help-27

[restive river]

I did one mistake

Here

I forgot to add → on Fq3q1

This should also have vector on it

People also write it as Fq2q3 and also Fq3q2 , i dont know which one is more correct

[light zinc]

ok i understand now thx for explaining

[restive river]

Ok

[light zinc]

.close

[devout snow]

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[idle stirrup]

A school district runs a gifted language program in fifth grade with a selection criterion of 96th percentile. That is, students who score at 96th percentile or above on a standardized language test are eligible for the gifted program.

If we use a Z distribution to represent all the test scores from the fifth-graders, what is the Z score that serves as the critical (or cutoff) value for determining eligibility for the gifted program?

[devout snow]

@idle stirrup Has your question been resolved?

[idle stirrup]

<@&286206848099549185>

[devout snow]

@idle stirrup Has your question been resolved?

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[bitter thistle]

how

[winter patrol]

gonna need more context

[rustic jetty]

Is there a picture of the full page or something

[devout snow]

@bitter thistle Has your question been resolved?

[bitter thistle]

well there is

[rustic jetty]

can you send it?

[bitter thistle]

but i just wanna know how -2x+5 = ax + 17
turns into -2

[bitter thistle]

hold on

[winter patrol]

gonna need more context

[rustic jetty]

I don't know either

That's why I would like to know what you are looking at

[winter patrol]

is there even more context?

is this supposed to be the work solution?

[bitter thistle]

yes

[winter patrol]

can you show the original question

[bitter thistle]

sure

this is the rest of the solution, but i dont think itll help

[supple knot]

original question doesn't mean the whole solution

original question means original question

[bitter thistle]

this is the question

[bitter thistle]

no yeah i know, but i just wanted to share the whole thing

[winter patrol]

is pointless to us if we don't know the context

anyway, now that we have the context

in the question you are told that f(-2) = -4

[bitter thistle]

yep

[winter patrol]

thus finding when -2x+5 is -2
can lead to finding out more information about a

f(value of x when -2x+5 is -2) = a * (value of x when -2x+5 is -2) + 17

which they applied in the work you showed afterwards

[devout snow]

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[high turret]

How does the k+1 summation become that??

I’m talking about step 2->3 under inductive step

[upper schooner]

Is the third step supposed to be that? I would think you’d substitute the inductive hypothesis, that the sum to k is (k + 1)! - 1, in that position, no(?)

[high turret]

Sorry I meant step 1 to 2

Step 3 is incomplete, pls ignore

[upper schooner]

Step 1 to 2 is taking the (k + 1) term of the sum separate to the sum from 1 to k

[high turret]

Oh

So I plug in the k+1 into i(i!)

How do i simplify it further ?

For step 3

[high turret]

This?

[upper schooner]

Yep use the fact you assumed what the sum from 1 to k was!

[high turret]

Ok!

[restive river]

rizza

[upper schooner]

That should hopefully get you pretty far!

[high turret]

:0

lmao

[high turret]

i am stuck on simplifying it further after subtituting (k+1)!-1 ;-;

[upper schooner]

When you put that in you should have your step 3 as $(k + 1)! - 1 + (k + 1)(k + 1)!$, right?

[woven radish]

@upper schooner

[high turret]

yes

thats what i have rn

i tried to factor out (k+1)!

[upper schooner]

What did you get when you did?

[high turret]

k(k+1)!

[upper schooner]

[high turret]

uhhhhhhhhhhhh

[upper schooner]

As one tiny rearrangement before you do

[high turret]

yes

[upper schooner]

You can write it as $(k + 1)! + (k + 1)(k + 1)! - 1$

[woven radish]

@upper schooner

[upper schooner]

Which should hopefully make the factoring a bit better to see what you should get

Remember that the -1 is separate

[high turret]

ohhh

ok ill try !

[high turret]

ok i got (k+2)!-1

so i see that it works

but how do i show that its correct (?)

idk how to word it

but how do i show that thats the answer im supposed to get

[upper schooner]

That’s basically what you did in the first line, right? If I’m understanding you right, after all that’s what happens when you replace k with k+1 in the formula

[high turret]

OHH

yes the first line

thank u chartbit

[upper schooner]

A pleasure

[high turret]

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[light zinc]

how did u get the 9 and 3 in value of R??

[light zinc]

I think ur formula is different from mine

cuz the value of k is 8.988×10^9 for me

[restive river]

8.988 ≈ 9

So generally its allowed to take it as 9

But if u want very precise calculation, idk if calculator is allowed for you

If it is them yes u have to to take precise values

[restive river]

That 9 and 3 is the 9 height and 3 length of (x,y) coordinates of point b

Simple Pythagoras theorem

I thought of it as a triangle , perpendicular = 9 and base = 3

So i calculated hypotenuse

To find the distance between point b and c

Formula i used is

F = ( K q1 q2 ) / r²

r = hypotenuse = √(perpendicular ² + base²)

Or

U can simply use distance formula

From coordinate geometry

But Pythagoras theorem is easier

(√(9² + 3³))² = 90

[light zinc]

but how do u get the r in ur formula cuz we just use the distance meter

[restive river]

Wait it wasnt in meters?

I think i didn't read the question carefy

Carefully

[light zinc]

wait the formula has meters

k value is

[restive river]

Can u send the question again

[light zinc]

k=8.988×10^9 Nm2/c2

[light zinc]

btw its Fc to b

[restive river]

Its already in meters

The coordinates

(3m,4m) , m : meter

So u can directly put the distance

In formula

r

Which is 90 meters hmm but 90 meters sounds too much

It should be like 0.9 meters

[restive river]

But it shouldn't matter

Because its a question

[light zinc]

wait i stilll dont understand how did u get the 9 in r😥

[restive river]

Ok let me

Draw something

And explain

[restive river]

We need to find r

[restive river]

∆BCD

Is right angle triangle

So we can apply Pythagoras theorem:
BC ² = CD ² + DB ²

BC = r

r² = CD² + DB² 

CD = 3 , DB = 9

r² = 3² + 9²

r = √( 3² + 9²)

@light zinc

[restive river]

Orange a ,b ,c are charges

Purple are Points to show geometry

Green is charge C

Red are 1st and 2nd charges

[restive river]

Wait....

Its (3,4)

Sorry my bad

I misread it as 9

[devout snow]

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[ocean tulip]

Is there a way to turn this into closed form:
$\sum^{log_2(n)+1}_{i=1}(2^{i-1})$

[woven radish]

ChaoticLord

[ocean tulip]

$\sum^{log_2(n)}_{i=0}(2^{i})$

[woven radish]

ChaoticLord

[supple knot]

only if n is a power of 2

otherwise the sum doesn't make sense

[ocean tulip]

yes, it's given that n is a power of 2

i'm kind of stuck

i'm thinking 2n-1 rn

[supple knot]

well simplify the log

then use geometric series

[ocean tulip]

if we set n as 2^k

we simplify the log to k

[ocean tulip]

how do i use geometric series? sorry i'm a bit rusty

[supple knot]

https://www.cuemath.com/geometric-sum-formula/

[ocean tulip]

thank you

.close

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[digital granite]

• I’ve been able to solve parts of this equation using a gdc. However my teacher says I need to solve all these algebraically, and I have no idea how to even begin with this

[ashen dagger]

ok

well we know parabolas are symmetrical

and if we want the truck to fit we want the center of the truck to be under the vertex of the parabola

so whats the vertex of the parabola

[digital granite]

So what I did was I plotted the graph onto my gdc and used the maximum tool to find the highest point

But I’m not sure how to go about it algebraicallu

[ashen dagger]

do you know how to complete the square

[digital granite]

Not exactly

Oh: I forgot to add additional info. This was actually part of a group project and the rest of my group mates had solved it using the quadratic formula

[ashen dagger]

yeah

you can use quadratic formula too

and then just get the midpoint of the roots

but you should know how to complete the square because the quadratic formula comes from completing the square

[digital granite]

You are right

My teacher got pregnant mid semester and went on leave so my knowledge is a little out of order

I apologize

However I’d like to know how it is solved using the quadratic formula because it’s the way my teacher has approved it

[ashen dagger]

ok

so we have a = -0.0625

b= 1.25

c= 5.75

you wanna plug those into -b (+/-) sqrt(b^2-4ac)

then all over 2a

[ashen dagger]

the whole thing is over 2a

including the -b part

[digital granite]

my bad I didn’t catch that

[ashen dagger]

yeah

you just gotta do the arithmetic from there

here theres a trick actually

youre gonna do the average of the roots right

one of the roots is gonna be -1.25/(2*-0.0625) + square root stuff

the other one is gonna be -1.25/(2*-0.0625) - square root stuff

when you take the average

the square root stuff will cancel

and you will just get -1.25/(2*-0.0625)

[digital granite]

Oh thanks for the trick

So since this was a group project a classmate already completed this part and the zeroes are -3.856 and 23.856

and using that information they were able to solve all 3 of the given problems

I just don’t understand how they applied that to get to their answers

[ashen dagger]

there is a different way to think about it then the zeroes

so the truck is 11 feet tall and 7 feet wide

we want to place the middle of it in the middle of the arch

but the height still has to be at least 11 feet on either side of the arch

it has to be 11 feet for at least 3.5 feet on each side

(since the truck is 7 feet long)

so we just need to check if f(13.5) and f(6.5) are greater than or equal to 11

and because of the symmetry we really only need to check one

cuz they will actually be the same value

[digital granite]

I’m kind of not understanding

[ashen dagger]

its kinda hard without a graph to explain

here ill make one

[digital granite]

Here’s how my classmates did it

My teacher expects me to do it the same way

[ashen dagger]

yeah so the way they are doing it

is checking if, given a height of 11, we can actually enclose a 7 foot long space

the way im talking about rn is if, given a 7 foot long space we have a height of at least 11 feet

they are just kinda reverses

lowkey i think the one im talking about is easier cuz the other one you have to use quadratic formula

[digital granite]

Yeah I appreciate you trying to make it easier to understand but for our project we have to present and explain each other’s work

And I have no idea what they’ve done

[ashen dagger]

ok

when the set the function equal to 11

they are looking for these two points

because the distance in between these points will be the maximum width we can have

so they have 11 = 0.0625x^2 + 1.25x + 5.75

they subtract 11 from both sides

0 = 0.0625x^2+1.25-5.25

and then they use the quadratic formula to get x = 6 and x = 14

the distance between x = 6 and x = 14 is 8

thats more than 7, so the truck will be able to fit

[digital granite]

THANK YOU

What about parts b and c?

[ashen dagger]

This is part B

[digital granite]

Ah sorry, c

Specifically that one because I was called to answer that on the board and I couldn’t

[ashen dagger]

part c is actually the one i was talking about before

so now we have this setup

we know the width of the truck is 7

the maximum we can have will be at these points

so we just have to find the height of the parabola at one of those points

which is just plugging one of them into the function

the one on the left is 10-3.5 = 6.5

so if we calculate f(6.5)

that will be our maximum height if the width of the truck is 7

tell me if youre confused

[digital granite]

a little bit

[ashen dagger]

which part

[digital granite]

wait nevermind I think I might have gotten it ill check my work

yeah I might have gone wrong somewhere: what’s the final equstion used to solve this?

[ashen dagger]

Uh

Max height = 0.0625(6.5)^2+1.25(6.5)+5.75

[digital granite]

Thank you very much

[devout snow]

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[brave pilot]

This will be undefined right? since y =/= 0 in such cases or?

[brave pilot]

Also this is just a pratice sheet

[stone prism]

hi could u help me after?

[brave pilot]

Me?

I dont think I an help anyone-

[timber pebble]

cosine isnt really undefined

its not one of those functions

these would be like, tangent

or cosecant

[brave pilot]

what

[timber pebble]

cosine just wiggles between -1 and 1

have you drawn a picture?

[brave pilot]

I tried

I mean

Lemme show

[timber pebble]

okay

[floral ridge]

cosine is continuous over all real numbers

[timber pebble]

oh im not really sure here

[brave pilot]

I drew it on the wrong side

Oops

[brave pilot]

I see

[timber pebble]

so (8,0) is at x=8 and y=0

[brave pilot]

Yes, I think

[floral ridge]

yeah

on the x axis

[timber pebble]

so, your picture is kind of nonsensical

P is a point thats just on the x axis yea?

so what angle does that make?

[floral ridge]

from positive x axis

[brave pilot]

Right

Sorry 😭 Idk what I am doing

[timber pebble]

youre good

the picture should just look like this

lemme see

[floral ridge]

points on x axis have no y value

[floral ridge]

aka y=0

[brave pilot]

Mhm

[timber pebble]

https://www.desmos.com/calculator/aco5zy2x4k

[brave pilot]

Right

Yeah

Okay

[timber pebble]

so what angle does it make with the x-axis?

[brave pilot]

0

[floral ridge]

https://www.desmos.com/calculator/qmzx2skkzy

[timber pebble]

yea exactly

so whats cos0?

[floral ridge]

this should help

[brave pilot]

0?

[brave pilot]

Ohh

Tysm

[floral ridge]

that's sine

[timber pebble]

,tex .unit circle

idk if youve seen this before

if you havent its gonna seem like a lot

[woven radish]

jan Niku

[brave pilot]

1

[timber pebble]

yea

[brave pilot]

Cos0 is 1 😭 sorry

[timber pebble]

ur good

[floral ridge]

i gave her that in desmos

[floral ridge]

nice job

[brave pilot]

Yeah thanks, I'll use that

Also

Do ypu guys know what terminal arm is

[floral ridge]

no i don't study math in english, so unfortunately cant explain

[brave pilot]

because of this the terminal arm is on like 270 and you dont even need to calculate but I doubt they would give that answer

[brave pilot]

It's okay

Tysm

For helping before

[timber pebble]

terminal just means end here, basically

[brave pilot]

So the end of the

[timber pebble]

you usually think of the Initial Arm being the x-axis

[brave pilot]

Angle arm

[timber pebble]

and then the terminal arm being the end

[brave pilot]

Mhm

Okay

[timber pebble]

so the angle goes from the initial arm, the start, to the terminal arm, the end

[brave pilot]

So tje blue line

Is terminal?

Yeah-

[timber pebble]

yea

[brave pilot]

Okay

[timber pebble]

and green is initial

[brave pilot]

But there is no terminal in

Hold up

Is the terminal arm length 5?

[timber pebble]

youre not drawing these right

[brave pilot]

Oh

[timber pebble]

you should draw the point first

then connect the point to the origin

that gives you the terminal arm

the origin is the middle, where the x and y axis cross

[brave pilot]

Okay

Oh okay

The 5

Would be up not down

[timber pebble]

im not sure what you mean

[brave pilot]

I’m not sure either

When I make a point

Like x is -2 so it would be

To the left at -2

Ahhhh

Is this wrong too?

[timber pebble]

yea

sorry

you wanna go to the left -2

[brave pilot]

Maybe I’ll just

Dial the test

[timber pebble]

and up 5

[brave pilot]

Yeah

Oh

Yeah

Nvm

Got it

[timber pebble]

sorry, i guess to the left 2

-2 means go to the left 2

[brave pilot]

Yes

I know that, I’m not sure what’s up with me

[timber pebble]

youre good it trips people up

,w plot point (-2, 5)

[woven radish]

Results provided by WolframAlpha

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

Requested by janniku

[floral ridge]

try helping in desmos

[brave pilot]

I tried

It confused me even more 😭

Idk how to use it

[timber pebble]

does this help?

[floral ridge]

bruh

[brave pilot]

Yeah

[timber pebble]

do you see where the point should be?

[brave pilot]

I am not the brightest

So

It takes time 💀

[timber pebble]

so, you wanna draw that point, then connect the point to (0,0), the origin, where the lines cross

[brave pilot]

Yeah it’s making sense now

Okay I think I get it

[timber pebble]

no one is dwai

[floral ridge]

same

[floral ridge]

like a radius of a circle

with a center of (0,0)

[brave pilot]

Yeah I get it now 😭

[floral ridge]

everyone struggles once

everyone improves

[brave pilot]

It’s definitely easier for those who understand stuff

Being dumb means extra work and also troubling other around so 💀 that’s not fun

BUT ANYWAYS

Thanks tho

[timber pebble]

yea but no one in math does stuff they know

[brave pilot]

😭 this sever helps me a lot

[timber pebble]

it doesnt matter if you study your whole life

[brave pilot]

Hm

[timber pebble]

even the best mathematicians died with lists of things they still wanted to understand

[floral ridge]

you will never be able to grasp "everything"

[brave pilot]

Mhm

[devout snow]

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[frigid yacht]

help

[frigid yacht]

how do i do number 46

its improper integrals btw

[crisp niche]

to show that the integral in 46) is divergent, apply g(t) = 1/t

[frigid yacht]

how would i do that

oh split it into two

[crisp niche]

that is , to use the limit comparison test

no, not split

[frigid yacht]

g(t) = 1/2 and f(t) = 1/sqrt (1-t^3)?

huh

[crisp niche]

i repeat, you must use limit comparison test

[frigid yacht]

?

im not exactly sure where to proceed

[crisp niche]

i write, moment,

that is theorem:

write it in your exercise book and next we wil see the solution

[frigid yacht]

im still confused

this is what i have so far

[crisp niche]

i said, g(t) = 1/t

$\text{in your case:}\text{ }f\left( t \right)=\frac{1}{t\sqrt{1-t^{3}}},\text{ }g\left( t \right)=\frac{1}{t},\text{ }t\in \left( 0,\frac{1}{2} \right]$

[woven radish]

Joanna Angel

[crisp niche]

ok?

[frigid yacht]

yea

[crisp niche]

$\lim_{t \to 0^{+}}\frac{f\left( t \right)}{g\left( t \right)}=_{\cdots }etc$

[woven radish]

Joanna Angel

[crisp niche]

we compute limes for t reachgn zero from right side, since

0 is yoru singularity

[frigid yacht]

yeah

[crisp niche]

not infinity but zero

[frigid yacht]

ohhhh i see yeah

[crisp niche]

thwn

$\lim_{t \to 0^{+}}\frac{f\left( t \right)}{g\left( t \right)}=\lim_{t \to 0^{+}}\frac{1}{\sqrt{1-t^{3}}}=1>0$

[woven radish]

Joanna Angel

[crisp niche]

great )

and now remind the p-test

for improper integrals for yoru acse

[frigid yacht]

huh

[frigid yacht]

for this one wouldnt f(t) just be 1/sqrt 1-t^3

or is it the ful lthing

[crisp niche]

$\int_{0}^{a}\frac{dt}{t^{p}}\text{ }\text{ is }\text{ }\left{ \begin{array}{cl}
\text{convergent for }p<1\
\text{divergent for }p\ge1
\end{array} \right.\text{ where }a>0$

[woven radish]

Joanna Angel

[crisp niche]

write it in

and then you write:

$\int_{0}^{\frac{1}{2}}\frac{dt}{t}\text{ }\text{ is divergent since }p=1$

[woven radish]

Joanna Angel

[crisp niche]

finally you write:

$\int_{0}^{\frac{1}{2}}f\left( t \right)dt\text{ }\text{ is also divergent due to limit comparison test}$

[woven radish]

Joanna Angel

[frigid yacht]

ohhhh that actuallt makes snese

[crisp niche]

write all in, to not lose it, that is such a scheme

[frigid yacht]

wdym write all in

[crisp niche]

i meant: write all in your exercise book

[frigid yacht]

ohhh gotchu

lemme write it

thank you for ur help

[crisp niche]

yw

[devout snow]

@frigid yacht Has your question been resolved?

[frigid yacht]

.close

[devout snow]

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[undone chasm]

yes, it is possible to know what 8x4 is

[devout snow]

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[smoky gyro]

Did I differentiate this wrong?

[quaint citrus]

,w differentiate t(t+1)^(-1)

[woven radish]

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[quaint citrus]

,w differentiate (t+1)^(-2)

[woven radish]

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[quaint citrus]

looks like it

for second derivative u didnt use product rule on first term

@smoky gyro

[smoky gyro]

ohh

thanks

[opal cloak]

honestly, i'd use the quotient rule as to avoid making mistakes like these

[smoky gyro]

dont really like quotient rule

[opal cloak]

it honestly saves time

[pseudo basin]

wait, what are we differentiating?

t/(t+1)?

quot rule and op's method BOTH suck for that one lmfao

[opal cloak]

quot rule takes like 2 steps?

[pseudo basin]

it's error-prone tho

$\frac{t}{t+1} = 1 - \frac{1}{t+1}$

[woven radish]

Ann

[pseudo basin]

and then differentiating once and even twice is just piss easy

[quaint citrus]

(1 + t^-1 ) ^ -1

-(1+t^-1)^-2 * -t^-2

u dont have to read that dw

[smoky gyro]

how do u look at it like that

[pseudo basin]

same way yo ucould write 8/9 as 1 - 1/9

[smoky gyro]

i guess

$A(3,-1)$ is a vertex of the rectangle ABCD whose side AD has equation $2x-y-7=0$. Find in general form of the equation of the side AB.

[quaint citrus]

its just long div @smoky gyro

see it now?

[smoky gyro]

for this one im not sure how to begin tho

[smoky gyro]

like

polynomial long div?

[quaint citrus]

yea

t / t+1

[smoky gyro]

i cant do that in my head haha

[quaint citrus]

dont gotta, just do it on paper

[pseudo basin]

PARENTHESES

but also like

t is (t+1)-1

[quaint citrus]

ye

[pseudo basin]

idk how else to say it

but it is on the surface

to me

[pseudo basin]

begin by making a graph

[quaint citrus]

the only time i ever use +c - c is as integration technique so i didnt think abt it here tbh lol

[pseudo basin]

that way you can at least orient yourself

also like

@smoky gyro better to close this channel and open a new one with your new problem

[woven radish]

water beam

[smoky gyro]

whats general form?

[pseudo basin]

ax + by + c = 0

same form as the equation of AD was given to you in

i also see you've chosen to ignore my directions

but ok

[smoky gyro]

i cant be bothered opening a new channel rn

[pseudo basin]

you cannot be bothered to type .close, look for a new unoccupied channel, and repost your problem there

ok

[smoky gyro]

So how do I proceed with this?

[pseudo basin]

erase this. this is labeled incorrectly.

AD is a side, not a diagonal; when naming a polygon, its vertices must be written down in order either CW or CCW.

the rectangle you drew would have been ABDC and not ABCD.

[smoky gyro]

Don’t think I understand this

[pseudo basin]

all polygons are named by their vertices

the vertices in a polygon's name must be listed in the order that they would appear if you were to go around the polygon in one direction.

do you understand this? [Y]es/[N]o/give me time to [D]igest this/[F]uck you im not answering (default = F)

[smoky gyro]

Quite a range of options there

I need time to digest this

[pseudo basin]

ok

ping me when you have done so

[smoky gyro]

Is this what you mean?

[pseudo basin]

i mean yeah that's better i guess.

axes are nowhere to be seen tho.

but at least it should now occur to you AB is perpendicular to AD.

from which you can find the gradient of AB.

[smoky gyro]

I’m not sure I understand the side AD 2x-y-7=0 thing and how we can use it to find it

[winter patrol]

have you previously determined the slopes of lines before?

[devout snow]

@smoky gyro Has your question been resolved?

[smoky gyro]

like y = mx+b

?

yes i have

[winter patrol]

rearrange the equation to that form (or otherwise) and determine its slope from that

[smoky gyro]

oh okay

so m = 2

for the slope of AD

but we are looking for AB

how does this help?

[winter patrol]

apply relation between slopes of perpendicular lines

[devout snow]

@smoky gyro Has your question been resolved?

[devout snow]

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[half wedge]

Can anyone help me with this, im lost

[half wedge]

i know absolute a equals a*a

[pseudo basin]

do you have the definitions of dot and cross products handy

[half wedge]

yh

[pseudo basin]

show them

[cold bough]

lol i didn t know that, thx

[pseudo basin]

ok

[half wedge]

yh same

[pseudo basin]

so then what is $(\bd{a} \cdot \bd{b})^2$ based on this?

[woven radish]

Ann

[pseudo basin]

also i believe that the second of your two definitions should be spelled out more explicitly: in it, n is the unit vector perpendicular to both a and b, its direction determined by the right-hand rule.

[half wedge]

[a]^2[b]^2costheta

[pseudo basin]

|a|^2 |b|^2 cos**^2**(θ).

you forgor to square the cos(θ).

[half wedge]

oh right

[pseudo basin]

anyway:

[hot steeple]

hi ann

[half wedge]

once ive done that, how do i get the cross product though

[pseudo basin]

now, what is $\nrm{\bd{a}\times\bd{b}}^2$?

[woven radish]

Ann

[pseudo basin]

note you do not need the cross product itself

[half wedge]

its (a x b)^2/sin^2

[pseudo basin]

only its squared length

[half wedge]

ok

i dont get the last part

is it because cos^2 + sin^2 = 1

[pseudo basin]

why did the sine go into the denominator...

wait ok no

i don't get what you wrote there at all

[pseudo basin]

yes that's what you will need eventually

[half wedge]

because of the cross product definition

[half wedge]

i dont get that part

[hot steeple]

$||a \cross b||$ is the length of the cross product aka magnitude aka size

[woven radish]

Katharine

[hot steeple]

normally the cross product is a vector

but you want the length of that vector

[half wedge]

ok, im still stuck at $(a*b)^2/cos^2theta$

[woven radish]

Hector

[half wedge]

i dont get how to introduce the cross product with that

[hot steeple]

You found the dot product squared

$|a|^2 |b|^2 \cos^2(\theta)$

[woven radish]

Katharine

[half wedge]

ok

doesnt that bring us to the beggining though

[hot steeple]

you need to show that the dot product squared + the cross product squared

is equal to the left hand side

and so you find the dot product squared

and the cross product squared

which is where you were stuck right?

[half wedge]

no, i tried going from the LHS to the RHS

and i got $[a^2b^2]cos^theta$

[woven radish]

Hector

[half wedge]

because of the scalar definition

[hot steeple]

lets go back to the definitions of dot and cross products

you found the first term of the rhs

now the second

what do you think the second term is?

[half wedge]

cross product squared

[hot steeple]

right and what is that using the definition?

[half wedge]

its 0

no thats not right

a^2b^2sin^2

[hot steeple]

exactly

so now you have $|a|^2|b|^2 \cos^2(\theta) + |a|^2|b|^2 \sin^2(\theta)$

[woven radish]

Katharine

[hot steeple]

on the rhs

and $|a|^2|b|^2$ on the left

[woven radish]

Katharine

[hot steeple]

1 more step

[half wedge]

sin^2+cos^2=1

[hot steeple]

yes

so now rhs and lhs are exactly the same

[half wedge]

theyre not though

[hot steeple]

so you started with $|a|^2|b|^2 = (a \cdot b)^2 + ||a \cross b||^2$

[woven radish]

Katharine

[hot steeple]

and you found that $(a \cdot b)^2 = |a|^2|b|^2 \cos^2(\theta)$

and you also found

$||a \cross b||^2 = |a|^2|b|^2 \sin^2(\theta)$

[hot steeple]

the question is to show that this is the case

so the next step is to fill in the dot product and cross product

[woven radish]

Katharine

Katharine

[hot steeple]

which results in $|a|^2|b|^2 \sin^2(\theta) + |a|^2|b|^2 \cos^2(\theta) = (a \cdot b)^2 + ||a \cross b||^2$

[woven radish]

Katharine

[hot steeple]

which is the rhs

i hope that makes sense?

[half wedge]

yh, so at the beggining its like a double magnitude

i get it

thanks

.close

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[ivory bluff]

I have never seen the notation $\nabla f(x)$, could someone explain it?

[woven radish]

bigpufik

[stone stump]

the gradient

"derivative"

but higher dimensional

[ivory bluff]

How do I calucalte it?

[stone stump]

put all the partial derivatives into a vector

[ivory bluff]

I have only learned lagrange multitpliers where I just take partial derivatives and equal zero, and then get a point in Rn

Same thing?

[stone stump]

well thats the first part

[ivory bluff]

and then you have some weird matrix stuff yeah

[stone stump]

"either nabla g=0 or ..."

[ivory bluff]

What does the nabla f = lambda nabla g mean

weird

[stone stump]

well that one vector is a multiple of the other

[ivory bluff]

Yeah i guess

true

[stone stump]

the first partial of f is equal to lambda times the first partial of g

etc for all the other partials

[ivory bluff]

Btw how will this just not be a point?

dont we need certain condititons on g

[stone stump]

why should there only be one x with g(x)=c

well sometimes of course it could just be one point

but for example g could be x^2+y^2

and if c=1 then S is the unit sphere

[ivory bluff]

Yeah

[stone stump]

S could also be empty

[ivory bluff]

and along S just means on S in R^n right?

[stone stump]

yes

[ivory bluff]

thats a cool theorem

okay thanks

.close

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[rain garnet]

good morning

[rain garnet]

this is my integral

and they wanted me to switch the order

this is my drawing

and i concluded that

it can be lie that

but

the integral suck

and dont want to be calculated

[devout snow]

@rain garnet Has your question been resolved?

[upper schooner]

Shouldn't that be $\sqrt[3]{x^7}$ as the upper limit?

[woven radish]

@upper schooner

[rain garnet]

why

(i dont understand that switching order thing this is the first example im doing)

[upper schooner]

Cause if you had $x = \sqrt[7]{y^3}$ then you'd have $y = \sqrt[3]{x^7}$ right

[woven radish]

@upper schooner

[rain garnet]

wow that makes sense

and the rest is correct?

[upper schooner]

Also you may need to have the order being as $\int_{\sqrt[3]{x^7}}^x \ldots \dd y$ because it's first the curve $y = \sqrt[3]{x^7}$ you meet, then $y= x$ after

[woven radish]

@upper schooner

[upper schooner]

Like if you draw it out from left to right

[rain garnet]

right i just understood it very badly

but now i think its gonna work

thank uu

[upper schooner]

[rain garnet]

.close

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[steel gazelle]

How should i proceed after i written the matrix

To get it to upper echelon form?

[elfin atlas]

do you not know the gauß jordan elimination? i can show it by solving this question

[devout snow]

@steel gazelle Has your question been resolved?

[steel gazelle]

yeah i know sorta how to do it

like the first 3 examples it was easy

im not sure how to do it with the a's

[elfin atlas]

okay so do you have your final matrix? (after all the eliminations)

[steel gazelle]

The only thing ive managed to do is subtract row 2 from row 3

[devout snow]

Channel closed

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[steel gazelle]

then i just kinda stopped cus not sure how to continue

.reopen

[devout snow]

✅

[elfin atlas]

so you're not able to reach the echelon form?

[steel gazelle]

yes

i can try again im not sure how can i get 0s when all i have pretty much are As

[smoky nimbus]

Can you show your first step?

[steel gazelle]

sure

1 sec

[elfin atlas]

perfect, this is how you should proceed

now eliminate the elements below the pivot value (1) using the first equation, you can write eq 2 - (eq1 x a) and for eq(3) write eq3 - eq1

[smoky nimbus]

This step is not ideal

[elfin atlas]

yeah completely useless

[steel gazelle]

Oh

I understand

[smoky nimbus]

You want 0s here first

[steel gazelle]

oh so always do it under the pivot right?

[smoky nimbus]

Yes

So start with the second row, how would you make element (2, 1) to be a 0?

Using the first row, as the first row is your base, where ideally you should manipulate that first to do operations on the other rows

[steel gazelle]

Subtract from row2 , row 1 * a?

[smoky nimbus]

Meaning R2 - a * R1?

[steel gazelle]

yyes

[smoky nimbus]

Yes that's right

[steel gazelle]

Doing that right now

For the last row i can just do R3 - R1 right?

[smoky nimbus]

You aren't limited to what I'm saying, but as mentioned it's more ideal to manipulate R1 rather than R2. It just depends on how difficult it makes it

[smoky nimbus]

Yes

[steel gazelle]

Oh i get it now

[smoky nimbus]

Like you can do R2/a - R1 if you wanted to but you'll have horrible fractions doing that. In the end, you'll end up with the same answer or the answer, just in a different form

[steel gazelle]

Noted

Thats what I wanted to know, if there were multiple ways of doing so

Thanks, thats all i needed i think

.close

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[tranquil horizon]

Stuck on the inverse of e to an exponent...

[cold bough]

ln?

[tranquil horizon]

At this point, I'm not sure how to pull the exponent off of e in order to manipulate y

Yeah, it'll be the natural log, but...

Against what?

[hidden dragon]

Both sides

[thin inlet]

...natural log= log base e

ln=log_e

[tranquil horizon]

I know the natural log is base e.. that's not what I'm asking

[cold bough]

like all step you have done ...you have to act to both sides

[hidden dragon]

Against both sides

[thin inlet]

do whatever operation you do to LHS to RHS

[tranquil horizon]

ln y?
ln y-7?
ln x?
ln AGAINST WHAT?

[thin inlet]

left hand side right hand side

[hidden dragon]

Both sides of equation

[tranquil horizon]

.close

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[thin inlet]

think of doing ln to an exponent e like doing a square root to a square

.ok nevermind

[devout snow]

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[sage willow]

Need help with question 4. Solving previous year papers.

I have solved till finding eigen values

Need help with finding eigen vectors.

Can anyone help me with this?

[devout snow]

@sage willow Has your question been resolved?

[devout snow]

@sage willow Has your question been resolved?

[devout snow]

@sage willow Has your question been resolved?

[sage willow]

No

[devout snow]

@sage willow Has your question been resolved?

[devout snow]

@sage willow Has your question been resolved?

[devout snow]

@sage willow Has your question been resolved?

[plush vault]

Hey @sage willow

Why you are having problem with eigan vector

[sage willow]

I'm getting x1 and x2 as 0 when solving the equations.

[plush vault]

You get a equation in x1 and x2 otherwise your eigan value wrong

[sage willow]

Yeah I get 2 equations in x1 and x2. But I don't know how to solve them.

[plush vault]

2 equation are multiple of each Other ??

[plush vault]

I think slight calculation mistake

But you get your eigan vector [ -0.06365 1.0785] and its multiples

[devout snow]

@sage willow Has your question been resolved?

[devout snow]

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[smoky gyro]

I’m not sure how I use this formula, any ideas?

[devout snow]

@smoky gyro Has your question been resolved?

[eager lodge]

write t instead of n

A is the amount (money after investment)
P is capital(money before investment)
r is interest rate
t is time

[devout snow]

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[smoky gyro]

I’m unable to identify what variables are which from the information I’m given

[wicked turtle]

r is your interest rate

n is the time, and since r is per annum, n is in years

[smoky gyro]

So what am I trying to find exactly? Is it n?

[wicked turtle]

"find the time" yes, n

[smoky gyro]

Do I know A and P?

[wicked turtle]

you know how they are related

P is your original investment amount
A is the amount you have after time n elapses

[smoky gyro]

so r is 0.05

[wicked turtle]

yea

[smoky gyro]

I don’t understand the time it takes for an investment to double thing

[wicked turtle]

your original investment is P

A is your final investment

if your investment doubled, how are A and P related?

[smoky gyro]

2A?

[wicked turtle]

A = what

as a function of P

[smoky gyro]

2A = 2P(1+r)^n or is it just P

[wicked turtle]

only one of the two letters A or P should appear in your equation

either write P in terms of A or A in terms of P

[smoky gyro]

hm

[smoky gyro]

But to do this wouldn’t I need another equation relating the two variables? I only have one which is A = P(1+r)^n

[wicked turtle]

you can express A in terms of P

A is double the amount of P, right?

how do you express that as an equation

[smoky gyro]

2A = P?

[wicked turtle]

no

that would mean A = P/2

i.e. it would mean that A is half of P

it's supposed to be double

[smoky gyro]

oh

A = 2P

[wicked turtle]

yea

so now use that in your equation

write 2P instead of A

[smoky gyro]

okay so 2P = P(1+0.05)^n

[wicked turtle]

yea

now you can get rid of the P's

divide both sides by P

so now you just have an equation involving n

[smoky gyro]

this is assuming P can’t be 0 right

[wicked turtle]

yea we assume they invested something

otherwise it can't double

[smoky gyro]

alright

[wicked turtle]

well ig zero could double to zero

but it seems reasonable to exclude that case haha

[smoky gyro]

Yeah

alright I got the answer, thank you for the help

.close

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[modern badge]

how do i take integral of this?

[coral crescent]

take u=1-x

[modern badge]

okay i'll try now

[wicked turtle]

or just simplify d(1-x)

[modern badge]

can you give me more steps i got confused

[coral crescent]

if we took u=1-x, you'll have Integral x du, right?

we want our integral to be in terms of u

[modern badge]

if i take u=1-x doesn't it make -dx=du?

i think i'm havinga brain fart

[coral crescent]

right, but you'll reach the same form

[valid iron]

let u = 1-x
so x = 1-u
yeah?

[modern badge]

yea let me try it that way

OH!!

1-x^2/2

damn thanks i guess never tried it solving anything this way

thanks for help everyone! .close

.close

[devout snow]

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[frank zinc]

Is Circle C

C = (0,0) and r = 3 or r = 7?

[zinc flint]

c is (0,0) and r should be 4

[frank zinc]

Oh okay

[zinc flint]

howd you get 3 or 7?

[frank zinc]

It's meant to be 4

but I got 7 cuz I count the individual lines

Not the lines with numbers given with them

I guess I'm answering this entire worksheet again

🤦

[zinc flint]

stange but sure

[frank zinc]

oh god my worksheet is messy now

💀

Thanks though

You saved my score

I could've got ZERO points in the entire worksheet

[zinc flint]

all g

[frank zinc]

.close

[devout snow]

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[delicate arch]

how does this function have 2 x intercepts??

[winter patrol]

the equation you rate doesn't satisfy the range condition

[zinc flint]

its probably a truncus

[vital blaze]

the range is -1 to infinity

not negative infinity to -1 to infinity again

[delicate arch]

oh i see where I went wrong, thank you

[devout snow]

@delicate arch Has your question been resolved?

[devout snow]

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[marble quail]

Prove that the parts that is created by AP, AB and AQ divide the figure in four equal long parts.
Tips: choose the vector ā = OA and b = OB as base.

[marble quail]

The picture with the formula is a hint. Any ideas how this works?

[devout snow]

@marble quail Has your question been resolved?

[devout snow]

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[calm sky]

@devout snow

[calm sky]

1 fourth times X + 1 sixth times X + 1 half times X + 5 = X
The X is 60, but i dont understand why the x is 60
can someone help me?

@devout snow 1 fourth times X + 1 sixth times X + 1 half times X + 5 = X
The X is 60, but i dont understand why the x is 60
can someone help me?

[restive river]

The bot is just a bot. Pinging it won't do anything

[calm sky]

;-;

i dont understand how this works

[vast wing]

$\frac{x}4+\frac{x}6+\frac{x}2+5=x$

[woven radish]

onion

[vast wing]

you can work it out using lcm

[past scaffold]

Move all the X terms to the RHS
And then take X as common