precalculus

[vestal peak]

cool

[hard hornet]

imagine a rod going through the center of both shapes

and yea

[vestal peak]

dude im so confused

[hard hornet]

mmmm what are you confused about?

[vestal peak]

because on the question it says the method to find the volume of the cone is different

i understand the hemisphere

[hard hornet]

ok so the volume of a cone

[vestal peak]

but the method for the cone is different

[hard hornet]

is 1/3 pi r^2 h right?

[vestal peak]

it says 1/3 pi r sq h on the sheet

[hard hornet]

where r represents the radius

[vestal peak]

yeah

[hard hornet]

and h is the height right?

but

the height is 4x

1/3 pi r^2 h is just the formula for a one

cone

given the height and the radius

in this case

the radius is x, and the height is 4x

[vestal peak]

I understand that you substitute it

why does it say 4/3 pi r sq h on the mark scheme

[hard hornet]

gimme a sec

$ 1/3 * pi * x^2 * 4x$ this is what you get right?

[obsidian monolith]

MemesPlease:

[hard hornet]

r = x, and h = 4x

all they really did

was pull out the 4 into the front

so it becomes

$(4/3) * pi * x^2 * x$

[obsidian monolith]

MemesPlease:

[hard hornet]

$(4/3) * pi * x^3$

[obsidian monolith]

MemesPlease:

[vestal peak]

still confused

[hard hornet]

im confused on what you're confused about

[vestal peak]

i understand that the height is 4x

and the r is x

[hard hornet]

yup

[vestal peak]

but why does the mark scheme use 4/3 pi r sq by h instead of the actual formulae 1/3 pi r sq by h

[hard hornet]

its not

[hard hornet]

its using the formula 1/3 pi r^2 h

[vestal peak]

q 11

[hard hornet]

yes

the second chunk right?

in the first equation

[vestal peak]

Yeah

[hard hornet]

its not

its using the formula $(1/3) * pi * r^2 * h$

[obsidian monolith]

MemesPlease:

[hard hornet]

but we know that the radius is x

and the height is 4x

so it becomes

$(1/3) * pi * x^2 * 4x$

[obsidian monolith]

MemesPlease:

[hard hornet]

r = x, and h = 4x

what happens next

is that they take the 4

and move it to the front

$(4/3) * pi * x^2 * x$

[obsidian monolith]

MemesPlease:

[hard hornet]

multiplicative property, ab = ba, you can move things around when you're multiplying

its NOT using the (4/3) formula, its using the (1/3) cone formula

but after plugging in numbers

it happens to turn out to be (4/3) * pi * x^3

[vestal peak]

How do you know when to move it

and why do you move it

[hard hornet]

just experience, and its good practice to move numerical constants to the front

like if you get something like

$3 * x^2 * 5 * y * z * 3$

[obsidian monolith]

MemesPlease:

[vestal peak]

so just switch it like a fraction

[hard hornet]

that looks incredibly messy

so move out all the constants

$4*(\frac{1}{3}) = \frac{4}{3}$

[obsidian monolith]

MemesPlease:

[hard hornet]

does that make it a bit clearer?

[vestal peak]

nope

[hard hornet]

what are you still confused about

[vestal peak]

how to move it

and when you know

you have to

[hard hornet]

its kind of experience, the more problems you do, the more you know when to do something.

Also, because it looks a lot cleaner

remember your multiplication property?

$a * b * c = c * b * a = c * a * b$

[obsidian monolith]

MemesPlease:

[hard hornet]

if its multiplicative

you can move them around

[vestal peak]

Do you not have a mathematical explanation?

[hard hornet]

cuz it still represents the same thing

[vestal peak]

yeah

[hard hornet]

its practice, in higher maths

when you do derivatives, integrals, and all that nast ystuff

moving out constnats make problems easier to solve

it looks cleaner

its almost like organizing

it doesn't get so cluttered

$3 * x^2 * 5 * y * z * 3 = 45 * x^2 * y * z$

[obsidian monolith]

MemesPlease:

[hard hornet]

by moving constants, the equation already looks a lot cleaner

its just a mathematical practice you should develop. think of it as an organizational skill

does that clear things up?

[vestal peak]

im getting closer

But i dont understand how you move the 4

because

usually only the bottom number in a fraction is replaced

Like when you switch them

[hard hornet]

what do you mean?

[vestal peak]

like 2a/3 =4

[hard hornet]

mhm

[vestal peak]

then you move the 3

over

[hard hornet]

2a = 12

[vestal peak]

so 4*3

then you can move the 4

so its 2a.4

2a/4

[hard hornet]

thats when you have an equation

$\frac{2a}{3} = 4$

[obsidian monolith]

MemesPlease:

[hard hornet]

basic algebra says that you can do something to one side, and that you have to do it to the other side too, because they are equal

you could add 1 to both sides

or subtract 1 to both sides

or you could

multiply both sides by 3

which gives you

$3 * \frac{2a}{3} = 4 * 3$

[obsidian monolith]

MemesPlease:

[hard hornet]

the fraction in the bottom cancels out

and you get

2a = 12

at this point, it just comes down to basic algebra

[vestal peak]

But that doesnt explain the

[hard hornet]

if you were given an expression

[vestal peak]

cone calculation

[hard hornet]

not an equation

an expression

$(1/3) * pi * r^2 * h$

that is an expression

you can move things freely because multiplication

[obsidian monolith]

MemesPlease:

[hard hornet]

you are getting equations and expressions mixed up

you could probably have something like $(1/3) * h* r^2 * pi$

[obsidian monolith]

MemesPlease:

[vestal peak]

so you can move them

[hard hornet]

and it represents the same exact thing

[vestal peak]

but they always remain

[hard hornet]

for an expression yes

an expression

a+b

a*b

a divided by b

a+b can also be re-written as b+a

a*b can be re-written as b * a

thats an expression

[vestal peak]

so it can be used for equations and stuff

[hard hornet]

if you're given an equation, then thats a different story

but for now

you're only worried about expressions

expressing the volume

[vestal peak]

Ok

thank you

im heading to sleep

[hard hornet]

yea np, feel free to ask anytime

[vestal peak]

ty

[vestal peak]

@hard hornet Yo

[hard hornet]

hey

[vestal peak]

Could you assist me again today?

[hard hornet]

ye can do

[vestal peak]

What is wrong about the mean

I know there is a different number of people

[hard hornet]

13 or 14

[vestal peak]

13

[hard hornet]

ok so you're given

10 girls average a 70

and 15 boys average an 80

[vestal peak]

yup

[hard hornet]

average really means, the total score of all 10 girls divided by 10 girls

[vestal peak]

yup

[hard hornet]

I think in order to approach this problem, you should work backwards

so if we know the average of 10 girls is 70

what is the total score of the 10 girls

(you'll see where I'm getting at)

if you were to take every score of a girl, and add them all up

what would you get

Another important thing to note, is that the individual score of each girl doesnt matter, the only thing you need to compute the total score of girls is the average and the number of girls

i think this is a bit too much, lets simplify it a bit

lets say there are two people

Alice and Bob. Now, you know that the average score of Alice and Bob is 10

what is the TOTAL score, of Alice and Bob?

[vestal peak]

idk

[hard hornet]

Alice and Bob got some score

but their average is 10

[vestal peak]

okay

above 10

[hard hornet]

what is Alice's score + Bob's score

yes its above 10

but there is an exact value

[vestal peak]

I know it

but cant seem to work it out

in my head

[hard hornet]

think about it

Like I said

[vestal peak]

20

[hard hornet]

CORREC

How'd you get that

[vestal peak]

*2

because

there is two people

working backwards

like you said

[hard hornet]

you don't need to know their scores exact

you just know their total is 20

this is important

the scores could be 1/19, or 10/10

but you know the average is 10

so the total MUST be 2

20

now, going back to the original question

if 10 girls had an average of 70

what is the TOTAL score of the girls?

[vestal peak]

700

OH

btw are you a teacher

[hard hornet]

nop

college student

now

[vestal peak]

Oh k

[hard hornet]

for the bos

boys

what is the total score of the boys

[vestal peak]

Tyvm for you time then

[hard hornet]

ye ye np

[vestal peak]

80 *15

[hard hornet]

happy to help

[vestal peak]

1200

[hard hornet]

yup

now, the total score of both boys and girls would be 700 + 1200 right?

[vestal peak]

Yup

1900

[hard hornet]

1900 is the total score of both boys and girls

now, what is the average of the whole class then?

[vestal peak]

Divide by the amount of girls and boys

[hard hornet]

mhm

[vestal peak]

so 1200/ 25

[hard hornet]

no

1900/25

[vestal peak]

oh mb

yeah 1900/25

[hard hornet]

and what do you get

[vestal peak]

idk

1 sec

mental maths

is a bit of a problem for me

[hard hornet]

just use a calculator

[vestal peak]

80

[hard hornet]

no

[vestal peak]

70

[hard hornet]

no

bro calculator

[vestal peak]

its non calculator

[hard hornet]

whats 2000/2

2000/25

and then subtract 4

1900/25 = (2000/25) - (100/25) = (2000-100)/25

[vestal peak]

76

then

[hard hornet]

and boom, you proved Nick is hella wrong, so you can go beat em up

[vestal peak]

lol

[hard hornet]

the average isnt 75

if you want a more intuitive explanation

its because the number of boys and girls aren't the same

the average should be weighted more towards teh boys

[vestal peak]

Yeah it said that on the mark scheme

[hard hornet]

because there are more boys who scored higher

thus the score should skew more towards 80

but ye, you can show it via math or logic

hope that helped!

[vestal peak]

ty

Could you also help

with the surds question

[hard hornet]

surds question?

[vestal peak]

14

[hard hornet]

ah

[vestal peak]

Isnt it

expand and rationalize

[hard hornet]

think about it

what does

[vestal peak]

i just seemed to forget how to do it

[hard hornet]

(a+b)(a-b) become?

[vestal peak]

a sq - b sq

[hard hornet]

correct

now do the same thing

where a = 4, and b = root(3)

[vestal peak]

thats a great method to do it tbh

errr its

13

[hard hornet]

whats 13 divided by root(13)

[vestal peak]

1

[hard hornet]

no no nooooo

13 divided by 13 is

1

what is 13 divided by squareroot of 13

[vestal peak]

rationalize?

[hard hornet]

$\frac{13}{\rt{13}}$

[obsidian monolith]

MemesPlease:
Compile Error! Click the reaction for details. (You may edit your message)

[hard hornet]

welp fuck

[willow bear]

$\frac{13}{\sqrt{13}} = ?$

[obsidian monolith]

Ann:

[hard hornet]

thank you

i still need to learn latex dammit

[vestal peak]

times bottom and top by square root 13

[hard hornet]

that works too

and then what do you get?

[vestal peak]

then you get 13 at the bottom

and at the top you get

169

over 13

[hard hornet]

nop

the top is not 169

[vestal peak]

oh

[hard hornet]

check your math again

[vestal peak]

its 13 square root 13

[hard hornet]

yup

cancel the 13's

[vestal peak]

what am i doing

[hard hornet]

and you get your answer

[vestal peak]

square root 13

[hard hornet]

tada

[vestal peak]

I dont understand why the mark scheme never gives good working out

[hard hornet]

all of 15?

[vestal peak]

yeah

[hard hornet]

i got time to burn, so why not

alright lets start with 15a

are you familiar with squareroots and cube root?

[vestal peak]

yup

[hard hornet]

ok

my recommendation

break that number into its prime factors

$8 * 10^6$

[obsidian monolith]

MemesPlease:

[hard hornet]

break this up into its prime factors

let me know when you're done

what im going to teach you is a general method that works for almost every root problem

[vestal peak]

2

[hard hornet]

write it all out

[vestal peak]

2,2,2 and 10 is 5,2

mb mb mb

[hard hornet]

that looks correct

im going to work ahea

[vestal peak]

4,2,1 and 10 is 5,2,1

[hard hornet]

so what you should've gotten is

$2^9 * 5^6$

[obsidian monolith]

MemesPlease:

[hard hornet]

when you prime factorize

you break a number up into its primes

if you have 24, for example, prime factorization would become $2^3 * 3$

[vestal peak]

why did you do 2^9

[obsidian monolith]

MemesPlease:

[hard hornet]

oh i skipped a step

$8 * 10^6 = 2^3 * 5^6 * 2^6$

[obsidian monolith]

MemesPlease:

[hard hornet]

correct?

the 8 becomes 2^3

and the 10 becomes 2^6 * 5^6

[vestal peak]

Oh k

[hard hornet]

makes sense?

i did skip a few steps here and there

[vestal peak]

yup its fine

[hard hornet]

now

we want to take the CUBE root of that

ok before we do that

lets do something easier

lets say, you want to take the cube root of

2^3

what would you get?

[vestal peak]

wdym

8

[hard hornet]

take the cuberoot of 2^3

[vestal peak]

2^3

[hard hornet]

start with something easier

you'll c the pattern when we work through it

[vestal peak]

i dont understand

what im supposed to do

[hard hornet]

take the cube root

for now

i want you to

take the cuberoot of 2^3

[vestal peak]

thats 2

[hard hornet]

correct

now, i want you to take the cube root of 2^6

i want you to recognize some pattern

[vestal peak]

2

[hard hornet]

no

cuberoot of 2^6

[vestal peak]

idk

i cant work that out in my head

[hard hornet]

dont work it out in your head

look for a pattern

[vestal peak]

4?

[hard hornet]

correct!

[vestal peak]

does it go up in 3s?

[hard hornet]

or 2^@

2^2

[vestal peak]

and advance by 2

[hard hornet]

you're starting to see it

now

what if i asked you to take the cube root of 2^15

dont do the math

you'll kill yourself

find the pattern

[vestal peak]

10

[hard hornet]

im gonna assume you didnt learn how to multiply exponents

do you know how to multiply exponents?

its not 10

[vestal peak]

what do you mean

[hard hornet]

do you know how to multiply exponents

like if i told you

what is

(2^2) * (2^5) * (2^3), what would it be

as in combine it into one term

2^?

[vestal peak]

wait

was the answer

[hard hornet]

do you know how to multiply exponents? if not, i'll have to teach you some thing easier

[vestal peak]

32

[hard hornet]

or 2^

5

YUP

You starting to see it?

[vestal peak]

i see

you double it

ever 3

every 3

[hard hornet]

in higher maths, you'll learn how to do this formally with exponent multplication, but im assuming you dont know how yet

but yes

if you take the CUBE root

you only need to look for powers of multiple of 3

so cube root of 2^21 would be 2^7

cube root of 2^6 would become 2^2

make sense?

[vestal peak]

what would 5^6 be

same thing?

[hard hornet]

same pattern

[vestal peak]

so 563

5^3

125

[hard hornet]

not 5^3

cube root

what you just calculated was the square root

we are looking for the CUBE root

[vestal peak]

oh

oh

5^2

[hard hornet]

yup

now

[vestal peak]

so 400

[hard hornet]

multiplication and root rules

allow us to do something like this

$\sqrt{2^9 * 5^6} = \sqrt{2^9} * \sqrt{5^6}$

[obsidian monolith]

MemesPlease:

[hard hornet]

you'll learn this in higher maths, but just know that when things are multiplied, you can do this

[vestal peak]

can i divide the cube root by 3

for any

[hard hornet]

wdym?

[vestal peak]

Like for 9^12

to find the cube root

i divide by 3

and to find the square root

i divide by 2

[hard hornet]

you got it

😄

thats exactly what you do to compute roots

first thing, is break it up into its prime factors

[vestal peak]

Thats amazing

[hard hornet]

in this case

9^12 would be 3^24

and then if you take square root

it would be 3^12

if you take the cube root, it would be 3^8

heck, if you wanted to take the EIGHTH ROOT, it'd be 3^3

thats my general method for computing roots

1. Break it up into prime factors
2. Do math

[vestal peak]

wait what

how did you get from 3^12 to 3^8

arent you supposed to divide by 3

[hard hornet]

read it again

cube root would be 3^8
eighth root woudl be 3^3

[vestal peak]

how would the cube root be 3^8

isnt it supposed to be 3^4

[hard hornet]

3^24

whats 24 / 3

[vestal peak]

8

[hard hornet]

so its 3*8

3^8

i mean

[vestal peak]

But where did you get the 24 from

[hard hornet]

o it was just an example i pulled from my ass

to demonstrate

[vestal peak]

oh lmfao

[hard hornet]

LOL

but does this make sense

[vestal peak]

Yeah

so for 9^12

[hard hornet]

if we wnated to take the nth root, we just take the exponent and divide by n

[vestal peak]

it would be different

[hard hornet]

what would you like to do?

squareroot, cube root, or some other root?

[vestal peak]

cube

divide by 3

right

9^4

[hard hornet]

mhm

[vestal peak]

okay

[hard hornet]

correct

[vestal peak]

and square root is 9^6

That makes sense

what about for the other questions

b and c

[hard hornet]

have you learned fractional exponents?

[vestal peak]

kind of

Im in high school

[hard hornet]

what is something to the power of 1/2

or what does it really mean

[vestal peak]

forgot how to do it though

isnt it square root

[hard hornet]

yup

[vestal peak]

and 1/3 is cube root

[hard hornet]

now how about negative exponents

what does that mean?

[vestal peak]

flip

then square

[hard hornet]

$x^-1 = 1/x$

yup

[vestal peak]

so -1/3

becomes 3/1

then cube

[hard hornet]

wait no

wait no

[vestal peak]

it

[hard hornet]

no no

[obsidian monolith]

MemesPlease:

[hard hornet]

GODDAMMIT

But

x^(-1) = 1/x

there

[vestal peak]

lol

[hard hornet]

negative exponents

mean you just make it 1/whatever it is

[vestal peak]

So what would - 1/3 become

[hard hornet]

x^(-1/3) = 1/(x^(1/3))

[vestal peak]

lol

the servers trolling you

[hard hornet]

i dont know latex i cri

[vestal peak]

np

[hard hornet]

ok getting back

what is the squareroot of 144

[vestal peak]

12

[hard hornet]

and what is the cuberoot of 64

or i should say

cuberoot of 1/64

[vestal peak]

wait

we didnt discuss

what happens with the

• 1/3

what does it become

1/3?

[hard hornet]

negative exponents

mean we just take 1 over it

let me try see if i can do it again

[vestal peak]

can you relate it to the question

[hard hornet]

x^(-1/3) = 1/(x^(1/3))

[vestal peak]

it would be more understandable

[hard hornet]

fuck can any honorable show me how to do exponents

on this shite

aaaHHHHH

<@&286206848099549185> knid of a weird question, how do you do exponents using latex

[willow bear]

$x^4$

[obsidian monolith]

Ann:

[willow bear]

like this

[hard hornet]

negative exponents

it dont work

it does weird things

[charred hull]

$x^{-3}$

[obsidian monolith]

xollouarfloride-a white stainimi:
Compile Error! Click the reaction for details. (You may edit your message)

[willow bear]

if you want more than one character do this: $x^{-4}$

[obsidian monolith]

Ann:

[hard hornet]

oh

thank you

as i was saying

$x^{\frac{-1}{3}} = \frac{1}{x^{\frac{1}{3}}}$

holy fuck

oh my god

that looks horrendous

there we go

thats what it means

when you take the negative exponent

[obsidian monolith]

MemesPlease:

[vestal peak]

soo

1/ 64 cube rooted

8

[hard hornet]

$64^{\frac{-1}{3}} = \frac{1}{64^{\frac{1}{3}}}$

[obsidian monolith]

MemesPlease:

[vestal peak]

yup

[hard hornet]

what is that

and do math

and ur done

[vestal peak]

so its 1/8

right?

[hard hornet]

mhm

[vestal peak]

last one

c

[hard hornet]

can u repost question

[vestal peak]

np

[short sorrel]

wait wait

1/8

did i miss some context

[hard hornet]

wait shit

its not 1/8

[short sorrel]

yeah its 1/4

[hard hornet]

thats square root

cube root is not 1/8

[vestal peak]

mb mb

[short sorrel]

the cube root of 64 is 4

[hard hornet]

shit im too tired thanks for catching the mistake

appreciate it

ok now

for part c

heres a hint

$3^{2x} = (3^{x})^{2} = \frac{1}{81}$

[willow bear]

wrong

this is false

[hard hornet]

wait shit no not like that

is something wrong?

[willow bear]

$3^{2x} = (3^x)^2$

[vestal peak]

you guys are harsh lol

[obsidian monolith]

Ann:

[hard hornet]

yes

thank you

i couldnt latex

i was gonna fix it

but ye think about it that way

ok gotcha

ye

think about the question like that

[vestal peak]

so 9 x

[hard hornet]

wat

[vestal peak]

after expansion

9x sqaured

[obsidian monolith]

MemesPlease:

[hard hornet]

how would you solve this

or solve for x

its an equation

so what you do to one side, you have to do to the other side

[vestal peak]

square root

[hard hornet]

uh huh

keep going

[vestal peak]

so you get (3x) = 1/9

[hard hornet]

$3^x = \frac{1}{9}$

yes

[obsidian monolith]

MemesPlease:

[vestal peak]

then divide by 3

x = 1/3

[hard hornet]

you can't do that

its saying

[vestal peak]

ok

[hard hornet]

3^x gives you 1/9

[vestal peak]

ok

[hard hornet]

you cant just divide by 3

[vestal peak]

idk

[hard hornet]

heres a hint

do this

$3^x = \frac{1}{3^{-x}}$

holy fuck

[obsidian monolith]

MemesPlease:

[hard hornet]

there

think about it like that

and then solve for x

that means

$\frac{1}{3^{-x}} = \frac{1}{9}$

[obsidian monolith]

MemesPlease:

[hard hornet]

solve for x here

hint, x is negative

[vestal peak]

dammit

idk

• by + x

both sides

[hard hornet]

think about it

given this $\frac{1}{3^{-x}} = \frac{1}{9}$

[obsidian monolith]

MemesPlease:

[hard hornet]

you should only be concerned about this part

$3^{-x} = 9$

[obsidian monolith]

MemesPlease:

[hard hornet]

in order to make the equation true

then the above must also be true

solve for x now

[vestal peak]

multiply - x by +x

moving it to other side

[hard hornet]

you dont need ot multiply anything

just think about it

what is -x supposed to be???

do i need to simplify it further?

[vestal peak]

to the power of something

[hard hornet]

$3^{-x} = 9 = 3^2$

[obsidian monolith]

MemesPlease:

[hard hornet]

there hope that helps

[royal gull]

3 to what power raised equals 9?

[hard hornet]

now

what is -x

[vestal peak]

-2

[hard hornet]

-x is 2

thus

x is -2

[vestal peak]

okay

i see now

[hard hornet]

but we had to do a LOT of simplifcation

the more you practice

the more you see it steps ahead

[vestal peak]

so what would the answer be

[hard hornet]

i just wrote it

[vestal peak]

3 ^ -2 = 9

[hard hornet]

x = -2

[vestal peak]

ty

[willow bear]

$3^{-2}$ is not 9

[obsidian monolith]

Ann:

[hard hornet]

im sorry if i messed up the problem, had to go through a LOT of hurdles to simplify the problem down, hope you understood it

if not, I can repeat it

[hard hornet]

17 use quadratic equation

[vestal peak]

I used that

but im having difficulties

[hard hornet]

what are you having difficulties with

[vestal peak]

when simplyifying

[hard hornet]

what did u end up with

take a picture of it

[vestal peak]

some crazy answer

[hard hornet]

send picture

[vestal peak]

How do I send it from my phone

Done mb

[hard hornet]

lolwat

hoold on

36 - 4(1)(-8)

check your math on that again

[charred hull]

,rotate -90

[vestal peak]

dont you just multiply -4 by 1 and -8

[hard hornet]

check your math carefully

36 - (4)(1)(-8)

you should NOT be getting 32 + 32

[vestal peak]

how do i do it then

[hard hornet]

check your math

[charred hull]

oh i can see where you went wrong here

[hard hornet]

do your math again

[charred hull]

when you multiplied out 4(1)(-8), for some reason you turned that into -4+32

[vestal peak]

whoops

im a bit of an idiot

68

then

[short sorrel]

making a mistake doesnt make you an idiot

[charred hull]

[short sorrel]

well

[vestal peak]

not realizing itdoes

[short sorrel]

at least, if it does

then im a superidiot

[charred hull]

an idiot is the kind of person who makes those mistakes repeatedly despite being told how to improve, refusing to learn

everyone makes a lot of mistakes, so just make sure you do your best to learn from them

[vestal peak]

No wonder you guys are helpers

you help not only mentally

but with maths aswell

[charred hull]

me helping

[vestal peak]

ty

👌

what do you do

after you get 68

6+- square root 68

over 2

@hard hornet

[hard hornet]

yup

but i recommend divindg everything by 2

[charred hull]

try to simplify it as best you can

[vestal peak]

ok

so 3+- square root 35

34*

[hard hornet]

not 34

carefu

[charred hull]

no, you wouldn't be dividing like that

don't simplify through dividing by 2 yet

so the term: 68 can be broken up into 2x2x17

[vestal peak]

WHAAAAAAAAT on earth

[charred hull]

which means the internals of the squareroot is $\sqrt{4*17}$

[obsidian monolith]

xollouarfloride-a white stainimi:
Compile Error! Click the reaction for details. (You may edit your message)

[vestal peak]

oh so we doing surds?

[charred hull]

which simplifies to $2\sqrt{17}$

[obsidian monolith]

xollouarfloride-a white stainimi:
Compile Error! Click the reaction for details. (You may edit your message)

[vestal peak]

how did you go from square root 4 *17 to 2 sqaure root 17

[charred hull]

so when you want to take terms out of a square root, you have to squareroot them

basically: $\sqrt{4*17} = \sqrt{4} * \sqrt{17}$

[vestal peak]

This is really helpful

[obsidian monolith]

xollouarfloride-a white stainimi:
Compile Error! Click the reaction for details. (You may edit your message)

[vestal peak]

how did you do 68 then

[charred hull]

well, you can find that 68 = 2*34 by just testing with the low factors

then you can find 2*34 = 2x2x17

[vestal peak]

oh k

Thats amazing

then you get 4*17

[charred hull]

yep

[vestal peak]

then simplify it

2* square root 17

[charred hull]

yes

now you can simplify as a whole

[vestal peak]

and you get square root 17

[charred hull]

yep

[vestal peak]

i see

[charred hull]

$\frac{6 \pm 2*\sqrt{17}}{2} = 3 \pm \sqrt{17}$

[obsidian monolith]

xollouarfloride-a white stainimi:
Compile Error! Click the reaction for details. (You may edit your message)

[vestal peak]

Can you help with 18?

1 sec

brb

[hard hornet]

thanks for taking over

now im back

can you repost question?

this might be my last one, i need to head to bed soon

oh yikes im not very good at geometry

you should as this in #geometry-and-trigonometry

[ruby otter]

when I say log

it generally means base ten or 'e'

i always use base 'e'

[willow bear]

good. you should continue using log w/o explicit base indication as natural log.

[ruby otter]

so log = ln ??

[royal gull]

Log = logarythm in base 10, ln=natural log

Any other usage is heresy

[ruby otter]

so when i say log it means base 10 ??

[royal gull]

Thats how I would interpriate it

And most people

[ruby otter]

but Ann :good. you should continue using log w/o explicit base indication as natural log.

[royal gull]

[willow bear]

in basically any higher math text that uses log

log is understood as natural log

all other logs differ from it by a constant multiplier anyway.

[royal gull]

How high is higher math text

I usually see ln

[ruby otter]

ya thats why i always treat log as ln but many of my frnds say its wrong -__-

[frozen needle]

Well the iso norm does say not to use log as ln

[royal gull]

[frozen needle]

[viscid thistle]

ln base 10 to express log

[ruby otter]

does iso norm say log is log base 10?

[viscid thistle]

just log is base 10

[frozen needle]

iso recommends lg for base 10 logarithm

I don't like it xd

[viscid thistle]

lg?

what

I've definitely never seen lg anywhere

[frozen needle]

and lb for base 2

[viscid thistle]

just the brand LG, life's good

well that's Bs

[ruby otter]

lb for base 2
lg for base 10
ln for base e
then what is log for ??

[viscid thistle]

to make confusion

lol

just use ln

[frozen needle]

"log is used when the base does not need to be specified"

[ruby otter]

this shit is confusing when log is used what base should i consider

[viscid thistle]

log's "default" base is 10

its that easy!

[frozen needle]

There is stuff to disagree with in the iso norm ¯_(ツ)_/¯

[ruby otter]

||||

[rare zephyr]

Log?

Yeah that confused me when I first learned it

I don't know how I consider this harder than derivatives

And it took me longer to understand it too

[heavy panther]

@viscid thistle actually it isn't.
As already pointed out, it is generally used to denote the logarithm base e

[leaden crow]

Default is natural log, base e, hence the name natural

[viscid thistle]

ln is used to denote the logarithm base e

[heavy panther]

That too

[viscid thistle]

but log is used to denote the logarithm base 10 right?

[heavy panther]

No

[viscid thistle]

Hmm

[leaden crow]

Not always

[viscid thistle]

I wonder why they told us this in high school

[leaden crow]

Because they did always use that as base 10

Not a hard rule outside of school though

[ruby otter]

coz in school we study applied math like thermodynamics ( phy, chem ) so while doing numerical we would consider base 10

[deft flume]

Can someone help me with this please? Thanks

[dapper geyser]

1 + tan^2 = sec^2

@deft flume

use that once

then u have a quadratic with x = sec(theta)

[deft flume]

i used that to get sec^2θ -1 = -3secθ -3

do i set both of the expression to equal to 0?

[dapper geyser]

move all terms to 1 side

[deft flume]

done

[dapper geyser]

solved?

[deft flume]

then i treat it like x^2+x+2 =0 ?

[dapper geyser]

yeah

or whatever the correct polynomial is leme check

[deft flume]

-2 , -1

[dapper geyser]

do u mean x^2 + 3x + 2

ok yeah

[deft flume]

oh yea

typo my bad

then i use the unit circle

[dapper geyser]

yeah

[deft flume]

now im a bit lost

how do i find the value for -2

[dapper geyser]

sec = 1/cos

cos = 1/sec

cos = -1/2

@deft flume

[deft flume]

can you elaborate the steps on utilizing the x values

I know you used the reciprocal identities but im still confused

nvm i understand now

thx for the help

[atomic urchin]

Can someone help me with this. I am so lost.

[limber bone]

zoom please

cant see

or just write out question'

[wise kelp]

Which part do you need help with?

[atomic urchin]

Part A and part C

Please and thank you

[wise kelp]

Do you understand the definitions at the top of the page?

[atomic urchin]

Honestly barely

[wise kelp]

@atomic urchin So starting with the first one

[atomic urchin]

Ok

[wise kelp]

Generic: No numbers (variable coefficients)

This means something like ax^2 + bx + c

[atomic urchin]

Right

[wise kelp]

As opposed to something like 3x^2 + 5x + 2

So you got that

You know what a coefficient is, right?

[atomic urchin]

-8

Or 8

[wise kelp]

?

[atomic urchin]

It’s -8

[wise kelp]

Oh

I see what you’re saying

Not quite

[atomic urchin]

🤔

[wise kelp]

A coefficient is the number part of a term in a polynomial

The part that a variable is multiplied by

So if one of the terms is 3x^2, the coefficient of that term is 3

[atomic urchin]

So -12

[wise kelp]

That’s one of the coefficients

[atomic urchin]

Ahhh ok

[wise kelp]

Each term has a coefficient

So the coefficient of the third term is 33

[atomic urchin]

1, -12,33,8

[wise kelp]

Yes

I’m not sure the last one would be considered a coefficient

Usually coefficients are numbers that variables are multiplied by

[atomic urchin]

Ok

[wise kelp]

Standalone numbers are typically just called constants

And in this case the constant would be -8, not 8

[atomic urchin]

Gotcha

[wise kelp]

But that’s irrelevant right now

“Leading coefficient” means the first coefficient

[atomic urchin]

-12

[wise kelp]

As in the leftmost one

[atomic urchin]

33

[wise kelp]

The coefficient of the highest-degree term

Which in this case is x^3

[atomic urchin]

Ohhhhh

So it’s x^3

Ding!!!

[wise kelp]

So what’s the coefficient of that term?

[atomic urchin]

1?

Or x?

[wise kelp]

It’s 1, yes

Coefficients are always constants

[atomic urchin]

Got you

[wise kelp]

So this polynomial is monic

[atomic urchin]

Because it’s one

[wise kelp]

Yes

[atomic urchin]

That’s

Cool

Ty

[wise kelp]

Np

“Reduced polynomial” is kind of a weird term

[atomic urchin]

I was confused by that

[wise kelp]

I’ve never heard of it and I can’t find any reference to it on the internet

Based on their definition I’m not sure when the term would be useful

(I’ve also never heard of the term “monic” but I at least found references to that on the internet)

But I suppose it means that for example, if it’s a fourth-degree polynomial, there’s no term with x^3

Like 2x^4 + 5x^2 + x + 7

So an nth-degree polynomial with no (n-1)th term

[atomic urchin]

Ok

That’s is weird

[wise kelp]

I guess

Yeah I’m not sure what the significance of that is

But do you know whether the polynomial they’re asking about is one?

[atomic urchin]

I’m sure it’s the x^3......-8

Honestly I’m not confident in my answer

This instructor is difficult

[wise kelp]

Do you know what the “degree” of a polynomial means?