help-13

[crimson sedge]

Yes sir

doing that

[kindred storm]

Then you can fill in different as to get the expanded terms.

Oh, sorry, the top are all like (2x + a)³.

[crimson sedge]

no problem I got it

transforming all into (ax^2 + bx + c)(2x + a)

One question (forgot if I can):
4x^2 + 8x + 4 equals to x^2 + 2x + 1

[kindred storm]

Well, that's one of its factors, but we don't want to factor yet.

[crimson sedge]

Alright

Everything done

So the result came as

(4x^2 + 4x + 1)(2x + 1) + (4x^2 + 8x + 4)(2x + 2) + (4x^2 + 12x + 9)(2x + 3)

[kindred storm]

No.

[crimson sedge]

no?

[kindred storm]

First, get (2x + a)³ expanded.

What do you get for that?

[crimson sedge]

quadratic formula multiplied by (2x + a

it's the way I said for you

[kindred storm]

Which is what?

[crimson sedge]

or is it wrong?

[kindred storm]

Expand that.

[crimson sedge]

(dx^2 + bx + c)(2x + a)

[kindred storm]

No, you have (4x² + 4ax + a²)(2x + a)

You don't have d, b, or c.

[crimson sedge]

but I thought

ooooh

I did the opposite of what you told

if I expand (2x + a)^3

yes I get

(4x^2 + 4ax + a^2)(2x + a)

[kindred storm]

Right, now expand that.

[crimson sedge]

even more?

[kindred storm]

Yes, completely expand it so there are no parentheses.

[crimson sedge]

oh my god labor work

doing it

[kindred storm]

Yes, but that's why you do it once with a instead of a number.

[crimson sedge]

oooooh

[kindred storm]

Then you can apply your result by filling in a and getting that work done for free for each a value.

[crimson sedge]

I see it

efficient

instead of doing each one at a time

alright

[kindred storm]

Right.

[crimson sedge]

so after doing it

I get

a^3 + 6a^(2)x + 12ax^2 + 8x^3

correct?

[kindred storm]

(2x + a)³
(2x + a)(2x + a)(2x + a)
(4x² + 2ax + 2ax + a²)(2x + a)
(4x² + 4ax + a²)(2x + a)
8x³ + 8ax² + 2a²x + 4ax² + 4a²x + a³
8x³ + 12ax² + 6a²x + a³

Yes, that's what I got too.

[crimson sedge]

perfect

and then I apply 3 of the (2x + a)^3

[kindred storm]

Well, there are more than 3.

[crimson sedge]

3 of (2x - a)^3
3 of ( 2x + a)^3
& 1 of (2x)^3

[kindred storm]

Oh, no.

If you do it that way, then you also need to expand (2x - a)³.

That will leave you doing that work all over again.

But instead, you can just do a = -3 through a = 3.

a = -3: ⋯
a = -2: ⋯
a = -1: ⋯
a = 0: ⋯
a = 1: ⋯
a = 2: ⋯
a = 3: ⋯

If you do it for those a values, you can use your one expansion 7 times.

Does that make sense?

[crimson sedge]

Just a bit

because I thought we were expanding everything

[kindred storm]

We are.

[crimson sedge]

and then canceling

[kindred storm]

Yes, after simplifying.

[crimson sedge]

but you said we weren't expanding (2x - a)^3?

[kindred storm]

No, we are.

We use our (2x + a)³ expansion with a = -3.

Does it make sense that that works?

[crimson sedge]

yes

so I don't to go through all the work of calculating the base again

instead I can simply replace as -3

[kindred storm]

Right.

You made up your own math law that works for any a.

Now you can use it.

[crimson sedge]

Now my task is to apply to everything

and remove every parentheses

[kindred storm]

Right.

[crimson sedge]

I'm going to simplify (2x)^3 as 8x^3 no secret, correct?

[kindred storm]

Right.

You can skip the law you discovered for that one.

[crimson sedge]

okay

one question

8x³ + 12ax² + 6a²x + a³

for (2x + 1)

x = 2 and a = 1

[kindred storm]

No, x is 32768.

[crimson sedge]

ooooooh

[kindred storm]

But we're leaving it as a variable.

[crimson sedge]

so the real constant here is a

[kindred storm]

If you fill it in now, it will make the division not work out.

Yes, the a is the one you fill in.

[crimson sedge]

okay

ooh so I will get

a

3rd degree polynomial

[kindred storm]

Yes, that's right.

8x³ + 12ax² + 6a²x + a³
8x³ + 12(1)x² + 6(1)²x + (1)³ for a = 1.

[crimson sedge]

yes

(2x + 1) = 8x^3 + 12x^2 + 6x + 1

(2x + 2) = 8x^3 + 24x^2 + 24x + 8

[kindred storm]

Don't forget the ^3.

(2x + 2)^3 = ⋯

[crimson sedge]

oh yes

[kindred storm]

But yes, you got it.

[crimson sedge]

and finally

(2x + 3)^3 = 8x^3 + 36x^2 + 54x + 27

[kindred storm]

OK, did you get (2x - 3)³ and so on as well?

[crimson sedge]

I'm going to do the first one so I can tell you

1 second

(2x - 1)^3 = 8x^3 - 12x^2 + 6x - 1

correct?

[kindred storm]

Yes, that's right.

[crimson sedge]

Okay, so I got all of them

[kindred storm]

Notice how it looks almost like (2x + 1)³?

[crimson sedge]

yes

almost equal

[kindred storm]

The only difference is the x² and constant term signs.

[crimson sedge]

So after doing everything I just need to sum it up

[kindred storm]

So, if you have (2x - 3)² and (2x + 3)², the x² and constant terms cancel out each other.

[crimson sedge]

and get the simplified version of it

[kindred storm]

Does that make sense?

[crimson sedge]

completely

the fast way

[kindred storm]

So, the final sum will have 0x² and 0.

Because they'll match up and cancel.

[crimson sedge]

oh my god

why didn't I see that

now that makes completely sense

they cancel each other

[kindred storm]

So, everything has 8 x³.

[crimson sedge]

yes

[kindred storm]

And there are 7 things.

[crimson sedge]

56x^3

BOOOMmm

[kindred storm]

And then the x term will have the sum of the first three squares times 2.

2(1² + 2² + 3²)

Because it's (-3)² + 3² = 2(3²) and so on.

[crimson sedge]

wait I didn't get that

[kindred storm]

Oh, it was a bit off.

6a²x

[crimson sedge]

You changed from 56x^3 to 2(1^2 + 2^2 + 3^2)

confused me

[kindred storm]

Oh, no, not x³, x.

6a²x is the x term for each a.

8x³ + 12ax² + 6a²x + a³

[crimson sedge]

you're still on the numerator?

[kindred storm]

Yes.

[crimson sedge]

okay

[kindred storm]

But it's a trick for the x term.

[crimson sedge]

but didn't they cancel each other?

and the only left is 8x^3

[kindred storm]

No, because of this: (-3)² = (3)².

[crimson sedge]

and there are 7 as we said

[kindred storm]

The x² and constant terms cancel because their signs disagree.

But a²'s sign is never negative, so they can't cancel.

[crimson sedge]

9 = 9

[kindred storm]

Right.

So, 3² counts twice.

Once for -3 and once for 3.

2² also counts twice.

1² also counts twice.

Does that make sense?

[crimson sedge]

A bit

[kindred storm]

Sorry, once for a = -3 and once for a = 3.

You get 3² for both of those.

Because squaring gets rid of negative signs.

[crimson sedge]

Because I thought 56x^3 was the numerator

[kindred storm]

No.

The x² term cancels.

8x³ + 12ax² + 6a²x + a³

[crimson sedge]

yes

[kindred storm]

12(-3 + -2 + -1 + 0 + 1 + 2 + 3)x²

And -3 + -2 + -1 + 0 + 1 + 2 + 3 is 0.

[crimson sedge]

yup

[kindred storm]

Same with the constant term.

8x³ + 12ax² + 6a²x + a³

-27 + -8 + -1 + 0 + 1 + 8 + 27 = 0

Does that make sense so far?

[crimson sedge]

Still confused but got most part

Because let me explain

what im having conflict with

(2x - 1) + (2x - 2) ( 2x - 3) + (2x + 1) (2x + 2) (2x + 3)

we showed that

they are going to cancel each other

and okay

[kindred storm]

Yes, for the x² and constant terms.

But not for the x³ and x terms.

[crimson sedge]

Yes

they are going to still remain

8x^3 for 7

[kindred storm]

Right, and the x³ term is:

8x³ + 12ax² + 6a²x + a³
8(1 + 1 + 1 + 1 + 1 + 1 + 1)x³

[crimson sedge]

56x^3

that's where I lost

[kindred storm]

The x term is:
8x³ + 12ax² + 6a²x + a³

6((-3)² + (-2)² + (-1)² + (0)² + (1)² + (2)² + (3)²)x

[crimson sedge]

OOOOOOOH

[kindred storm]

And (-3)² = 3² and so on.

[crimson sedge]

so 6a^2 + a^3

are still there

[kindred storm]

No.

a³ cancels, as we showed.

[crimson sedge]

oh yeah sorry

8x^3 and 6a^(2)x

[kindred storm]

The odd degree terms remain.

Right.

[crimson sedge]

yeah got it

[kindred storm]

So, the trick with the x terms is this:

(-3)² = 3², right?

Same with (-2)² = 2².

[crimson sedge]

exactly

[kindred storm]

So,

6((-3)² + (-2)² + (-1)² + (0)² + (1)² + (2)² + (3)²)x
6(2(1² + 2² + 3²))x

Since -3 and 3 work the same, we can just do what 3 does twice.

[crimson sedge]

Yes

[kindred storm]

6(2(1² + 2² + 3²))x
6(2(1 + 4 + 9))x
6(2(14))x
168x

[crimson sedge]

Okay so you basically

did the math of the remaining

[kindred storm]

Yes, and we have 56x³ + 168x as the whole thing on top.

[crimson sedge]

beautiful

[kindred storm]

Now we can factor a bit.

[crimson sedge]

now I understood it perfectly

56x(x^2 + 3)

[kindred storm]

Good.

Now we do the bottom.

Unless you already got that.

[crimson sedge]

It's the same process

Remove the parentheses

[kindred storm]

Yes, now we have (x + a)(x + a + 1)

[crimson sedge]

I'm going to do it really quick

[kindred storm]

OK.

But making a law that works for any a has less of a work savings here perhaps.

[crimson sedge]

Yes, using the same process

Doing each one is way harder

[kindred storm]

OK, then you have (x + a)(x + a + 1).

Where a is the lesser of the two things added to x.

[minor meteor]

Hi

Can i ask a question

[crimson sedge]

Okay but how do I get the general formula? I have (x - 3)(x - 2) + (x - 1)x + x(x + 1) + ( x - 2)(x + 3)

[minor meteor]

Or occupied?

[kindred storm]

@minor meteor Occupied. If the channel name has | with a person's name, it's occupied. Ask in a #help channel without a name in it near the top of the channel list.

[minor meteor]

Oo

[crimson sedge]

(x - a)(x - a)

The same pattern does not repeat as the last time

[kindred storm]

Are you using (x - a)(x - b) for the bottom?

Oh!

[crimson sedge]

Shouldn't I? There's 3 and 2

[kindred storm]

Well, for the bottom, notice that it's like (x - 3)(x - 2) or something where the second one is one higher.

So, it's like (x + a)(x + a + 1)

[crimson sedge]

ooooooooh

I see

[kindred storm]

The lower one is a, then the other one is a + 1.

[crimson sedge]

I see

Got it

[kindred storm]

Now you can make your law.

[crimson sedge]

Roger that

but how do I work with

(x - a)x

(x + a)x

[kindred storm]

a = -1.

Or a = 1

Sorry, or a = 0.

x(x + 1) has a = 0.

(x - 1)x has a = -1.

You can check that.

[crimson sedge]

(x + a)(x + a + 1) + (x + a)x

[kindred storm]

(x + a)(x + a + 1) with a = 0:

(x + 0)(x + 0 + 1) = x(x + 1)

[crimson sedge]

oooh

That's efficient

[kindred storm]

(x + a)(x + a + 1) with a = -1:

(x - 1)(x - 1 + 1) = (x - 1)x.

But the idea is that you know what a represents: the lower number you add to x.

[crimson sedge]

(x + a)(x + a + 1)

[kindred storm]

x(x + 1) has 0 and 1 added, the lower is 0, so a = 0.

[crimson sedge]

x^2 + 2ax + x + a + a^2

[kindred storm]

Good, now combine like terms.

Pretending that a isn't a variable.

[crimson sedge]

Wait how?

Combine them?

[kindred storm]

Well, like terms are those with the same variables and powers on those variables.

We don't care about a, though.

So, we look at x.

Which terms have x² in them?

Just x², right?

[crimson sedge]

Yes

[kindred storm]

So, x² + ⋯

Which terms have x in them?

[crimson sedge]

2ax + x

[kindred storm]

So, we factor out the x.

(2a + 1)x

[crimson sedge]

Because we already know that the other's cancel themselves

[kindred storm]

x² + (2a + 1)x + ⋯

[crimson sedge]

Oookay

[kindred storm]

See how we can easily see the coefficient of x?

We just calculate 2a + 1 and we have the x coefficient.

What are the constant terms?

[crimson sedge]

x² + (2a + 1)x

A

[kindred storm]

What else?

[crimson sedge]

hummm

[kindred storm]

There's two terms without an x in them.

[crimson sedge]

a^2 + a

[kindred storm]

Good.

[crimson sedge]

a(a + 1)

[kindred storm]

x² + (2a + 1)x + (a² + a)

Have you done the quadratic formula yet?

[crimson sedge]

Oh doing on all of them?

[kindred storm]

No, you won't use the quadratic formula in this problem, I just was wondering if you'd learned it.

[crimson sedge]

I'm still a bit lost on it

[kindred storm]

You know how you get a, b, and c for it?

[crimson sedge]

Yes

[kindred storm]

Like the x², x, and constant coefficients.

Well, if you have something with two variables in it.

You can get a, b, and c this way too.

[crimson sedge]

But I thought (2a + 1)x

• x^2

was the final point here

because wont the rest cancel themselves?

[kindred storm]

No.

It won't always work the way it did in the top.

[crimson sedge]

aaaah I see now why

[kindred storm]

We have to check first whether anything nicely cancels out.

So, we have x² + (2a + 1)x + (a² + a).

[crimson sedge]

(a^2 + a); per example a = -2
(-2^2 -2 ) = 2

[kindred storm]

So, what are the a values we need to work with?

[crimson sedge]

The ones are

-3

-2

-1

+1

+2

+3

[kindred storm]

Nope.

[crimson sedge]

what

[kindred storm]

Remember that a is the lower of the two numbers.

(x - 3)(x - 2) has a = -3.

[crimson sedge]

oooooh

[kindred storm]

(x - 1)x has a = what?

[crimson sedge]

-2

[kindred storm]

Nope, what are you adding to x in each factor there?

You're adding -1 and 0, right?

[crimson sedge]

oh mg

yes 0

[kindred storm]

And a is the lower one.

So, a = -1.

[crimson sedge]

Yes

[kindred storm]

What about x(x + 1)?

[crimson sedge]

a = 0

[kindred storm]

What about (x + 2)(x + 3)?

[crimson sedge]

a = 2

[kindred storm]

So, -3, -1, 0, 2.

[crimson sedge]

yep

[kindred storm]

Now our x² term is easy.

You just multiply the count of terms by x².

[crimson sedge]

9 + 1 + 0 + 4

[kindred storm]

What is that for?

The x term?

[crimson sedge]

A

[kindred storm]

No, I mean, what are 9, 1, 0, 4 for?

[crimson sedge]

I kinda got lost

for x?

I just did -3 ^ 2

etc

[kindred storm]

x² + (2a + 1)x + (a² + a)

a = -3: 1, (2(-3) + 1), ((-3)² + (-3))
a = -3: 1, -5, 6

So, when a = -3, you have x² - 5x + 6.

Does that make sense?

[crimson sedge]

Completely

[kindred storm]

OK, but we don't want to do it that way exactly.

a = -3: 1, -5, -6
a = -1: ⋯

What do you get for a = -1?

[crimson sedge]

x^2 - x

[kindred storm]

Yes, but we don't want to do it that way.

We want to get the x² coefficients first.

Let's do it that way.

So, it's 1 for each of them, right?

[crimson sedge]

right

[kindred storm]

So, 4x² in the final result on the bottom.

What about the x coefficients for each?

2a + 1 for each.

[crimson sedge]

...

[kindred storm]

2a + 1

2(-3 + -1 + 0 + 2) + 4

[crimson sedge]

I'm going to use the -3, -1, 0 & 2

for all?

[kindred storm]

Yes, for the a values.

Fill them in to 2a + 1 for each.

Add up the results.

[crimson sedge]

its 0

[kindred storm]

OK, now the constant term is a² + a.

Do those for each a, then total them.

[crimson sedge]

oh no sorry its 8

wrong answer

[kindred storm]

For x, it's 0.

[crimson sedge]

2(2) + 4

oh ok yes it's 0

[kindred storm]

OK, the constant terms are a² + a. Total those.

[crimson sedge]

Oh damn, I think I'm so lost

We are using the general formula and using the values we got

-3, -1, 0 and 2

[kindred storm]

-3, -1, 0, 2.

[crimson sedge]

yes sorry

and the formula is

[kindred storm]

x² + (2a + 1)x + (a² + a)

[crimson sedge]

yes

and all I need to do is apply every number on it?

[kindred storm]

Yep.

[crimson sedge]

-3, -1, 0, 2

[kindred storm]

But instead of getting a polynomial and adding the polynomials, use this trick.

You can add numbers in any order you want.

3 + 2 + 1 = 1 + 3 + 2 and so on.

[crimson sedge]

so I would get a similar result as in the top

[kindred storm]

So, instead of getting the polynomials and adding, get the x² terms in all the polynomials and get the x² in the final polynomial by adding them.

[crimson sedge]

Okay

[kindred storm]

We got the sum of the x² terms as 4x², right?

[crimson sedge]

Yep

[kindred storm]

And we got the sum of the x terms as 0x, right?

[crimson sedge]

ah

oh no yeah right

the x = 0

[kindred storm]

What do we get for the sum of the constant terms?

[crimson sedge]

considering x as 0

[kindred storm]

No, we didn't solve for x.

It's 0x, not x = 0.

[crimson sedge]

So you're saying solving a?

[kindred storm]

No.

We know the a values.

No need to solve for them.

[crimson sedge]

(2a + 1)x

[kindred storm]

What we're doing is finding the coefficients of the bottom polynomial.

Yes, and 2a + 1 for all of our as added together gives 0.

[crimson sedge]

Okay

[kindred storm]

-5 + -1 + 1 + 5 = 0

[crimson sedge]

But where -5, -1, 1, 5 came from?

[kindred storm]

2a + 1 when a = -3 is 2(-3) + 1 = -5.

2a + 1 when a = -1 is 2(-1) + 1 = -1.

[crimson sedge]

Oooh okay

[kindred storm]

And so on.

[crimson sedge]

So we've solved each one

for -3, -1, 0, 2

[kindred storm]

Right, then added them.

[crimson sedge]

Got it

[kindred storm]

Now we do the same for the constant term.

a² + a

[crimson sedge]

25 - 5 = 20

[kindred storm]

Why are you using a = -5?

[crimson sedge]

...

[crimson sedge]

because of this

[kindred storm]

Yes, that's not a, that's 2a + 1.

Notice how it says that's 2a + 1.

[crimson sedge]

Oh ok

a^2 + a

(-3)^2 - 3 ?

[kindred storm]

Yes.

[crimson sedge]

Then we get

6, 0, 0, 6

[kindred storm]

Good.

So, that sums to 12

[crimson sedge]

Yep

[kindred storm]

So, we have 4x² + 12 on the bottom.

Factor that.

[crimson sedge]

4(x^2 + 3)

Isn't 4x^2 + 12x?

Oh not nevermind

[wraith dagger]

Chai T. Rex

[crimson sedge]

Yep

[kindred storm]

Notice how x² + 3 is multiplied by the ENTIRE top and the ENTIRE bottom.

[crimson sedge]

And now we can cancel everything

Yes

[kindred storm]

Right.

Oh, wait.

[crimson sedge]

56/4

oh wait

its

[kindred storm]

Made a slight error.

56x(x² + 3) on top.

[crimson sedge]

56x(x^2 + 3)

For me

yeah

[kindred storm]

Right!

56x/4

[crimson sedge]

56x/4

Yup

[kindred storm]

So, what does 56/4 reduce to?

[crimson sedge]

fourteeeennn

[kindred storm]

So, the answer is 14x.

x = 32768 = 2¹⁵.

So, 14 · 2¹⁵.

And 14 can be further factored.

[kindred storm]

Yes, now factor 14.

[crimson sedge]

2 * 7

[kindred storm]

OK, so what's 14 · 2¹⁵ simplify to?

[crimson sedge]

7 * 2^16

[kindred storm]

Good.

[crimson sedge]

Oh so the problem was expecting me to know 2^15

[kindred storm]

Well, they say 2¹⁶ a lot in the answers, so you can find out what that is.

[crimson sedge]

finally done, a beautiful problem

sorry for the trouble

Now I understand it

[kindred storm]

2² = 4
2⁴ = 4² = 16
2⁸ = 16² = 256
2¹⁶ = 256² = 65536

[crimson sedge]

Kind of a memorization thing

[kindred storm]

No, you can do it on paper.

16 is a power of 2.

So, you can do repeated squaring.

Each time you square the last number, the power doubles.

2¹ = 2
2² = 4
2⁴ = 16
2⁸ = 256

And so on.

[crimson sedge]

oh okay

[kindred storm]

You just do the 2 times 2 on paper, and so on.

[crimson sedge]

thank you chai

such big help

[kindred storm]

You're welcome.

[crimson sedge]

the chat is not going to be deleted right:

?

[kindred storm]

If you remember one thing, remember that making your own math law thing.

No, it'll still be here, just filled with other help sessions after, so right click and get a message link.

[crimson sedge]

Making the my own math law makes things efficient

got it

[kindred storm]

The message link will lead you back to the message you want to start looking at.

[crimson sedge]

Good night because im sleepy

1:20 am

[kindred storm]

Have a good sleep.

[crimson sedge]

Finally solved it

Thank you and good night

[kindred storm]

You're welcome.

[crimson sedge]

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[thin roost]

How is it

110°

Tell me pls

[crimson sedge]

i would assume it's because 180-110=70

maybe a property of cyclic quadrilaterals is that opposing angles must sum to 180

[thin roost]

That's the reason

Can anyone pls properly make me clear

How ABC + ADC = 180

[glad kestrel]

property of a cyclic quadrilateral

opposite angles are supplementary

[thin roost]

Ok what does supplementary mean

[glad kestrel]

angles add to 180

[thin roost]

Oh

I see

That means I'm write

Ok

property of a cyclic quadrilateral opposite angles are supplementary

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[cosmic zodiac]

how would i find this

[warm vector]

Trigonometric ratios

How about you watch videos on them on YouTube

You will get it

[cosmic zodiac]

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[cedar kiln]

@fiery summit Has your question been resolved?

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[glad fox]

What exactly is an argument here? Is it a random element of the domain?

[cedar kiln]

@glad fox Has your question been resolved?

[fallen heath]

Suppose you're taking a function: $f : \bR \times \bZ \to \bZ$ defined by $f(x,y) = \floor{x} + y$

[wraith dagger]

Ansh

[fallen heath]

the 2nd argument of f, i.e, y in this case, must be an element of Z

and the 1st argument of, i.e., x, must be an element of R

[glad fox]

Oh

Ok thx

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[crimson obsidian]

Just want to confirm something

[crimson obsidian]

ln(a) - ln(b)= ln(a/b)?

Yes/No?

[inland tartan]

correct

yes

[crimson obsidian]

Ok ty

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[grave jay]

How do you go from line 2 to 3, why does it go from -(1-cos) to -1+cos

[upper abyss]

Because -(a - b) = -a + b

[bright surge]

when you subtract by something in parentheses, you remove the subtraction sign and flip the sign of anything inside the parens

basically what kaynex said :\

[grave jay]

Oh ok, thank you both

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[fallen heath]

uhh? your work on it?

hint: work on $\prod_{r=2}^n \frac{r^2+r+1}{r^2-r+1}$

[wraith dagger]

Ansh

[cedar kiln]

@shy elbow Has your question been resolved?

[shy elbow]

@fallen heath how did you cancel (r-1) by (r+1)?

[fallen heath]

I didn't, that's a hint

the other product is simple enough, and can be evaluated separately

[fallen heath]

@shy elbow Any progress?

these kind of questions heavily rely on your intuition/previous practice with such patterns

[shy elbow]

Unfortunately, this is the first product series question, I have come across. I was reading up on the properties of capital Pi

[fallen heath]

Oh

[shy elbow]

I know that the first part will become some sort of factorial and cancel out.

[wraith dagger]

Ansh

Ansh

[shy elbow]

Yes

[fallen heath]

$\prod_{r=2}^n \frac{r^2+r+1}{r^2-r+1}$
This part, however, unlike the first part doesn't end with a convenient factorial and instead requires you to see a trend between the numerators and denominators going forward

[wraith dagger]

Ansh

[fallen heath]

So you can start with maybe writing the first few terms?

And seeing how their product cancels out or reduces or something

[shy elbow]

I don't think that, kind of a thing seems to be happening here... although, my intuition is super bad

If you have already solved the problem, can guide me in the direction of the topics I need to know, I didn't find anything like this in sequence and series or integration.

[fallen heath]

Uh

[shy elbow]

Oooo, got it, thanks a lot

[fallen heath]

Don't thank me... just ... next time listen to people if you're asking for advice

Doing that yourself would've had way better positive impact on your learning and thinking process than me just giving away the crux of it

[shy elbow]

. close

[fallen heath]

.close

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[crimson sedge]

How 343 power 2/x is became 2/x × 343

[dense wing]

it doesn't

[crimson sedge]

Okay

what are the steps for this question

[storm thunder]

Take the log of both sides

Then the power comes down

[dense wing]

343 and 49 are both powers of 7

[crimson sedge]

This question asked for a 8th grade student , they didn't teached log those kind of math ,, is there any other way ??

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[dense wing]

yeah, respond to what I said

they're both powers of 7.

[storm thunder]

Yeah thats also a way

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[left meteor]

howdy

[left meteor]

hey brp

does this make sense to you

from which table do i simplify

9?

[obsidian coral]

What do you mean table?

[left meteor]

what times what

like simlifying

[obsidian coral]

It want's you to write the mixed fractions as improper fractions first

[left meteor]

yes i did

[obsidian coral]

Then multiply the two fractions

[left meteor]

11/6 x 9/2

99/12?

[obsidian coral]

Yes

[left meteor]

but then he asked to simplify

it cant be 99/12

@obsidian coral

[obsidian coral]

Then simplify

[left meteor]

how

[obsidian coral]

Find the prime factors of 99 and 12

[left meteor]

no dude

like

[obsidian coral]

Cancel out the numbers that are the same

[left meteor]

i know how to simplify-

im just asking

how does 9 come in 11, 6 or 2?

my teacher told

its this way

[obsidian coral]

This makes no sense

[left meteor]

bruh

he told

6 and 9 are common

so 6 will get cancelled for 2 and 9 for 3

NOW HOW DO I SIMPLIFY 13 WITH 2 AND 5 WITH 2

bruh this makes no sense

school is dumb

tell me why this is used for

[obsidian coral]

It just wants you to simplify the answer

That's all

[left meteor]

how tf do i do that

like

bruh

2 and 5 dont go together

and 13 and 2 dont like

they are not even in the same table

[obsidian coral]

Also, is that 64 or 65 typed?

[left meteor]

64

bruh

nvm

[obsidian coral]

That's not even right

[left meteor]

ikr

my bad

how can

an even number come in a 5 table

[obsidian coral]

It isn't so the answer that you have, which is 65/4, is the most simplified it can be

[left meteor]

yeah i got the answer to that

but

6/6 = whole number form

[obsidian coral]

Yes

So write the simplified answer

[left meteor]

what is the whole number form of 6/6

[obsidian coral]

What's 6/6?

[left meteor]

u mean 6 divided by 6?

1

=1

[obsidian coral]

There you go

You got your answer

[left meteor]

yes

im dumb

Such numbers which on multiplying with each other give 1 as the product are called reciprocals of each other.

What is the reciprocal of 4/7? then @obsidian coral

[obsidian coral]

https://www.math.net/reciprocal

[left meteor]

hwo

i got 1 as the answer

but its wrong

like

[obsidian coral]

What's the fraction multiplied by 4/7 to get 1?

[left meteor]

bru

its 7/4

[obsidian coral]

Yes

That's what the question was asking

[left meteor]

that dividing a number by a given number is the same as multiplying it by its reciprocal.

Hence to divide a number by a fraction, we have to multiply it by its reciprocal.

[left meteor]

bruh

that was taught in like

2nd grade

so ez

[left meteor]

^ right?

[obsidian coral]

If it's "so ez", why are you asking help on that stuff?

[obsidian coral]

Yes

[left meteor]

i thought it was something diffrent

i need help

[obsidian coral]

Based on this info you said

[left meteor]

but how does it change the sign

wait

multiply by its reciprocal?

[obsidian coral]

Yes

[left meteor]

so also written as 1/3 x 5/4?

i mean

5/2

[obsidian coral]

Yes

[left meteor]

cool

[cedar kiln]

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[sharp surge]

Okay so im going to translate this word for word and explain what i know

[sharp surge]

It wants the surface area of sector EFD and gives the length of an arc (BC) it is equal to 6pi centimetres.

So im assuming BC must be equal to ED, but i have no reasoning to back it so im already lacking a portion of the answer but lets assume that they are equal

If they are, since the arc of BC is 6pi cm, you just find the radius of the circle through arc [length of arc = 2pi•r(m/360)]

And rearrange to 2pi•r(45/360) = 6pi cm

Pir(45/360) =3pi

r(45/360)=3

r=0.375cm?

Then the area flops and equals to 0.21pi so idk what to do

I know that 72 is correct but i wouldnt had the question not been multiple choice

<@&286206848099549185>

[cedar kiln]

@sharp surge Has your question been resolved?

[cedar kiln]

@sharp surge Has your question been resolved?

[gray condor]

apologies, i fail to understand the info given in your question. could you give me a literal translation without omitting any info, like what’s the 45 degrees for instance marked in the diagram but not on your translation

[sharp surge]

45 degrees is the central angle of sector EFD

it must be the same as BFC right?

[cedar kiln]

@sharp surge Has your question been resolved?

[gray condor]

ah 45 = pi/4
radius bisecting chord, those triangles are congruent
6pi = r* pi/4
r = 24
area of sector = r^2 theta /2
6pi24/2 = 72 pi i think

@sharp surge

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[hollow garden]

Find the equation of a line passing through the point (2.3), and perpendicular to the line with equation 3y-6x=4?
The answer in the book is 2y- x-8=0 and I don't know how I am always getting this wrong. My working out has been rearranging the first equation and finding that the gradient is -2.
Having a -2 gradient means, m1 (in this case, -2) x m2 = -1, thus m2 must be 1/2.
Then doing, y-3= 1/2(x-2)
thus gibing me 2(y-3)=1(x-2)
2y - 6 = x -2
rearranging to give me: 2y = x + 4
I understand if rearranged to other form it would be, 2y -x -4 = 0 but I don't know how it is -8?

[obsidian coral]

My working out has been rearranging the first equation and finding that the gradient is -2.
That statement there was the cause of the issue

The gradient isn't -2, it's 2

Because 3y-6x=4 becomes 3y = 6x + 4, then $y = \frac{6}{3}x + \frac{4}{2}$

[wraith dagger]

dldh06

[obsidian coral]

And 6/3 = 2

Not -2

[hollow garden]

omg thank u!! i couldnt pick that up thank u sm!!

[obsidian coral]

If you're good now, you can close the channel using .close

[cedar kiln]

@hollow garden Has your question been resolved?

[hollow garden]

actually, i need help with another q if thats ok

Calculate to a decimal place, the angle made with a positive direction of the x-acis by the line that passed through the points (2,1) and (8,-4)

Calculated the gradient anf got -5/6 then did tan inverse (5/6) and got 39.8 deg. but in the book its 140?

im so confused

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[hollow garden]

.reopen

[cedar kiln]

✅

[hollow garden]

Calculate to a decimal place, the angle made with a positive direction of the x-acis by the line that passed through the points (2,1) and (8,-4)
Calculated the gradient anf got -5/6 then did tan inverse (5/6) and got 39.8 deg. but in the book its 140?
im so confused

[cedar kiln]

@hollow garden Has your question been resolved?

[cedar kiln]

@hollow garden Has your question been resolved?

[cedar kiln]

@hollow garden Has your question been resolved?

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[calm sierra]

.close

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[crimson sedge]

When can you conjugate gradient method ?

[calm sierra]

@crimson sedge pls dont multipost

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[cedar shell]

Hello

[cedar shell]

I am trying to figure a percentage out but I am the worst at math

In my county, our corona cases were at 253 approx 3 weeks ago

Now they are at 984

How much % increase is this?

Tryna figure this out because I need to make a school article about it

[keen elbow]

0.98% and 0.25%

[cedar shell]

So a 98% increase in total??

Sorry I’m lost

I am getting 288%

I think it’s wrong though

[cedar kiln]

@cedar shell Has your question been resolved?

[cedar shell]

Yes thanks

[cedar kiln]

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[cosmic ibex]

how would i solve this without a graphic calculator

[wise perch]

dan someone explain how this is correct

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[wise trail]

why aren't this answer E?

[cedar kiln]

Please don't occupy multiple help channels.

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[shadow plume]

the first two terms of an arithmetic progression are given by k and p-2. Given that the 10th term is 2k+p. Express p in terms of k

[shadow plume]

T10=2k+p

a=k

d=p-2-k?

then i dont know how to express p into k

[crimson sedge]

nth term of an AP is given as A{n} = a + (n-1)d where a = first term, d = common difference

[shadow plume]

yes

any ideas on how can i express p into k

[crimson sedge]

find d firstly then second term which is p-2 would be a+d

[shadow plume]

what?

@crimson sedge

channel ocupied

sorry mate

bruh

cmon dude

channel is occupied

[last compass]

srry

[shadow plume]

i dont get it

i got d right

i just listed

is it correct

[crimson sedge]

no that's not correct

[shadow plume]

wait wait

how

2 first terms

what does it mean

i thought its like first time =k

then 2nd is p-2

[crimson sedge]

yeah that's correct sorry

[shadow plume]

owh

[crimson sedge]

you can another equation for d in terms of p and k

[shadow plume]

what?

[crimson sedge]

using the fact that 10th term is 2k+p

[shadow plume]

t10=2k+p?

[crimson sedge]

yes

[shadow plume]

what should i do next

[crimson sedge]

t10 = k+ 9d

as k is the first term

[shadow plume]

wait how did u get k+9d

i thought its 2k+p

[crimson sedge]

tn= a+ (n-1)d, here a= k (as k is first term) and n = 10

[crimson sedge]

and t10 is 2k+p, equate 2k+p with k+9d

find d in terms of k , p

[shadow plume]

what does that mean

[crimson sedge]

2k+p = k+9d

from here d = (k+p)/9

[shadow plume]

owh

i think i got it from here

thanks alot mate

rlly helpful

+rep

[crimson sedge]

okkk

[shadow plume]

.clsoe

.close

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[worthy snow]

i need help

[livid hound]

what have you done so far?

[worthy snow]

nothing

idk what to do

[upper abyss]

Can you calculate y"?

[worthy snow]

find the second derivative ? yes

$y'' = 12x^{2} - 12$

[wraith dagger]

breadiculous

[worthy snow]

for a

[livid hound]

and then solve y'' = 0
i.e.
12x^2 - 12 = 0

[worthy snow]

oh

so for a it would be x = 1 or x = -1

[livid hound]

yes

[worthy snow]

ok ty!!

.close

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[heavy seal]

Towards the bottom of the page, it says that f* must be one-one in order for s to be a subset of f*(tau). I don't understand why f* must be one-one. Please can someone try to explain. This is the identification topologies section in Mendelson's introduction to topology.

[cedar kiln]

@heavy seal Has your question been resolved?

[heavy seal]

I understand that the initial statement: f*^-1(s) is a subset of tau, means that each open set in Y has an open preimage under f*, since f* is continuous (this follows from a theorem a page or two ago). But why does f* need to be 1-1 in order to justify the second statement, which is essentially just the first, but we have applied f* to both sides. I tried showing f*(f*^-1(s)) = s only when f* is 1-1, but couldn't get anywhere.

<@&286206848099549185>

fyi, I might not be able to check my phone for a while, so if anyone wants to free up this channel for something go ahead and I'll come back later.

[cedar kiln]

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[cedar kiln]

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[static sonnet]

hey

[static sonnet]

help

@ocean scroll

[ocean scroll]

yea

[ocean scroll]

do you know soh cah toa?

[static sonnet]

im shit at math😭

8th grade

[ocean scroll]

nah bruh it's fine

[drowsy robin]

This doesn't really need soh cah toa

Basically pythagoras theorem

[ocean scroll]

oh that's right

haha

[drowsy robin]

5²=4²+x²

[ocean scroll]

a^2+b^2=c^2

[versed wedge]

hi

[versed wedge]

its Pythagreom

[static sonnet]

wait

u have to pick an answer

[drowsy robin]

@static sonnet could you do it from here?

[static sonnet]

from what it shows

[versed wedge]

a^2 +b^2=c^2

[static sonnet]

my math = shit

[versed wedge]

he cant give you answers

[drowsy robin]

Don't worry... I'll help you

[static sonnet]

fr ty

[drowsy robin]

X²=25-16

[static sonnet]

one of these

[drowsy robin]

X²=9

X=3

[ocean scroll]

you are in exam rn?

[static sonnet]

nope math homework

khan academy

[ocean scroll]

i see

[drowsy robin]

X²=5²+3²

[static sonnet]

wha-

[drowsy robin]

Soo for pythagoras theorem... for you to get the hypotenuse you find the square root of the base square added to height square

In this case, x is our hypotenuse

So x²=5²+3²

[static sonnet]

ooooooh

what?

[livid hound]

ooooooh
suggest some sort of realisation
what's with the what? after that

[obsidian coral]

If this is Khan Academy, there should have been videos about Pythagorean Theorem

[static sonnet]

yes

skipped it

[obsidian coral]

That should have been in that lesson

You shouldn't skip it

[livid hound]

skipped it
ctrl+z

[obsidian coral]

That's probably why you don't understand the concept

[static sonnet]

pls help

[obsidian coral]

I suggest watching the lesson first

To understand it

And if you still don't, come back

[static sonnet]

i dont

so hard ahh

[obsidian coral]

Did you even watch the video?

[livid hound]

you can't have finished the lesson within 1 minute of saying that you skipped it

[obsidian coral]

Khan Academy has good videos, and teaches pretty well

[cedar kiln]

@static sonnet Has your question been resolved?

[cedar kiln]

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[cedar kiln]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[crimson sedge]

How would I solve something like this using SOHCAHTOA?

[crimson sedge]

<@&286206848099549185> Ik I haven’t waited 15 mins but this is kind of urgent sorry

I’m on it

It’s 2 similar triangles

2:1 ration

Ain’t it 3 triangles stuck to each other?

[crimson sedge]

I mean the large triangle and the top triangle

Oh yea

cos32=6/x

x=6/cos32

6/cos 32

=

Put it into scientific calculator

X

Yea

👍

Wait

Y is a side here right

Wdym?

y is a side not an angle right