one" [proceeds to locate and respond to the question that i DID ask]

okay so

are you familiar with the concept of intervals as geometric objects on a number line

[neon iron]

Yes

[drifting swift]

please do not reply-ping me so often.

but ok

[neon iron]

Ok

[drifting swift]

and you know that rather than speaking of individual 'solutions' to equations and especially inequalities, one should really speak of the solution set (which may contain only one number, or may be empty, or may even be infinite), yes?

[neon iron]

Yes

[drifting swift]

okay

in solving some inequalities, it is helpful to partition the number line into several subsets - typically these subsets are intervals and hence described by simple inequalities
then, on each interval, we reduce the inequality to one that is easier to solve, solve it, and take the intersection of the resulting solution set with the interval covered by the case

and afterwards, we take the union of all of those intersections, and that's the solution set of the original inequality

you are familiar with this principle, yes?

[neon iron]

Yes

[drifting swift]

ok

it applies here just fine

our second case covers the interval [-4, 4)

[neon iron]

Yes

[drifting swift]

and the inequality we got for case 2 is true unconditionally, i.e. its solution set is R

or (-∞, +∞) if you prefer

so the 'contribution' of this case to the solution set is (-∞, +∞) ∩ [-4, 4) = [-4, 4)

the whole solution set is [-4, 4) ∪ <something else>

[neon iron]

Ok

4+x+x-4

Same thing

Hmm

This is for x<=-4

What did i do wrong?

[sweet shard]

can you write the whole inequality

[neon iron]

4+x+x-4<=10

[sweet shard]

do this

[neon iron]

Ok

[sweet shard]

try this again

[neon iron]

4+x+4+x<=10

2x<=2

x<=1

That’s a bit controversial

Because we had x<=5 and x in [4,-4)

I’m stuck

[sweet shard]

ok. start with $x \le -4$. simplify individually $|x+4|$ and $|x-4|$ separately

[thorny flame]

riemann

[sweet shard]

use this definition

[neon iron]

|x+4| is -x-4

|x-4| is -x+4

[sweet shard]

plug in x=-5

[neon iron]

-x-4-x+4

0

Hmm

[sweet shard]

this is also wrong

[neon iron]

I hate this problem tbh

[sweet shard]

everyone hates math at some point

[neon iron]

-2x<=10

[sweet shard]

looks good

[neon iron]

X>=-5

This makes the answer [-4,4) U [-5,5]

Right?

[sweet shard]

looks good, just simplify

[neon iron]

Simplify what?

[sweet shard]

your interval

[neon iron]

In what way?

I don’t know how to simplify it

[sweet shard]

draw a picture

[neon iron]

I know they overlap

[sweet shard]

great. what are the endpoints of which they overlap

[neon iron]

Except one is nonstrict

The -4)

[sweet shard]

do you know the definition of the union?

[neon iron]

Oh wait

[neon iron]

Yes

[sweet shard]

use the definition and you'll get it

[neon iron]

This right?

[sweet shard]

no i wouldn't tell you to simplify if it were

[neon iron]

I do y really know the definition then

[sweet shard]

https://www.sangakoo.com/en/unit/union-intersection-and-complementary-of-intervals#:~:text=Union of intervals,belong to the second one.

[neon iron]

[-5,4)

Did I simplify it

[sweet shard]

write this in inequality notation

[sweet shard]

and spend 3 minutes reading the first half of this before guessing again

[neon iron]

-4<=x<4 and -5<=x<=5

[sweet shard]

"and" is different from "or". the former refers to intersection $\cap$ and the latter refers to union $\cup$

[thorny flame]

riemann

[neon iron]

So it’s or

I’m really confused

[sweet shard]

do the first two examples here

ask questions about this

[neon iron]

I mean

[sweet shard]

there's at least one mistake in there, can you find it?

[neon iron]

Our intervals overlap

[neon iron]

How is x both less than 9 and 11

[sweet shard]

your lesson is to understand unions right now

[neon iron]

Which is it

[vital relic]

Have you seen set notation?

[mortal thunder]

Isn't 8 less than both 9 and 11 ._.

[neon iron]

[sweet shard]

tattoo this on your hand

[neon iron]

I@don’t see any mistakes

[vital relic]

$$A\cup B := {x:x\in A \textnormal{ OR }x\in B}$$
$$A\cap B := {x:x\in A \textnormal{ AND }x\in B}$$

[neon iron]

I’m not allowed to get tattoos

[thorny flame]

Shuri2060

[sweet shard]

it was simply a joke. the point is to do more sanity checks

[neon iron]

Well

In our case

Is it the first or the second

I think the first

Do our union is ok

I also have to go soon

[sweet shard]

fix this with the correct word that corresponds to union

[neon iron]

Or

[sweet shard]

do you see how $(3, 9) \cup [7, 11] = (3, 11]$ from the definition? 10 isn't in the first interval, but it is in the second interval, and that's all you need to be in the union.

[thorny flame]

riemann

[sweet shard]

in english, the union represents all numbers in at least one of the intervals.

[neon iron]

Ok

So what did i have to simplify

[sweet shard]

re-write this as a single interval

but or instead of and

[sweet shard]

using the definition in this article

[neon iron]

I can’t

Since it’s or

[sweet shard]

this was an or and it was simplified to a single interval

you should really understand that example

[vital relic]

Can you see how these 2 intervals can be combined into one, because they overlap?

Terrible drawing, but hope it expresses the idea 🤔

https://i.imgur.com/Aevvkjw.png

[neon iron]

Ok [-5,-4)

Or

[sweet shard]

.

[sweet shard]

.

[neon iron]

[-4,-5]

[sweet shard]

.

[neon iron]

What am I not understanding about them

Hmm

[sweet shard]

.

[neon iron]

[-5,5]

I think that’s it

[sweet shard]

proof?

[neon iron]

I@ certainties

[neon iron]

Proof?

What proof?

[sweet shard]

it's about as long as this proof

[neon iron]

It was correct

Gotta go

Thanks for the help

[sweet shard]

hope you learned unions and intervals

[neon iron]

👍

.close

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[neon iron]

This is what I got going so far

Not sure if much of it is right. I'm not entirely sure how to go on from here

[topaz sinew]

@neon iron Has your question been resolved?

[topaz sinew]

@neon iron Has your question been resolved?

[topaz sinew]

@neon iron Has your question been resolved?

[vernal vale]

@neon iron still around?

[neon iron]

hii there!

sorry for the late reply (๑•﹏•)

@vernal vale nw if you can't help rn, but if you're not free within ~10 minutesish I think I'll head to bed since it's nearly 1am lol

[vernal vale]

oh

tight

nah im here

[neon iron]

oh thank you

[vernal vale]

so

just in case you were gonna come back

i made a thing

[neon iron]

whoa

[vernal vale]

https://www.desmos.com/calculator/ltngu6lph3

[neon iron]

thank you!!

[vernal vale]

might help illustrate idk

if its just confusing ignore it

well this is just part a i guess

[neon iron]

yeah, it's okay I've tried using desmos to visualise it before

[vernal vale]

and you maybe already had it

[neon iron]

no I didn't get this result though

[vernal vale]

that its just like

its just finding f(x) = f'(r)(x-r) for some root?

[neon iron]

is that it?

[vernal vale]

yea

f'(r)(x-r) is the equation of the tangent line at a root r

so where its equal to f(x) is the other intersection

youll get at least 2

at least for distinct

maybe only two

[neon iron]

so f'(r)(x-r) = f(x) will give me 2 solutions?

[vernal vale]

yea, the first is obvious

the second is the one a is asking for

[neon iron]

alright yeah well the first one is just the root, right?

[vernal vale]

yup 😄

since its a tangent line

[neon iron]

coolio

[vernal vale]

we expect it be equal there

as far as colinearity

[neon iron]

alright can you stay here whilst I work this out because I spent a good chunk of my time on part a x)

[vernal vale]

yea lol

[neon iron]

worried I'll mess up somewhere

ty!!

[vernal vale]

im not 100% sure what the easy way is for colinearity

i think youd just show like

pick 2 of them, up to you

since a unique line goes through 2 unique points

[neon iron]

oh I've a question

[vernal vale]

show the third point is on that line

given this choice

ok

[neon iron]

f'(r)

that's not f`(x) right

[vernal vale]

no

f'(x) is the derivative

[neon iron]

yep

[vernal vale]

f'(r) is a constant, its the slope of the function at point x=r

well it will be a constant, if you elect a value for alpha or beta or gamma

[neon iron]

oh, I see

[vernal vale]

otherwise its a function of alpha or beta or gamma

but not x

[neon iron]

hmm well could you go about showing me how you'd calculate f`(r)?

since I'm not really too sure

[vernal vale]

like you have that $f(x) = ax^3 + bx^2 + cx + d$ then

[thorny flame]

jan Niku

[vernal vale]

$f'(x) = 3ax^2 + 2bx+c$

[thorny flame]

jan Niku

[neon iron]

yup

[vernal vale]

so for example $f'(\alpha) = 3a\alpha ^2 + 2b \alpha + c$

[neon iron]

mhm, got that

[vernal vale]

and thats it

[thorny flame]

jan Niku

[vernal vale]

then the tangent line for root alpha is $(3a \alpha ^2 + 2 b \alpha + c)(x-\alpha)$

[thorny flame]

jan Niku

[neon iron]

oooh I see

[vernal vale]

oh wait

is solving this equality not super trivial

huh

lol

so i think

i think the trick is

you know (x-a) is a root

i think

then you can divide it out

[neon iron]

yeah it's a root

[vernal vale]

and know there wont be a remainder?

[neon iron]

yup, according to the factor theorem

[vernal vale]

then its a quadratic

which more or less trivializes finding the roots

i think

[neon iron]

so... would it be like (x - beta)(x - gamma) = f`(r)

[vernal vale]

yup 😄

here r would be alpha

[neon iron]

yup, I didn't see that first lol

and uh, then what?

[vernal vale]

so you get what like

i may just use symbolab

[neon iron]

whatever is easiest for you!

just thankful you're helping out lol

[vernal vale]

frig man

its still not super easy it

well

its a quadratic

a = 1

specifically a as the coefficient of the x^2 term

not to get confused

since running out of letters

[neon iron]

lol

[vernal vale]

a=1 since x hits x once

then

b = -beta-gamma

and c = beta * gamma - f'(alpha)

[neon iron]

woah...

[vernal vale]

so if there is some simplification that happens idk

you can state it generally in that way

i am curious about the two solutions you get though

you only want one right soo

[neon iron]

yuh

[vernal vale]

idk lemme see

OH

lol

[neon iron]

hm?

[vernal vale]

you still get the root out

duh

so youll get two solutions, one will be just

like if you do this for alpha like we are

the two solutions will be alpha, then the actual solution

[neon iron]

the actual solution>

[vernal vale]

yea

so two solutions for the quadratic

the quadratic being like

our difference function right

between the tangent line and the original function

we wanna find where difference is 0 and they intersect

youll still get two roots since its a quadratic

and theyre the values you expect

im trying to think how to isolate the one thats not just the root you already have

does that make sense though

[neon iron]

hmmm

I think I'll sleep on this one ngl

[vernal vale]

word

[neon iron]

maybe another day when I have a clear head x)

[vernal vale]

well

[neon iron]

thanks for the help1

[vernal vale]

if you made it this far

youre pretty close

[neon iron]

yeee

tysm

[vernal vale]

remember what i said about colinear 😄

less to figure out later

[neon iron]

yee! I'll be rereading this!

[vernal vale]

i think that method will work

[neon iron]

gotchu

[vernal vale]

gn

[neon iron]

gn!1

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[crimson crater]

If the question asked what the slope of the inverse function was at -1, -1/2, how would we find that?

[topaz sinew]

@crimson crater Has your question been resolved?

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@crimson crater Has your question been resolved?

[narrow torrent]

slope of tangent of inverse function will be the reciprocal of slope of tangent of function itself

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[stoic elm]

7x^2 + 21x + 14 = 0

[stoic elm]

can you explain how to solve? ^ im a bit confused on all the steps

[wild inlet]

First we want to make it so we have 1x^2

How do we achieve that

[stoic elm]

divide 7?

[wild inlet]

Yeah and then we have ?

Write the equation now

[stoic elm]

x^2 + 3x + 2 = 0

[wild inlet]

Ok we don’t need to write 1 btw

That’s implied

[stoic elm]

ok

[wild inlet]

Ok so now you have

x^2 + 3x + 2 = 0 right

[stoic elm]

yes

[wild inlet]

Now what two numbers can you multiply to get 2. But also those 2 numbers have to be able to sum up to 3

[stoic elm]

i dont understand

[wild inlet]

For example if it was x^2 + 9x + 20
Then it’d be (x + 4)(x + 5)

Because 4 times 5 is 20

And 4 + 5 is 9

x^2 + 3x + 2 = 0 right
So for this what do we put?

First off what times what is 2

Think of the simplest thing

[stoic elm]

1

[wild inlet]

Right

[wild inlet]

So now what will it be written like this

(X ….) (x….)

Because you know 2x1 = 2

[stoic elm]

(x + 1) (x+2)

[wild inlet]

And 2+1 = 3

Yup

Now we equate that to 0

X + 1 = 0

X = ?

And next to it you put an “or” x + 2 = 0

X = ?

@stoic elm

[stoic elm]

-1 + 1 = 0

im so confused

-2+2=0

[wild inlet]

Yep

So you put

X = -1 or X = -2

As answer

Do you want to practice with one more?

[stoic elm]

one sec

writing notes

do we always divide by the x^2 first?

on all?

and why?

[wild inlet]

Because then we can use the (x…)(x…)

If we have 2x^2 in the formula we can’t use the (x…)(x…) notation

Because x times x is just x^2

It’s not 3x^2 or 2x^2

[stoic elm]

give another example please

im still lost

thanks a lot though

just type the question and ill solve

[wild inlet]

2x^2 + 10x + 12 = 0

[stoic elm]

just correct if its right or worng

[wild inlet]

There is also a method called the ABC formula which is used if you can’t solve it with the (x…)(x…) notation

With the abc formula u never have to divide the x^2

[stoic elm]

x+3 x+2?

[wild inlet]

Yep

X = ??

[stoic elm]

0

[wild inlet]

X = …. Or x = ….

[wild inlet]

These should be equal to 0

X + 3 = 0 OR x+2 = 0

[stoic elm]

-3 -2

[wild inlet]

Yep. Do u wanna check if it works?

[stoic elm]

u plug in right?

[wild inlet]

X^2 + 5x + 6 = 0

Plug in -3 or -2 and u will get a true equation

[stoic elm]

example^

[wild inlet]

Bye i gtg

[north field]

bruh cant u just use the quadratic eq

[stoic elm]

im confused sir, been a while

[north field]

wait a sec
doing it

[stoic elm]

.close

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[drifting swift]

that sounds like something to bring up with the professor before plunging into any computations

[drowsy dagger]

Would we proceed with the calculations with a normal piecewise fourier expansion, except keep period as 2pi?

[drifting swift]

i don't think we should proceed with any calculations at all

[drowsy dagger]

sorry I'm very confused, are there no calculations required for the question?
Because I was going to approach this using a piecewise fourier expansion with 2 integrals with limits 0-0.5pi and 0.5pi-pi.

but wasn't sure if my approach is correct

[drifting swift]

the question itself is messed up

because, as you said, we aren't told at all what the function is between pi and 2pi

[drowsy dagger]

ah fair enough

ty

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[stiff pollen]

How come x^2^0 is just x? I understand that 2^0 = 1 and x^1 = x but isn't zero to the power of any number defined as just 1? Like if I took any arbitrary integer a^0 it would be defined as one and if x is an arbitrary number then x^2 is just an arbitrary number so I would expect x^2^0 to be 1?

[mortal thunder]

$x^{2^0} \neq (x^2)^0$

[thorny flame]

Ansh

[stiff pollen]

Okay so the correct way to interpret x^2^0 is to interpret it as saying x to the power of (2 to the power of zero) rather than (x to the power of 2) to the power of 0?

I dont know how else to phrase this distinction

.close

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[neon iron]

Any way I can solve this?

[neon iron]

<@&286206848099549185> Anyone?

[rich fossil]

Triangle Proportionality Theorem

Side Splitter Theorem

Idk what it's called

There's a few other ways you could do this too

[neon iron]

Wait but if I 16/8 I get two, and then I 18 / 2 and get 9, does that make them similar?

[rich fossil]

Yeah

[neon iron]

But if the question says to round my answer to 3 s.f , that doesn't make sense because I have 9

[rich fossil]

9.00

[neon iron]

What about the 3rd side, is there any way of finding that?

BC

[rich fossil]

hmmm

We could if we could have the values for BE or EC

[rich fossil]

Then we apply this

[neon iron]

so there is no way of finding the 3rd side in this problem?

[rich fossil]

check 'Triangle Proportionality Theorem', 'Side Splitter Theorem', 'Angle Bisector Theorem'

https://en.wikipedia.org/wiki/Angle_bisector_theorem
https://en.wikipedia.org/wiki/Intercept_theorem
https://en.wikipedia.org/wiki/Apollonius's_theorem
https://en.wikipedia.org/wiki/Similarity_(geometry)
https://www.varsitytutors.com/hotmath/hotmath_help/topics/triangle-proportionality-theorem

@neon iron

[rich fossil]

Don't think so

[neon iron]

ok thanks

.close

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[pulsar wyvern]

I know the first digit has three possibilities (0, 1, or 2), but I'm not quite sure where to continue from there since the second digit may have 6 possibilities or 5 possibilities depending on whether the first digit was 0 or not, and the digits after that are also dependent on the previous digits since the last digits must be a 0 or a 5.

[topaz sinew]

@pulsar wyvern Has your question been resolved?

[mortal thunder]

Hiya

$\boxed{\cdot}\boxed{\cdot}\boxed{\cdot}\boxed{\cdot}$

[thorny flame]

Ansh

[mortal thunder]

Need to fill those 4 places with digits 0, 1, 2, 3, 4, 5, 6

such that, the number is always divisible by 5

And the thousands place digit is, at max 2

And the divisibility rule of 5 suggests another condition on how this number should look like

[pulsar wyvern]

That the last digit must be 0 or 5, right?

[mortal thunder]

Yes, exactly!

[pulsar wyvern]

Count?

[mortal thunder]

I'm sorry, I mean... Have you learnt the counting method in whatever lesson this question is from?

[pulsar wyvern]

Ah, combinatorics?

[mortal thunder]

Yeah

[pulsar wyvern]

Yes, I've learned about combinatorics and permutations.

[mortal thunder]

Oh nice!

$\boxed{\cdot}\boxed{\cdot}\boxed{\cdot}\boxed{\cdot}$

[thorny flame]

Ansh

[mortal thunder]

So you have these four places and.

conditions are :

Last two digits are 0 or 5

First two digits are at most 2

No digit must be repeated

It can be either a one digit number, a 2-digit, a 3, or a 4-digit

Can you count the number of ways of forming such a number?

[pulsar wyvern]

I'm not sure how. D:

[mortal thunder]

Aww

[pulsar wyvern]

This is as far as I have gotten.

[mortal thunder]

okay, I'll walk you through

First

how many ways can you fix a digit to the unit's place?

[pulsar wyvern]

Two.

[mortal thunder]

Right..

so whether it be a 1, 2, 3 or 4 digit number.. these two digits must always be fixed

Now..

1. how many 1 digit number?

[pulsar wyvern]

Two.

[mortal thunder]

ehh? 0 is positive? 👀

[pulsar wyvern]

Oh.

Oops, one.

[mortal thunder]

:p

next

how many "2" digit numbers?

[pulsar wyvern]

6 * 2 = 12?

[mortal thunder]

wow, how'd you do that-

[pulsar wyvern]

Wait, but then there's 0 being the first number.

[mortal thunder]

Mhmm

[pulsar wyvern]

11?

[mortal thunder]

*6( when 0 is the unit's digit ) + 5( when 5 in unit's digit )

Yes

,calc 6+5

[thorny flame]

Result:

11

[mortal thunder]

oops

[pulsar wyvern]

:o

[mortal thunder]

next, how many 3 digit numbers?

[pulsar wyvern]

Uhhh

[mortal thunder]

5 at unit's place -

[pulsar wyvern]

5 * 6 * 1 - 1 = 29 with 5 as the last digit?

[mortal thunder]

oops

nope

[pulsar wyvern]

D:

[mortal thunder]

since 5 is fixed

you're left with 0, 1, 2, 3, 4, 6

Now, since 0 can't go to hundreds digits, the number of ways of fixing the hundreds digit is?

[pulsar wyvern]

Ohhh, I subtracted the wrong number.

5 * 5 * 1 = 25 with 5 as the last digit?

[mortal thunder]

Yep!

[pulsar wyvern]

And then with 0 as the last digit,
6 * 5 * 1 = 30?

[mortal thunder]

5( because 1, 2, 3, 4, 6 ) times 5( because 0 plus 4 of the remaining digits )

[mortal thunder]

perfect!

[pulsar wyvern]

:D

[mortal thunder]

so total number of 3 digit numbers?

[pulsar wyvern]

25 + 30 = 55.

[mortal thunder]

(ヘ･_･)ヘ┳━┳

what about 4-digit numbers?

[pulsar wyvern]

Hmm

With 5 as the last digit...

5 * 5 * 4 * 1 = 100?

And then with 0 as the last digit,
6 * 5 * 4 * 1 = 120?

[mortal thunder]

okay I think you can do this...
the idea is to be able to write:

Answer = 1 + (5 + 6) + (5 x 5 + 5 x 6) + (5 x 5 x 4 + 6 x 5 x 4)

in one go

[pulsar wyvern]

:o

[mortal thunder]

Yep, correct!

:O

[pulsar wyvern]

So 1 + 11 + 25 + 30 + 100 + 120?

,calc 1 + 11 + 25 + 30 + 100 + 120

[thorny flame]

Result:

287

[mortal thunder]

mhmm

[pulsar wyvern]

Ak, that's not an answer choice. D:

[mortal thunder]

means we did sth stupid (ヘ･_･)ヘ┳━┳

you know what?

[pulsar wyvern]

?

[mortal thunder]

the conditions, can you review them?

conditions are :

Last two digits are 0 or 5

First two digits are at most 2

No digit must be repeated

It can be either a one digit number, a 2-digit, a 3, or a 4-digit

[pulsar wyvern]

Oh.

Oh.

[mortal thunder]

👀

[pulsar wyvern]

Oops.

[mortal thunder]

thousands digit can be either 1 or 2

that's it

[pulsar wyvern]

Then with 5 as the last digit.
2 * 5 * 4 * 1 = 40.
With 0 as the last digit.
2 * 5 * 4 * 1 = 40.

,calc 1 + 11 + 25 + 30 + 40 + 40

[thorny flame]

Result:

147

[pulsar wyvern]

It's there. :o

Thank youu :D

[mortal thunder]

happy nitrogen :D

[pulsar wyvern]

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[lost lynx]

would the answer be -6.5cos(pi/3x)+10

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[fossil stirrup]

I am in the mathematics voice channel. I need help with a calculation problem that I can not figure out.

[fossil stirrup]

It is annual interest calculations

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@fossil stirrup Has your question been resolved?

[fossil stirrup]

<@&286206848099549185> ?

[sweet shard]

Just screenshot or take a photo here

[fossil stirrup]

.close

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[hot nova]

Hello! I need to use a formula for the area of an n sided regular polygon

[hot nova]

P stands for the area

I got it from my book

I could just use it but I really want to know how to come up with it from scratch

I am a little confused by this tg (tangens)

[topaz sinew]

@hot nova Has your question been resolved?

[acoustic tangle]

If you want the derivation

What you could do is connect the center of the polygon with its vertices

Therefore dividing the n-sided regular polygon into n identical triangles

So if you work out area for just one such triangle, you can multiply it by n to get the entire area

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[hot nova]

I found answer on geeks for geeks website. Programming helps a lot :p

[acoustic tangle]

Can't see how programming is involved in the derivation of the formula but alright

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[iron harness]

What is your question?

[worthy storm]

Well you certainly can't get infinitely many solutions, the first three equations already constrain you to at most one solution.

[iron harness]

Not sure about that

[worthy storm]

So the fourth one either is or is not consistent with the first three.

[iron harness]

Oh yeah sorry, yes @worthy storm is right.

[worthy storm]

Yes, that's forced by the first three equations.

So in fact for any choice of a,b,c, you'll also have (0,0,0) as a solution to the fourth equation.

Hence the fourth one is always consistent with the first three

Well 0,0,0 is always a solution as we argued above, so "no solution" is not a possibility

And the first three equations rule out any solution other than 0,0,0

Hence 0,0,0 is always the unique solution, regardless of a,b,c

Strange question IMO. 😁

Maybe it gets more interesting in the subsequent parts.

Yes, 0,0,0 is always a solution, plug it in and see!

To use linear algebraic terms, the null space of a matrix is a subspace, so it always contains the origin.

Pleasure, good luck!

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[tight dragon]

What quadrant would it be in on the cartesian plane if tan theta is positive and sin theta is positive as well?

[vital relic]

https://www.desmos.com/calculator

[tiny wolf]

all students take calculus

[tight dragon]

so it would be first quadrant

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[neon iron]

Hi can someone help me? I will really appreciate it thank you

<@&286206848099549185>

[tiny wolf]

it hasn't even been close to 15 minutes!

[neon iron]

I'm sorry.

[tiny wolf]

set up two equations, one for each scenario, the solve the system

[neon iron]

Wdym sorry? english is not my first language so I am afraid of making mistakes. Could you please elaborate on that a little more? Thank you

[tiny wolf]

yeah so just write an equation for the first scenario

this one

thats step 1

[chrome glen]

radio ratio

[tiny wolf]

yeah they definitely could have picked a better product haha

[neon iron]

Tysm , how?🥲

Sorry if I'm bothering you

[tiny wolf]

youre not

do you know what ratio is?

[neon iron]

Thx so kind

[neon iron]

yep

[tiny wolf]

so if $x : y$ = $\frac{3}{4}$, this is the same as $\frac{x}{y} = \frac{3}{4}$, right?

[thorny flame]

Migillope

[neon iron]

Yeah , makes sense

[tiny wolf]

ok. so the first sentence says if the price for each is increased by 20, the ratio is 5:2

i.e., $\frac{x +20}{y+20} = \frac{5}{2}$.

[thorny flame]

Migillope

[tiny wolf]

Does that make sense?

[neon iron]

Yes

[tiny wolf]

ok, so set up a similar equation for the second scenario

[neon iron]

x-5/y-5 = 5/1

[tiny wolf]

great

so now you have to equations with two unknowns. Do you know how to solve a system of linear equations?

If not: https://www.khanacademy.org/test-prep/sat/x0a8c2e5f:untitled-652/x0a8c2e5f:heart-of-algebra-lessons-by-skill/a/gtp--sat-math--article--solving-system-of-linear-equations--lesson

[neon iron]

but we have

2 equations

we solve

them both?

[tiny wolf]

click the link

[neon iron]

simultaneous

[tiny wolf]

yes, otherwise known as simultaneous equations

if you know how to solve those, like by using substitution, do the same here

[neon iron]

Idk how 🥲

Can you help me

[tiny wolf]

Ping @ helpers, I gotta go

[neon iron]

Alr tysm

<@&286206848099549185> hi

[mellow venture]

what

[neon iron]

This 😞

[topaz sinew]

@neon iron Has your question been resolved?

[topaz sinew]

@neon iron Has your question been resolved?

[tiny wolf]

well im back now

@neon iron still need help?

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[toxic crystal]

how does horizontal and vertical transformations effect points ?

[toxic crystal]

Im trying to figure out this problem :

along with this :

[toxic crystal]

that doesnt help me

[empty sail]

It should

[toxic crystal]

how can I shrink by 6 horizontally and then shift right by 5

[empty sail]

A shrink is a compression

[toxic crystal]

yea

but I dont know how to do the math with only points

[empty sail]

So a horizontal compression of 6 means you divide x/6

[toxic crystal]

Im not given a graph

after doing that how do I shift 5/6 by 5 ?

to the right

[empty sail]

That should be it

[toxic crystal]

so (5,-2) shrunk horizontally by 6 = (5/6,2) now if I shift that to the right by 5 what is it

[empty sail]

You shift the x right by 5, by adding 5

[toxic crystal]

right so 10/6 ?

[empty sail]

No

Common denominator

[toxic crystal]

ah I see

any chance u know discrete mathematics ?

[empty sail]

Depends

[toxic crystal]

Can u explain this in english

[empty sail]

That, I can not

[toxic crystal]

ok, thanks for earlier

.close

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[neon iron]

i need to find these for the function given above

[leaden tusk]

what

[leaden tusk]

what have you tried

[neon iron]

i tried using the f(a+h)-f(a)/h

since i am not given a point yet except a = 5

but i dont know how to plug it in as a fractional

i belive im supposed to use this derivative formula

[leaden tusk]

nothing changes with the fraction

other than the fact that it's a fraction

you do it the same as any other time

[neon iron]

so assuming im doing this correct this is what im getting for the next step

one sec

Feel like im mixing up too many variables though from what i seen

[leaden tusk]

it's a tad confusing because there's two limit definitions, one normally uses x and h, one normally uses x and a

and yours seems to be combining the two

[neon iron]

i understand im supposed to use this one since im not given a point

but when i have to plug it in with it being a fraction and having a + h and x in my function it gets confusing

[leaden tusk]

replace (a+h) with (x+h) and continue from there

then once you actually get your derivative, find f'(a)

[neon iron]

i got 0 lol

[leaden tusk]

can't divide by zero

[neon iron]

do i have top part right or am i approaching it worng

[leaden tusk]

top part is right, but you can't plug 0 in for h yet

you need to cancel the h

or at least get it to where you're not dividing by zero

[neon iron]

i got -1/(x+5)^2 for f'(a)

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[fiery ocean]

hi does anybody know how to solve this? it seems impossible to me

[fiery ocean]

the rules are each number from 1-16 may only be used once

[empty sail]

Just a lot of trial an error

Trying out every sort of combination until you get the right one

[fiery ocean]

wouldnt that take hours

[languid cove]

if this is a puzzle you could just attempt solving for 1 answer at a time using brute-force method, for example on the left column a possible answer is 5 - 5 - 5 - 5 = -15

[fiery ocean]

a number is only allowed once

so like they all have to be different

[languid cove]

it's possible to sidestep that fact if you can use numbers with more than a single digit

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[tawdry mist]

Help

[tawdry mist]

Those are the questions

[empty sail]

What exactly don't you understand?

[tawdry mist]

I don't understand jow to turn them into algebraic expressions

[empty sail]

Do you understand that when it says 'a ...' that just means that it is the unknown?

Meaning a variable

Like a value, a number, etc

[tawdry mist]

For the first I guess it would be 2n + 5 but idk about the rest

[empty sail]

That's exactly it for the first one

Because the unknown, you declared as n, twice n is 2n, then 5 more that number so 2n + 5

For the next one, what does one quarter of the value mean?

[tawdry mist]

1/4?

[empty sail]

Yes

So what does that mean for the value?

[tawdry mist]

It's a variable

So like v

[empty sail]

But applying the one quarter part too

[tawdry mist]

Ok

1/4v

[empty sail]

Yes

Then it says three less than that

[tawdry mist]

So subtract 3

[empty sail]

Yes

So the expression is?

[tawdry mist]

1/4v - 3

[empty sail]

Yes that is right

[tawdry mist]

Wow

Can you help me with the next please

[empty sail]

Do you think you can do the rest on your own?

Try it yourself first

[tawdry mist]

Ok

[empty sail]

See what you get, then say it and I'll check that

[tawdry mist]

N + N × 7?

[empty sail]

Close

You have some things swapped

What does it mean when it's a product?

[tawdry mist]

It's a variable?

[empty sail]

Also, it's suggested to use two different variables if you have number unknowns

But in general, what is a product?

[tawdry mist]

Variable or letter?

Wait the total

Like the final number

[empty sail]

Like mathematically speaking, what is a product?

[tawdry mist]

Result

[empty sail]

What math operation is involved with product?

[tawdry mist]

Multiplication

[empty sail]

So then, the product of a number and another number means?

[tawdry mist]

A × B + 7?

[empty sail]

Exactly that

[tawdry mist]

Ok thank you so much! Do you mind if I figure the rest and send what I get to you?

[empty sail]

I probably won't be on much longer but if you post it, someone around can check it

Just recall that there are synonyms for the math operations

You discovered a few like increases, more than, etc is adding

Product, twice a number, etc is multiplication

[tawdry mist]

Ok thanks

[topaz sinew]

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[short wedge]

Hiii, can someone help me with this question? Whenever I do this question, I get 2.92 and -2.12 and that physically does not make sense for this problem

[sudden burrow]

I think u use 2.92 as the width then find the length no?

[short wedge]

oh wow why didnt i think of that

omg

THAT MAKES SO MUCH SENSE, TYY

[sudden burrow]

no probs

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[vestal skiff]

Trying to help?

[vestal skiff]

I’m stuck I don’t know what to do

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[alpine mist]

You're equations are wrong, the first one should have -y

But once you have\ $3x-y = 91$\ $2x+y = 89$\ you should try adding the two equations together

[thorny flame]

Zybikron

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[blazing kelp]

is this available

[blazing kelp]

how do you find the area of the colored section

radius is 2cm

[rich fossil]

there's a few ways we could do this

i would take a more well technical approach

So we know that triangle AOB=COB

Have you been taught how to find the area of a sector? sine rule? cosine rule? or even heron's formula?

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[pure frigate]

How do I find vc help?

[iron harness]

What is vc? I think you should just go ahead and ask the question.

[vapid lichen]

vc=voice chat

vcs are at the bottom of the channel list, but we dont have nearly as many vc helpers as text channel helpers so ur better off asking here

[pure frigate]

@vapid lichen where at the bottom can I find the vc option for help?

all I see is

Hidden math help

[vapid lichen]

see the "voice" section

[pure frigate]

alright

I typed it in Mathmatics-voice

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[south girder]

hi there i was wondering if someone could help me find my way to come up with an answer for this

[south girder]

i have tried different alternatives but they seem to make both statements true

be a lot better if someone could discuss on vc but it's fine on text as well

<@&286206848099549185> hate to be a bother but i would appreciate it someone could guide me. been at this for more than 3 hours and i just wanna know where i am going wrong cause i am losing my sanity 😩

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[south girder]

.close

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• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[keen kindle]

Determine A and v in the function y = A * cos(x-v) if y_min = -4 and y(0) = -1.

[keen kindle]

A * cos(-v) = -1

How do I find A?