help-28

[bronze magnet]

the function is above the x axis at (infinity, -6) and (4, infinity) ?

[quick fox]

correct!

some teachers like it when you use notation like that and some prefer you use inequalities (use whichever your teacher has been using but thats the correct answer)

[bronze magnet]

and then less than or equal to?

the equal to confuses me

[quick fox]

ah okay, let me write it out for you to make it more clear since i think you understand the math already anyways and just need help with notation

[bronze magnet]

yup

the function is less than the x axis between (-6,0) and (4,0) ?

[quick fox]

greater than 0 : $$(\infty , -6) and (4, \infty)$$ \
less than or equal to 0: $$[-6, 4]$$
this can also be written as
greater than 0 : $$x \le -6 and x \ge 4$$ \
less than or equal to 0: $$-6 \le x \le 4 $$

[bronze magnet]

ohhhhhh

so brackets mean "or equal to"

[glossy valve]

celeste16

[quick fox]

yesss

if its parentheses, then that value is not included but the square brackets means that value is included

[bronze magnet]

gotcha thanks

could you tell me how to get the domain of this function?

[quick fox]

the domain of the function is basically all the values except the ones that are "invalid"

so in this case we want to consider the values that would make the denominator 0

because we know we can divide by 0

so what values of x would make the denominator equal to 0?

[bronze magnet]

6 and -9?

[quick fox]

correct

so now we want to express the entire domain of this function and remove those 2 values

(like we did in the previous problem)

[bronze magnet]

-9<x<6 ?

[quick fox]

hmm yes that one of them but theres more
so you basically want to express all the times that the function is greater than and all the times that it is less than (what you have only considers when it is less than)

[bronze magnet]

so like

infinity to -9

6 to infinity

[quick fox]

greater than 0 : $$(\infty , -9)$$ and $$(6, \infty)$$ \
less than 0: $$(-9, 6)$$

notice that all of them are parentheses this time instead of square brackets because we dont want values that make the denominator 0 so we want to exclude the values -9 and 6

[glossy valve]

celeste16

[bronze magnet]

this makes more sense

my last question, how do i rewrite this without logs?

can the 1/2 go into the x

and the 3 go into the 3 by the 8?

[quick fox]

well the way it works is that it can go inside but it becomes an exponent

so basically, i can rewrite the left side

as log base 8 of x^(1/2)

and the right side as

log base 8 of 3^3

and since both sides are log base 8, we can set the insides equal to each other

so we have sqrt(x) = 27

so x = 27^2

[bronze magnet]

im not looking to solve just rewrite without logs

[rare dock]

x = -sqrt(27) doesn’t make sense as a solution

[quick fox]

oh right cuz we have logs here

[topaz valley]

you've done something weird

[quick fox]

have i? 🤨 maybe im trippin

[topaz valley]

you are very much trippin

[quick fox]

why?

[topaz valley]

consider this

[topaz valley]

and this

[quick fox]

ahhh notation, i miswrote it

i thought the math was wrong 😭 okay okay my bad, i rushed the end

[topaz valley]

uh no the math is wrong

[quick fox]

before that point the math is wrong?

[topaz valley]

we'll never find out what layla was gonna say

[quick fox]

is it right now? or is it still wrong?

[topaz valley]

ok good

[rare dock]

LOL i was going to make a joke and say she divided by sqrt but then realized that didn’t make sense

cus it would have been x = 27/sqrt in that case

not x = sqrt(27)

[bronze magnet]

do you guys know how to rewrite without logs?

[rare dock]

celeste explained it earlier

[bronze magnet]

what part? i think log is still there? im not solving for x

[rare dock]

rewriting without any logs and solving for x happen to be roughly the same problem here lol

but you could probably leave it as sqrt(x) = 27 if you wish

[bronze magnet]

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[torn jolt]

So.

1E6 = 1000000 = a million
Can i say 1E1 for 10? And can i say 1E0 for 1? (if i wanna show off?)

[rare dock]

lol sure

[light sonnet]

No one really says 1E0

[rare dock]

but i don’t know what it shows off haha

[torn jolt]

Now, what if i want to write something like 1.00000000000172652516191 and i really need the last part. Can i somehow contract it using E?

[light sonnet]

1E1 is fine because if you were dealing with sig figs and scientific notation

[gilded pulsar]

You can't really write exact values with E notation without having all of the decimal values in your base if I understand correctly

[rare dock]

1 + .172…*10^(something negative) maybe

i didn’t count the digits but that’s how you could determine the (something negative)

[torn jolt]

Btw, 6,6×10^-26 is less than 1, but isn't negative, isn't it?

[gilded pulsar]

Yeah it's not

[rare dock]

correct

[gilded pulsar]

It's just 1/(6,6^26) so it can't be less than 0

[torn jolt]

What's bigger? 6,6×10^-26 or 5,5×10^-26?

[gilded pulsar]

5.5

[knotty fossil]

no

[rare dock]

no?

[gilded pulsar]

1/(6,6^26) and 1/(5,5^26)

[rare dock]

that’s not what it is

[knotty fossil]

hmm you misunderstood

[gilded pulsar]

Oh yeah lmao I'm tripping

[torn jolt]

So, 6,6 is bigger?

[gilded pulsar]

Yes

[torn jolt]

Thanks for help btw

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[bronze magnet]

not sure how they want me to solve this

[gritty rose]

draw a triangle with hypotenuse length 1

[bronze magnet]

ok

does that tell me something?

[gritty rose]

the triangle should have side lengths that fit your given information for tan(theta)

[bronze magnet]

the problem doesnt give me side lengths

[gritty rose]

tan = opposite / adjacent

[bronze magnet]

so i put the 1 and 2 for those side lengths?

[gritty rose]

yes

[bronze magnet]

what does that tell me

[gritty rose]

did you draw the triangle?

[bronze magnet]

yes it has 1 1 and 2 for lengths

are these numbers used for something?

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[short gull]

For 41 I got f(x)=-2^x + 7 but its supposed to be -2(3)^x+7 where does the 3 come from?

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[short gull]

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[hollow moth]

let me see if i understand this correctly. the cosine of an obtuse angle will always be negative, and the cosine of an acute angle will always be positive?

[hollow moth]

pls @ me if anyone responds

[onyx glen]

@hollow moth yes

[hollow moth]

awesome, thank you

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[rapid valve]

for this problem, i would use permutations, but so would I do I know the total number is 45 for the digits, but how would i use the permutations

[empty sapphire]

no because you can have repeated digits

so for example 11111 is a valid code

sorry you edited your response lol.

[rapid valve]

Oh, so would the total amount of digits be 100,000?

[empty sapphire]

yep exactly! It's 10 * 10 * 10 * 10 * 10 = 10^5 or 100,000 permutations

[rapid valve]

Okay, thank you so much!

[empty sapphire]

np!

[rapid valve]

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[void arch]

I don’t quite understand the third step where they moved y from the outside of the first summation

[torn jolt]

this step?

[void arch]

No the one after that step

[torn jolt]

what part dont you understand with that step?

they are moving y into the second sum because y does not depend on the index x

brain fart hold up

instead of summing over the y, they are first summing over the x

and if they sum over the x, they have to change the inside sum to be define y with x instead

the first line is saying "sum over all y, and sum over all x such that g(x)=y"

the second says "sum over all x, and sum over all y such that y=g(x)"

so basically, theyve swapped the two sums

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@void arch Has your question been resolved?

[void arch]

For the first line

I don’t really get how the sum works

Take for example for when y=2

Is it then saying 2 multipled by the sum of all the P(X=x) such that g(x)=2

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[fading relic]

help

[fading relic]

does anyone knwo the answer to this?

[wide sundial]

do you know about supplementary angles?

[fading relic]

no

help

@wide sundial

pls

[wide sundial]

what's the angle on a straight line?

[fading relic]

its 154

180 - 26 = 1154

154*

.CLOSE

[wide sundial]

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[weak plaza]

hello

[weak plaza]

I have no idea how to solve this.

putting in 0 for the x makes it turn into 0/0 doesnt it?

[sharp vine]

it does

can u use l'Hospital?

[weak plaza]

6*0 for the top is 0 again isnt it

and i dont know how to calculate -sin(0) to be honest

not sure how this gives me a result

[weak plaza]

is there a way to know what -sin(0) is ?

[empty sapphire]

well yes sin(0) = 0

so you can do L'H again

[weak plaza]

but its not sin(0) but - sin(0) isnt it

like when you do the derivative thing of cos(x) its not sin(x) but -sin(x) isnt it

[empty sapphire]

right

[sharp vine]

it is, but what's the problem anyway?

[weak plaza]

i need to find the limit

[empty sapphire]

since sin(0) = 0, then -sin(0) = 0, just like if you had 10sin(0) = 0

sin(0) = 0 so whatever is multiplied by it will still be 0

[weak plaza]

oh okay :D... so another time doing the derivative thing right?

[empty sapphire]

yes exactly.
With the first round of L'H, you should get 6x/(-sin(x)), which when you plug in x = 0, you get 0/0

so do L'H one more time

then the answer will be clear

[weak plaza]

the bottom is -cos(0) then, is that = 1 ?

i know about cos(0) but not -cos(0)

for the top its 6 of course, so 6/1 , or is -cos(0) = -1 ? so 6/-1 ?

[empty sapphire]

yep exactly

cos(0) = 1, so -cos(0) = -1

so then you get 6/-1 = -6

[weak plaza]

thank you! so strictly speaking -sin(0) is also = -0

?

or does -0 not exist 😄

[empty sapphire]

-0 is the same thing as 0!

a bit weird I know

it's unique in that fact

-1 is not the same thing as 1, but -0 is the same thing as 0

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@weak plaza Has your question been resolved?

[weak plaza]

thank you, i was wondering if -0 is valid as a number even if its the same as 0

[empty sapphire]

typically you write -0 as 0

[weak plaza]

i tried looking on google if -0 counts as a number but i couldnt find anything

[empty sapphire]

0 doesn't have a sign

it's the only number on the real line with that property. 0 is special 🙂

[weak plaza]

i see

thank you:D

[empty sapphire]

yep np!

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[hazy pasture]

sin2@ =-1/2

[hazy pasture]

can anyone help me

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[ruby gorge]

Hey guys

[ruby gorge]

I was solving this, but I ended with another result

x + 5/2(ln |x²-4|) + C

Would it be correct?

[glossy valve]

Basudev

[ruby gorge]

Alright

So could you help me where I messed up?

Left corner is when I made the transformation

Nevermind

I got ir

But idk how to fix it?

I mean, once I've got this

How do I follow?

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[ruby gorge]

<@&286206848099549185>

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[cloud moon]

yo

what do you not get in that problem?

[fickle dock]

the answer?

[cloud moon]

well here you can't just get the answer

[fickle dock]

mine doesnt align with the one in the answer key

i can show u my working

[cloud moon]

what's your solut

yes show so i check

[fickle dock]

answer key says -0.08

[fickle dock]

is my working on track?

[cloud moon]

sorry was helping another channel

lemme check

[fickle dock]

no worries

[cloud moon]

unless you made arithmetic error the steps seem alright to me that's the approach

[fickle dock]

yea but i ve checked the calculations several times

its pretty straightfoward

however the answer says otherwise

[cloud moon]

ping helpers

[fickle dock]

<@&286206848099549185> could i get a second opinion on this thanks

@cloud moon could u assist me with another question

<@&286206848099549185>

[gritty rose]

This is separable

Rearrange and integrate

[fickle dock]

rearrange?

[fickle dock]

as in expand?

[gritty rose]

No

Have you solved separable equations before?

[fickle dock]

no

this seems to be a simultaneous eq question

[gritty rose]

Integrate both sides with respect to x

[fickle dock]

for which i ve simplified down to (k-1)^6 - 21k - 1 = 0

[gritty rose]

\int dy/dx dx = y

[fickle dock]

my antiderivative is ((kx-1)^6)/3k + c

[fickle dock]

is mine correct?

<@&286206848099549185>

[gritty rose]

Differentiate and check

[fickle dock]

correct

so where do i go from there

i have c and k variable

[gritty rose]

Use your point to solve for C

[fickle dock]

thats where im stuck

[gritty rose]

And your other point to solve for k

Set up two equations. One for each point

[fickle dock]

i have

[gritty rose]

Show it

[fickle dock]

sorry its a little disorientated

[gritty rose]

Was hoping you made an algebra mistake

[fickle dock]

feeling is mutual haha

[gritty rose]

,w solve (k-1)^6 -21k -1=0

[glossy valve]

Results provided by WolframAlpha

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

Requested by riemann#2529

[gritty rose]

Pain

Oh 0 and 3

0 should have been obvious

,calc 2^6 - 21*3-1

[glossy valve]

Result:

0

[fickle dock]

okay no this is way out of my league

[gritty rose]

The intent is for you to guess small integers

I can't explain beyond that

[fickle dock]

is there some other method that doesnt require the binomial expansion of that polynomial

[fickle dock]

as in brute force?

[gritty rose]

Yup

[fickle dock]

thats rather odd

[gritty rose]

If there is, I don't know it

[fickle dock]

alright thanks

[fickle dock]

also could u help verify my working and answer

apparently its wrong but im not seeing it

[fickle dock]

this is the corresponding question

<@&286206848099549185>

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[fickle dock]

<@&286206848099549185>

[fickle dock]

<@&286206848099549185>

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[tardy lotus]

A six digit number has 1 in the units place. When this 1 is changed from the extreme right to extreme left, without changing the order of any other digit, the new six digit number is one-third of the original number. Find the original number

[wild sleet]

it's an equation

XXXXX1, the unknown 5-digit number is the variable

[tardy lotus]

I know

I posted the answer as 599991 but it was wrong

[wild sleet]

okay

[tardy lotus]

Is it possible to have the last number and the first as same?

[wild sleet]

i don't exactly understand the question

awhy does it matter

just make an equation

and solve it

[tardy lotus]

Because there could be only two numbers as the last digit of the second number

1 and 7

[wild sleet]

that's too deep

you're doing too much

there's literally no trick

[tardy lotus]

If you put 7 as the last digit on the second number then you will have 7 as the first digit of the first number. Hence the first number will be bigger than 3x of the last number

[wild sleet]

If you put 7 as the last digit on the second number then you will have 7 as the second to last digit of the first number

[tardy lotus]

Again the last digit can't be any other number than 7 ( for some arithmetical reasons) so it is not possible not have the same number as the first and last digits of the two numbers

[wild sleet]

I don't know

you;re doing too much research, and it's gotta be false but it doesn't matter because it's unnecessary

[tardy lotus]

Why it is false?

[wild sleet]

because you said that if you put 7 as the last digit on the second number then you will have 7 as the first digit of the first number

so i misunderstood, or you misspoke or you're doing something wrong

[tardy lotus]

Yes?

According to do the question they should be equal

[wild sleet]

so, you're doing something that wouldn't actually happen

[tardy lotus]

Why?

[wild sleet]

sometimes it happens

[tardy lotus]

I gave the proof to you

[wild sleet]

i reject everything you said though

[tardy lotus]

I cam understand you though

You need full proof

[wild sleet]

no, i won't read it i promise

[tardy lotus]

Suppose the number is abcde1

Then why did you say am I wrong?

[wild sleet]

I'm like being totally transparent
I think you're wrong because of the statement

you made

that i quoted

i won;t read the proof if it's long, if it's like 4 lines then fine

[tardy lotus]

Leave It's 5 line long

[wild sleet]

i left

[tardy lotus]

👍

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[weak plaza]

hey

[weak plaza]

this is not the same as puttin the ^2 above the 7 and root seperately?

[mighty torrent]

No

[torn jolt]

No because

That can be rewritten as
[
(7-\sqrt{x})^2 = (7- \sqrt{x})(7-\sqrt{x})
]

[glossy valve]

♡LexQa♡

[torn jolt]

Foil this and you can find out what it really is

[mighty torrent]

Why your latex is white

$(7-\sqrt{x})^2 = (7- \sqrt{x})(7-\sqrt{x})$

[glossy valve]

sopinha

[mighty torrent]

'-'

[indigo ore]

Secret

[weak plaza]

what if there was a * symbol instead of -

[torn jolt]

Then yes it is what you described

[
(a \cdot b)^2 = a^2 \cdot b^2
]

[weak plaza]

what the frog man

[glossy valve]

♡LexQa♡

[weak plaza]

okay well

so we have to use the binomical forumale thing?

[weak plaza]

so its a^2-2ab+b^2 right

7 being a and the root being the b

[torn jolt]

Yes

[weak plaza]

math is so complicated

oh man i will fail the exam so badly

https://tenor.com/view/charlie-brown-the-harder-i-try-the-worst-i-get-failed-failing-quote-gif-12063964

[torn jolt]

It took me approximately 10 years to finally separate math from being a chore that I only associated with school into something more than that

Math is fun if you don't treat it as an arduous necessity that you have to undergo

[weak plaza]

may i ask you another one then maybe

if you do the ^2 thing with this

i assume you put a ()^2 around it

but if it would be multiplication instead of + and - then you could do ^2 for every single part of these right?

[torn jolt]

I mean you can always apply what I meant here, if that's what you are wondering about

But I didn't quite understand what you meant there

[weak plaza]

i mean like if on the picture it would be 49 multiplied with the rest and * x instead of + x at the end, could you apply the ^2 to all?

[weak plaza]

i wrote it down!!!

top is how its written and the bottom is what i meant with multiplication instead

<@&286206848099549185>

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[vivid pivot]

hey

[vivid pivot]

having trouble finding the answer for these

[wild sleet]

all of these numbers are rational

[viral jasper]

@wild sleet just giving out answers.

[wild sleet]

about half the time yeah

[vivid pivot]

lol im just trying to get this practice done as soon as possible

[wild sleet]

probably closer to 3/4 of the time

[viral jasper]

That's the fastest way to learn nothing

[vivid pivot]

.CLOSE

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[torn jolt]

How do I solve this logarithmic equation?

Tried searching for a video couldn’t find one

[brave blaze]

its not an equation

[torn jolt]

wdym

Logarithm* whatever

[brave blaze]

$\log_8 (\sqrt{2})$ isnt an equation

[glossy valve]

Ｈｅｒｅｌｓ

[torn jolt]

k

But trying to figure out how to solve it

[brave blaze]

not "solve" again but calculate

[torn jolt]

I cannot use a calculator

so have to figure out how to get the answer to those style problems

[brave blaze]

Who told u to use a calculator

[torn jolt]

just saying

[brave blaze]

brb irl

[torn jolt]

Here is the question again so don’t have to scroll all the way up

<@&286206848099549185>

[gritty rose]

find an a such that

$\sqrt{2} = 8^a$

[glossy valve]

riemann

[torn jolt]

What do I do after that

[gritty rose]

use property of log that says

$\log_x(x^y) = y$

[glossy valve]

riemann

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[bleak pier]

how do i convert a linear function to vertex form and factored form

[hollow sable]

Do you mean a quadratic?

[bleak pier]

no linear

the4 function is f(x)= 10x-0.06x^2

[torn jolt]

that's not linear

[bleak pier]

its a quadratic?

[torn jolt]

yes

[bleak pier]

when i put it in desmos it shows a line not any curve

[torn jolt]

then you are inputting it incorrectly

[bleak pier]

uh

can you show me a picture cuz its showing a line for me still

oh wait

oh i didnt add the x

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[last osprey]

Is this correct?

[full forum]

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[last osprey]

<@&286206848099549185>

[last osprey]

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[distant monolith]

can I get help?

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[karmic bronze]

.helpo

.help

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Commands:
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help: .help

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[karmic bronze]

.close

.solved

[boreal star]

.help

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[gilded thorn]

I am doing partial fractions and I am confused on why x+3A just disappears?

[hot herald]

where

[glacial pasture]

its because theyre equating the x coefficients on the left and right

on the left there are 0 and on the right is 6A+3B

[sharp turret]

It's because they only need the coefficient of X which one on LHS to equate in this question

[gilded thorn]

So they just get rid of x+3A?

[glacial pasture]

no

its 5=(6A+3B)x + 3A

then focus only on the x coefficients

there are no x terms on the left so 0, on the right the coefficient of x is 6A+3B, for the equation to work then 0=6A+3B must be true

[sharp turret]

No, it's like there are 2 terms on RHS they are
1.(6A+3B)x
2.3A
Now on LHS we have only single term that is a constant it's 5
Now actually what's happening we have to compare LHS and RHS with each variable ,in this question we only have constants on both side of quality we we'll compare 5=3A and we don't have the variable of X on LHS so it anemones 0=6A +3B
So you'll get A=5/3 from constants and when you'll substitute the value of A in to the variable equation you will get B as -10/3

Hope it helps

[gilded thorn]

Whats rhs stand for?

[sharp turret]

Right hand side

And LHS stand for left hand side

Of equality

[gilded thorn]

You said we have Constanta on both sides of the equation? I though 5 was the only constant?

[sharp turret]

Yes we a liner equation RHS with one variable of X and 3A is constant

In comparison when the LHS

OK let me explain you with an anology
You are standing on a road

[gilded thorn]

I swear it feels like constant and coefficient change

[sharp turret]

And you have two houses on both the sides do the road one on you left hand and another on you right hand side

No you a your parents living in house which is on right Hand and your friends parents live in the house in your left

Now

Suppose your father is a variable X and your mom is constant
Similarly your friends father went out of town for some urgent work so on his mom is at home and road signifies equality

So here what I am trying to say though your parents may have difference and similarity like different means different values in the equation but similarity means you have an equation on both the sides having parents

[gilded thorn]

So 3x is a constant? I thought any number in front of a variable was a coefficient

[sharp turret]

Have a look at this

[gilded thorn]

Ok I think I get it, 5 and 3A are constants so we group them together. Then divide x on both sides which is 0 since 5 is gone

Would've thought that 3A would become -3A though

[gilded thorn]

Still confused on what happened to the x variable tho?

Did you divide it by 0?

[gilded thorn]

Oh

[sharp turret]

Did you get it?

[gilded thorn]

Kinda, didn't know you could take a variable out of 0

[sharp turret]

Actually we cannot but it's only for better understand

Because X differentiating

If X is gone then how can we differentiate

[gilded thorn]

Whats differentiate mean?

[sharp turret]

Partial fraction is a part of integration and for integration you need to know differentiation

Actually we are just comparing a quantity with a variable

You can also do it like this

[gilded thorn]

Ok this actually makes sense

Now 5 ans 3A, do those just go together? Separate from the rest of the equation?

[sharp turret]

No bro

You are doing elimination method

To solve the value of A and B

So that you can right up go there and place the value og A and B in the question and solve further for integration

[gilded thorn]

How do we know it's 3A = 5 and not vice versa?

[sharp turret]

I guss you have understood

[gilded thorn]

I mean how do we know its not 5 = 3A

[full forum]

@gilded thorn Has your question been resolved?

[timber oriole]

@gilded thorn what's the problem here? catch me up to speed

[gilded thorn]

I'm mostly just trying to figure out how 5=(6A + 3B)x+3A became 6A+3B=0 and 3A=5

[sharp turret]

It's same, only representation is different

[gilded thorn]

Oh

Nvm then

[timber oriole]

well, I have no clue how they got from one to the other

feels like context is missing, you didn't screenshot the whole thing(or what's the problem)

but I need to go for now, so, sorry and good luck

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@gilded thorn Has your question been resolved?

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@gilded thorn Has your question been resolved?

[gilded thorn]

.close

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[crystal hare]

I have to use these to figure it out

There is a lot of material to work with, how do I know which one to use

[torn jolt]

which one?

[crystal hare]

.

[torn jolt]

i mean u gotta pick between rules of inferences and those simplification laws

[crystal hare]

Yes, but I don't know how to identify the correct ones

Where do I start?

<@&286206848099549185>

[crystal hare]

this is the problem

[full forum]

@crystal hare Has your question been resolved?

[crystal hare]

no

[carmine minnow]

yeah the SRi rules work with and/or

so looks like it makes sense to change the implications to and/or

@crystal hare

1 equivalent to not (P and Q) or R or S, which is equivalent to not P or not Q or R or S
2 equivalent to not S or not Q
3 equivalent to P or R
Q => R is equivalent to not Q or R

then the rest is just applying SR5 it seems

[crystal hare]

Thank you, I will try to do it

@carmine minnow

is my first step correct?

[carmine minnow]

no

[crystal hare]

Oh

[carmine minnow]

it makes sense to change the implications to and/or

[crystal hare]

How do i do that

Does this one work?

[carmine minnow]

yeah

[crystal hare]

I used 1.11.12 on 1.

[carmine minnow]

so far okay

[crystal hare]

I am here so far

I am worried, I have made a mistake

[carmine minnow]

so far seems fine

the not (P and Q) can be simplified?

[crystal hare]

which number

4.?

[carmine minnow]

yeah

[crystal hare]

Should i write this instead?

[carmine minnow]

no

that would be skipping a step

[crystal hare]

oh my bad

I think this must be what you meant?

But should i apply it to (r v s) too?

[carmine minnow]

just need to present it properly

next use the SR that are useful

[crystal hare]

Ok

SR5 on 6. and 7.?

or SR5 on 5. and 7.

I can't see it through

I know SR5 is useful, but I can't decide on which

[carmine minnow]

try out?

try both then and see which seems more promising to continue

[crystal hare]

Ok

I have these two

which one makes most sense

[carmine minnow]

7 is wrong

[crystal hare]

Oh

What is wrong?

[carmine minnow]

where did you get not P or not Q from?

[crystal hare]

From 4

[carmine minnow]

yeah but you threw R or S away

[crystal hare]

I don't know how to aplly (1.11.14) on 4. without throwing them away

[carmine minnow]

if A equiv B then (A or C) equiv (B or C)

[crystal hare]

Like this?

[carmine minnow]

why is there suddenly a not

idk what you are doing

[crystal hare]

It says i can add a not on Q, When i do that i get not(R v S)

Well it must be inocorrect but I'll try again

[carmine minnow]

why not rule 15

[crystal hare]

Let me check it

Ok if i use it on 4. how will it turn out

I am sorry if i am being dense, this topic is just not my strong side

[carmine minnow]

yeah next you can apply the SR5

[crystal hare]

Ok, so Which number do I apply SR5 on?

Let me test some numbers i guess

I can't figure the next step out

i have now seperated to be alone

Oh nvm

that won't work

I am aware that if i can get (not Q or R) then i'll be able to turn it into Q implies R. But getting there is the problem for me

@carmine minnow Sorry to bother you

But can you maybe point me in the right directio

direction*

if i can get rid of the Left hand side

on 10.

I can then apply sr5 on 5. and 10. because then I would get not Q or R

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[torn jolt]

what does this (5/3) part mean?

[neon meadow]

Combination

[torn jolt]

but like

what is that?

just all the combinations that contain 5 & 3?

[neon meadow]

Combination is a selection from a group of objects, regardless of order.
C(n,r)=n!/(r!*(n-r)!)

[torn jolt]

yeah ive got that

but theres a 2nd part to the question

it asks to deduce the value of 5 over 3

that's whati m asking

[heavy shoal]

deduce means find

[heavy shoal]

you have to use this formula

[torn jolt]

so it's just a numbers problem within a set problem?

[heavy shoal]

seems so

[torn jolt]

dang

alright thank you

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[heavy shoal]

oh btw

[torn jolt]

.reopen

[heavy shoal]

here’s a cool trick

[full forum]

✅

[heavy shoal]

for example

[torn jolt]

yes

[heavy shoal]

C(5,2) = 5*4/2!

what you do is

you start with 5 in the numerator

and multiply with (r-1) many numbers that come before it

since r=2

[torn jolt]

so 1 more aka 4

ic

[heavy shoal]

yep

[torn jolt]

and then just factorial r

[heavy shoal]

and just divide that by 2!

[torn jolt]

interesting

[heavy shoal]

here’s another example

C(9,3)

you can write this as

[torn jolt]

9x8x7 / 3!

[heavy shoal]

yep, exactly

[torn jolt]

danggg

[heavy shoal]

🙏

[torn jolt]

and 5,3 is gonna be

5x4x3 / 3!

[heavy shoal]

that’s how I memorized it

[torn jolt]

ayo

thank u

[heavy shoal]

yw

[torn jolt]

very helpful

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[heavy shoal]

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[dire laurel]

I'm not sure how to work this one out. My first instinct is to square 4.3 in order to remove the square root function, but then I don't know how to continue from there. I'm not even sure if that's a step I should be taking.

[limber flicker]

what would you get if you square both sides

[dire laurel]

18.49 = 2W / k

or 18.49 = 2W / 69.8

[limber flicker]

never put the numbers in before you have reached your goal

[dire laurel]

sooo 4.3^2 = 2W / k ?

[limber flicker]

just L

$L^2=\frac{2W}{k}$

[glossy valve]

Clarkie

[dire laurel]

gotcha

then what?

[limber flicker]

then solve for W

[dire laurel]

im not sure how to solve for W

[limber flicker]

if you multiply both sides by k what would you get?

remembering that $\frac{a}{a} =1$

[glossy valve]

Clarkie

[dire laurel]

( 2w / k ) * k = 2w

L^2 * K = 2W

[limber flicker]

niceee

now multiply both sides by a half

[dire laurel]

then I can divide L^2 * k by 2, right?

[limber flicker]

yep

[dire laurel]

or multiply by a half yee

Then I can plug 4.3 and 69.8 in. Sooo, I've got 645.3 foot-pounds as the answer. Is that correct?

[limber flicker]

yes

[dire laurel]

awesome 🙂 thanks for the help. I clearly forgot this formula

have a good one

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[mossy badger]

I am trying to get the angle between 2 vectors. One method I was was using the dot product:

cos θ = (a · b) / (|a| |b|)

θ = cos-1 [ (a · b) / (|a| |b|) ]

But I'm confused on how to apply that. Is that not a scalar dividing by a vector?

[fast peak]

no

|a| and |b| are both scalars

and a*b is a scalar

[mossy badger]

oh okay thank you i thought |A| was normalized vector

[fast peak]

that would be a/|a|

[mossy badger]

What does |A| mean then?

[fast peak]

the length of a

[mossy badger]

ohhhh

okay thank you

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[twin wolf]

Is this working correct

If f(x) = 5x^3 - 6x + 2, find f’(2)

f’(x) = 15x^2 - 6

f’(2) = 15(2)^2 - 6

f’(2) = 54

[safe trench]

,calc 15(2)^2 - 6

[glossy valve]

Result:

54

[safe trench]

So yeah

[twin wolf]

Alright

How would we do this

Do we first differentiate and then sub in 10 for x

[safe trench]

Yea

But not 10 for x

-1 for x

[twin wolf]

oh

[safe trench]

f'(-1) = 10

[twin wolf]

3(2x^2) - qx

f’(-1) =3(-2^2) - q(-1)

f’(-1) = -12 + q?

[twin wolf]

Would this be correct?

[worthy tree]

f ' (x) = 6x² - q

[twin wolf]

ohh

36(-1) - q?

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[grave juniper]

hi, i've been trying to figure this out. how can i think about a summation of many tiny sub-problems in terms of a single differential equation as n approaches infinity? i thought i was on the right track when i solved it for this first one:

[grave juniper]

when n=1, we just get:

but as n grows larger and larger, and approaches infinity, then this is what we get:

so that's all good for that first example. that is the exact correct equation that i wanted to turn that summation (which was just a brute force approximation) into.

but now i am trying to do the same thing for this (it is similar but slightly different), without any luck:

[full forum]

@grave juniper Has your question been resolved?

[grave juniper]

ok, so doing the same thing as before, when n=1, this is easy, just replace n with 1, and for this new one here, we get:

which is ultimately just 0, which is correct for n=1.

but for n->infinite, this is the answer, which i just arrived at now by continually guessing until it fits the curve of the summation/approximation. but i have no idea why this works, or how to arrive at this again without just guessing + curve-fitting. is there a concrete method to doing this?

[grave juniper]

🥲

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@grave juniper Has your question been resolved?

[covert goblet]

hard questions usuaally get more replies on one of the topic specific channels i think

[full forum]

@grave juniper Has your question been resolved?

[grave juniper]

i though this would be an easy question. also i have no idea what topic this is.

i don't know anything about math or the different types.

[covert goblet]

So you would need an expression for the sum of the square roots of the natural numbers up to n

https://www.quora.com/What-is-the-sum-of-the-square-roots-of-first-n-natural-number

which seems pretty difficult

[gritty rose]

You prove this with Riemann sums

https://math.stackexchange.com/questions/1157426/riemann-sum-of-sqrtx-with-uniform-partition

The sum approaches 2/3 as n goes to infinity.

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@grave juniper Has your question been resolved?

[full forum]

@grave juniper Has your question been resolved?

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[crystal hare]

I need to prove this by induction.

[crystal hare]

What can I do here, cause I am a bit confused

[torn jolt]

it should be log(j!)

[crystal hare]

Rhs or LHS?

[torn jolt]

start from the left, and use the hint about the log law

[torn jolt]

lhs

log(j!)+log(j+1)=log((j+1)!)

and your inductive step should have a sum

[crystal hare]

Like this?

[torn jolt]

yup

[crystal hare]

Okay so about the log law

Is my a = j and b = 1

[torn jolt]

b=j+1

[crystal hare]

Wait what is my a then?

Oooh

I see

I can rewrite to be log(a*b)

[torn jolt]

yup

[crystal hare]

As for my next step what options do i have from here?

Is there any rule I can use here?

maybe this?

[torn jolt]

yes

[crystal hare]

Now that I have this, I see that they are somehow similar, but how do I get the factorial in outside (j+1)

Should i multiply

I'll try it out

[torn jolt]

double check the definition of the factorial

j! = j(j-1)(j-2)(j-3)....2*1

[crystal hare]

Did I write the first term wrong?

why is there (j-3)?

In this example I found on the internet n! is n * (n-1) * (n-2) * ...(2 * 1)

[torn jolt]

do you understand what the factorial is?

[crystal hare]

Yes if I had the factorial of 3! for example it would be 3 * 2 * 1=6

[torn jolt]

do you understand what the "..." means?

[crystal hare]

No actually it confuses me a bit

[torn jolt]

what is 10!?

[crystal hare]

A really high number 3628800

I used maple for this

[torn jolt]

write it out as a product

[torn jolt]

like this

[crystal hare]

10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 =

[torn jolt]

right

the "..." is when we get lazy: 10! = 10*9*8*...*2*1

[crystal hare]

Oh I see

[torn jolt]

so "..." means fill in the gap with all numbers between 2 and (j-3)

[crystal hare]

But we don't know the numbers between them right?

[torn jolt]

yeah

so we just put the ... instead

[crystal hare]

Ok I see where that comes from. But what about 2 * 1?

Oh

I see

Nvm

It is always there at the end

That is why we used the "..." to fill the gap

[torn jolt]

yeah exactly

[crystal hare]

So what about (j-2)(j-3)

[torn jolt]

I just added them in for clarity

[crystal hare]

So I have to write (j-1)"..." * 2 * 1

[torn jolt]

yeah thats fine

[crystal hare]

I hope this is right

[torn jolt]

yup

[crystal hare]

Well how can i move on from here?

[torn jolt]

notice that the thing in the log looks like (j+1)!

[crystal hare]

Well yea

but I don't know what to do with it

[torn jolt]

thats the end

you've proved the inductive step

so by induction its true

[crystal hare]

Well i kinda see it but where does the ! come from in this situation

[torn jolt]

write out (j+1)!

[crystal hare]

Ok

[torn jolt]

and what is j+1-1?

[crystal hare]

j-1

Okay

Alright

it is the exact same therefore I have proven it

Thank you so much man. Your actually really good at explaining things

I can see how it works now. Thanks a lot

[crystal hare]

Wait that is j my bad

Wait sorry I am confused by just a little thing

[crystal hare]

if this equals j

Then i get (j+1) * (j * "..." * 2 * 1)

Well I guess nvm

Oh wait

I understand now xDD

.close

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[feral sorrel]

how do i find g(0) ?

[feral sorrel]

g(0) = 2cos (0) ?

[summer echo]

yes

and what is 2cos(0)?

[feral sorrel]

180,540?

[summer echo]

what's cos(0)?

[feral sorrel]

i looked at the cos graph and the point as 0 was, 90, 270, then doubled it?

[feral sorrel]

90,270?

[summer echo]

cos(0) just has one value

[feral sorrel]

how do i know that

[summer echo]

if you look at the graph where x = 0, what is y?

it's a function

[summer echo]

it has one output per input

[feral sorrel]

wait wait

[summer echo]

#❓how-to-get-help

[feral sorrel]

y is 1?

[summer echo]

yes, cos(0) = 1

[feral sorrel]

ohh

so now i double it, as 2cos(0) = 2

[summer echo]

yup

[feral sorrel]

and then multiply by 5

ohh

10

thanks so much!

.close

[summer echo]

yes f(g(0)) = 10

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[proven heron]

Have tried several times and still cannot find the inverse function of f(x), wanna get some helps from u guys

[hot herald]

stop innvading others' channels @empty monolith
and multiposting

[proven heron]

I have showed you the method in my previous channel bro

lmao

[hot herald]

note that no attempts should be made to find the inverse function for this

[proven heron]

Yeah, since I had tried and can't find out

[empty monolith]

Sorry

[hot herald]

note that the existence of an inverse for the specified interval indicates that there will be one solution to
2 = 2x + ln(x)
and that can be identified by inspection

[proven heron]

How could I inspect for this

[atomic blade]

Well it's kinda funny. Just find when f(x) = 2

And yeah

You kinda just have to like ... idk how to explain

Know what x is supposed to be just by looking at it

[summer echo]

for what values of x do you know exactly what ln(x) is?

[proven heron]

So all I need to do is let x = 2

[hot herald]

no

[sharp flame]

Just chuck in small integers

and see what happens

[hot herald]

you want the value of the expression
2x + ln(x)
to be 2,

it would help to think about simple case(s) where 2x + ln(x) would be an integer

[proven heron]

hmm