help-19

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[tidal jewel]

idk where "adjacent" is

could be two things

[cold sage]

my eyes cant comprehend those runes clearly

[warped arrow]

Can anyone help? I can’t figure out how to isolate theta from this derivative to find the minimum angle

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[warped arrow]

Sorry

[tidal jewel]

Sorry,

When we find cosine, can't it be 70/56
because 70 will be adjacent
if A/D is facing towards the top right (imagine)

Not sure which "adjacent" it is

[cold sage]

you can figure that out
CA is the longest so it must be the hypotenuse

from there you can draw the triangle quite easily

[tidal jewel]

wait let me draw it out to show

[tidal jewel]

oh right

shucks, good thing I somehow got it right

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[hexed sluice]

How do I algebraically determine the discontinuity of an Absolute Value function?

[hexed sluice]

Let's say for (x-2)/abs(x-2)

I know it's a jump discontinuity.

[quasi sparrow]

find the limit of the function where the denominator is zero

use one sided limits

[hexed sluice]

so I do lim x->2 ^+ and lim x-> 2^-

?

[quasi sparrow]

yes

[hexed sluice]

Okay

I got -1 and 1

[quasi sparrow]

sounds right

[hexed sluice]

Okay, got it

I understand now, thanks

[quasi sparrow]

,w plot (x-2) / |x-2|

[clever fjord]

Results provided by WolframAlpha

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

Requested by rie.mann

[hexed sluice]

Yep, and that's the graph

thx

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[sterile blaze]

would the G-equivalence classes of X here be {0} and R2\{0}

[sharp oak]

Yeah that sounds right. G being invertible guarantees 0 stays separate

[sterile blaze]

and the G-equivalence of L_H would be L_H itself as for each left coset of H in the form, gH, we can get any other left coset fH by multiplying (fg^-1) in G

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[sterile blaze]

thank you kay

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[fast ivy]

is this the proper process so far

[tidal jewel]

yo isnt that science lol

[quasi sparrow]

where's the question

[fast ivy]

sry it took long to load

idk why i put the =

[quasi sparrow]

your'e missing a g after 453.6

but yea looks right

[fast ivy]

oh oops

ok

so

here should be

in the numerator: meter?

then we go from meter to nm^3?

should the denominator be

100 cm?

or keep it 1cm^3

and put the amount of meter

in the numeraotr

good?

,flip

.flip

[forest sky]

,rcw

[fast ivy]

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[thin nimbus]

How did the book get that answer

[nocturne belfry]

seems like they used the product rule

fudged some numbers, though

[thin nimbus]

Oh

I did 2x-1

2x-5*

I was supposed to do 2x-1

[royal herald]

y = (x+2)^3 * (2x-1)^5
ln y = 3ln(x+2) + 5ln(2x-1)
y' = y * (3/(x+2)) + (10/(2x-1))

im literally the goat

[thin nimbus]

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[royal herald]

bruh 😂

why don't you use the goat's method

instead of urs

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[hoary marsh]

Hello

[hoary marsh]

Not sure how to start

[quasi sparrow]

Draw a right triangle with angle inverse sine x

[hoary marsh]

Pythagorean theorem, right?

Thanks

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[warped sphinx]

need help in the drawing part only

this is what the teacher told me but i feel like its wrong?

i can solve for the probability on my own if i just know how to draw it😭

[odd edge]

@warped sphinx Has your question been resolved?

[warped sphinx]

<@&286206848099549185>

[brittle beacon]

"A quadrant of a circle" implies a quarter of a circle, no?

The location of the quadrant itself isn't too important, just that it's contained within the rhombus

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@warped sphinx Has your question been resolved?

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[delicate charm]

I'm mainly confused here cuz I'm using trig sub, but my answer is slightly different from the book

i checked on symbolab and got the same answer

but even symbolab doesn't have the same answer as the book

LOL

[tacit haven]

what did you substitute?

[delicate charm]

that's what i started with

and that's my starting position basically

my final answer

the books final answer

[tacit haven]

i think you're correct on this one

[fluid tundra]

they are equivalent answers

notice + 3ln(1/2) is a constant

[delicate charm]

waaahhhhhh

[tacit haven]

oh yeah -ln2 can be added to C

[delicate charm]

OHHOHOHOHOHO

my god

my third eye

thank you so much

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[fast ivy]

So

[fast ivy]

for si prefixes

i find it ez to go b/w cm and like mm for example

but is it possible

and how hard is it

to go from like

liter to meter

liter to cm

or like kilometer to nanogram

is it like completely different process?

[forest sky]

you can always convert to the base unit first if you don't know the conversion factor between two given SI prefixes

[meager juniper]

Liter and meter are incompatible units though

[forest sky]

but the examples you listed are all incompatible units

[fast ivy]

wait wat

someone told me

I can go from

CM to liter

no?

[mental lotus]

Those units measure different things

[meager juniper]

Cm^3 to liter yes

[mental lotus]

1 Liter is 1000 cm^3

[fast ivy]

waaaaat

[forest sky]

you can go from cm^3 to liter, not cm to liter

[fast ivy]

so only cm^3

ok

but this cm^3 can work with things other than liter?

[forest sky]

because cm^3 and liters are both measurements of volume, whereas cm is a measure of length

[fast ivy]

Oh

[forest sky]

cm^3 can be converted to any measure of volume, e.g. liters, m^3, gallons, cups, tablespoons, olympic swimming pools, etc

[fast ivy]

not meters?

only m^3?

[mental lotus]

you can think of unit conversion in context. Like can you tell how far apart is London to New York in kilometers? Yes. But can you tell how many nanograms apart they are? No.

[fast ivy]

Oh

thts good example

ok

so meters is not a volume corfrect

is a measurement

m^3 is a volume

[mental lotus]

measurement of length

[fast ivy]

Ok

[mental lotus]

all units are for measurement

but what they measure is different

[fast ivy]

ah

[mental lotus]

You can convert units if they measure the same thing

[fast ivy]

so for example
if i want know how much cm^3 to get to 1 liter

what that process then

[forest sky]

you need to know the conversion factor between some si prefix of liters and some si prefix of m^3

[mental lotus]

You check first if both measure the same thing. They do, both measure volume. Now you can procceed

The rest of stuff is memory

you gotta remember 1L is 1000 cm^3

[fast ivy]

ohhh

just like remember how 3 ft in a yrd kind of thing

ok

[forest sky]

you can alternatively remember 1 mL = 1 cm^3 or 1 L = 1 dm^3

[fast ivy]

Ok i will start memorizing

thnx

[odd edge]

@fast ivy Has your question been resolved?

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[river lily]

can some one explain this

[river lily]

does it utilize the global error formula?

my prof said that you could never calculate global error

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[onyx cloak]

,rotate ccw

[onyx cloak]

try putting n=1,2,3...

[outer abyss]

Ahh, so that's how to get the answer

[static totem]

yeah, write the sequence

[outer abyss]

I couldn't understand the question lol

Lemme try

[onyx cloak]

what if u put n=-1

[static totem]

then they all appear i guess

[onyx cloak]

$a_{-1}$ doesnt make sense tho

[clever fjord]

Astar777

[static totem]

the intention is either that you start from 0 or from 1

[onyx cloak]

yeah

[outer abyss]

So I guess the answer was 0, letter a, thanks for the help... thats what I thought too, where the terms go negative and then the other. 3 choices would be correct lol but ig only positive

Thanks again for the help

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[vagrant orbit]

help me w diff cal

[odd edge]

Please don't occupy multiple help channels.

[vagrant orbit]

<@&286206848099549185>

[summer steeple]

Why are some crossed out

@vagrant orbit

[vagrant orbit]

i just need to answer the one w the red marks

[summer steeple]

Alr

Well tell me what you have done so far

[vagrant orbit]

in this? the graph of this is

[summer steeple]

Yh

[vagrant orbit]

?

[rustic canyon]

the most simple method is use graphing calculator... the most stupid mothod is to plug in the numbers from 0, 1, ...

[vagrant orbit]

im just learning to graph w/o the use of graphing calcu

[summer steeple]

Yh the equation is just -x-2

[summer steeple]

Most of them are pretty basic tbh

[vagrant orbit]

how about the domain n range?

[fringe bough]

almost...

[summer steeple]

Well u can’t get x=2 right?

[vagrant orbit]

yes

[summer steeple]

Well then what does this tell u about the domain

[rustic canyon]

for “root”, the number below the "root" sign can't smaller than 0

[summer steeple]

Also did u learn how to do inverse functions

[vagrant orbit]

d: (-infi, 2) (2, infi)?

[summer steeple]

Yh

And the range?

[vagrant orbit]

r: (-infi,-4) U(-4, infi)

?

[summer steeple]

Yh

[rustic canyon]

for fractions, the denominator cannot be 0

[summer steeple]

Alr

[vagrant orbit]

dom n rang?

[summer steeple]

What do u have for number 4

[vagrant orbit]

all real num?

[summer steeple]

No

The thing inside the root can’t be negative

[vagrant orbit]

ohh ok ok, i forgot

[summer steeple]

X^2-25>=0

Now solve that and u get domain

[vagrant orbit]

wait im gonna write it down

so factor this n it will be (-infi, -5)U (5,infi)?

for domain

or do i need to change it into (-infi, -5]U [5,infi)?

@summer steeple

[summer steeple]

Yh

[vagrant orbit]

how about range?

[summer steeple]

-5 and 5 included

[summer steeple]

This

Well the root is always positive

So just [0,inf)

[vagrant orbit]

correct?

[summer steeple]

Yes

Then for the graph just do point by point

[vagrant orbit]

dom n range?

u still there @summer steeple

[summer steeple]

Hmm

[vagrant orbit]

domain will be real num?

[summer steeple]

Well domain is all real numbers

Yes

[vagrant orbit]

range?

[summer steeple]

All real numbers also I think

[vagrant orbit]

ohh yes yes yes

[summer steeple]

Cus 3x-2<1 for x<1 and x^2>=0

[summer steeple]

Domain is all real

Range is {-3,-1,2}

[vagrant orbit]

wait domain?

[summer steeple]

Well it’s all real

U literally just wrote R

[vagrant orbit]

ohh ok ok

so the graph looks like this

wait idk what the graph looks like

@summer steeple

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[vagrant orbit]

is the graph correct?

[odd edge]

Please don't occupy multiple help channels.

[vagrant orbit]

@mystic saffron

@sage root

<@&286206848099549185>

[tacit haven]

you should graph in between the ranges too ( x < -2)

[vagrant orbit]

i did

[vernal yacht]

Can you please stop pinging helpers?

[odd edge]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[tacit haven]

well it should be a horizontal line where x < - 2

[vagrant orbit]

so what does the graph looks like

?

[vernal yacht]

,rotate

[tacit haven]

like that yeah

[vagrant orbit]

sorry for being slow, but what r the points?

[odd edge]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[vagrant orbit]

?

[tacit haven]

the same as what you answered

[shadow vessel]

I don't understand youre question

[vagrant orbit]

in this pic u just make a hori line for -3, -1, and 2?

[tacit haven]

yes

[vagrant orbit]

ohh

[shadow vessel]

Yes, because there are 3 constant fonction in a big fonction

[vagrant orbit]

how abt this

?

[vagrant orbit]

,rotate

[vagrant orbit]

?

[vagrant orbit]

so the domain is all real and range is {the constants}?

[vagrant orbit]

whats the domain n range?

all real?

[vagrant orbit]

whats the domainn range?

,rotate

[shadow vessel]

R i think

[vagrant orbit]

in this?

domain is all real and range is {the constants}?

[shadow vessel]

In the two questions

[vagrant orbit]

????

[shadow vessel]

For the first picture yes

[vagrant orbit]

ok ok, thanks

?

[shadow vessel]

You're welcome

[shadow vessel]

Did you understand ?

[vagrant orbit]

yes

[shadow vessel]

Ok, i'm sorry, i can't help you with this, i have to work too

[vagrant orbit]

its ok

.close

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[raven vigil]

mind if anyone can nudge me to the right direction?

[raven vigil]

I don't really know of what to do next

do I express BE and ED?

uh-

[odd edge]

@raven vigil Has your question been resolved?

[fervent hawk]

this is so hard!

[raven vigil]

it's an olympiad question

my brain fried rn

[fervent hawk]

i somehow proved (in my brain) that AB=AD, but is that useful?

[raven vigil]

w8 that actually helps everything

[fervent hawk]

really?!?

[raven vigil]

I think

[fervent hawk]

oh, then lemme draft my proof

actually i was showing triangle ABE~ACB
2side ratio + equal angle between the 2 sides

[raven vigil]

I thought ABE is congruent to EDC

my brain drunk rn

[fervent hawk]

Let x=AE
then AC:AB=2:√2=√2:1
and AB:AE=√2:1
also angle BAC=angle EAB
hence
triangle ABE~triangle ACB

[fervent hawk]

prob not 😦

[raven vigil]

thx for the help

.close

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[fluid tundra]

did u correctly solve

[raven vigil]

kind of

I got 2 root 3

[fluid tundra]

how "kind of"

u got answer but unsure if it is right?

2sqrt3 is indeed ans

[raven vigil]

no idea how, I assign random stuff to random sides that I don't know exist

[fluid tundra]

this problem has very misleading diagram

[raven vigil]

w8 fr?

[fluid tundra]

note:

[fluid tundra]

yes point E is very misplaced

and if BD is horizontal line, then this problem is much easier to see

[raven vigil]

I just realized

thx

[fluid tundra]

yw

[odd edge]

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[urban ridge]

need someone to check my answers plz

Rate of P4 = 6 x 10^-4
Rate of H2 = 1.8 x 10^-3

[viscid flint]

UNITS

[viscid flint]

numbers mean nothing without units

[urban ridge]

oh shi

i forgor xdd

is it m/s

[viscid flint]

idk

is it?

[urban ridge]

no idea

😢

[viscid flint]

then i have no idea if it's right

not sure where m would have come in, meters don't show up at all here?

[urban ridge]

no meters sorry

[viscid flint]

?

[urban ridge]

i think its moles per second

[warm blade]

Hi

[urban ridge]

Yo

[warm blade]

Do you remember me Hayley!

[warm blade]

I think rate of reaction is in mole per decimeter cube per second

[viscid flint]

that would appear to make physical sense. does the number of phosphorus atoms in / out balance properly?

[warm blade]

Mol dm^-3 s^-1

Basically concentration per second

@viscid flint

[urban ridge]

Yes

[viscid flint]

seems right then

[urban ridge]

lets go

thank you

[viscid flint]

no

[urban ridge]

,end

,close

hmm

,ticketclose

[viscid flint]

.close

[odd edge]

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[clever fjord]

Couldn't find a member matching cketclose!

Sep 22, 2024, 11:59 AM

[odd edge]

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[mystic saffron]

hi

[mystic saffron]

I got 6 rank

out of 400 students

[onyx cloak]

in what

zamn

[mystic saffron]

I got

[onyx cloak]

nice

[mystic saffron]

13

there are two papers:
6 in second and 13 in first

So what is my average rank

In total

out of 400 students

[onyx cloak]

oh its a question

[urban ridge]

💀

[mystic saffron]

No

A question + actuality

the final results aren't out yet

[urban ridge]

u are 6 out of 400 i think you can do it

[mystic saffron]

fine

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[novel venture]

How do i go about this

[fluid tundra]

is this the original problem

well w/e doesn't matter

[fluid tundra]

treat (sinx) like a variable, then use series expansion

i.e.
first thing here follows by geometric series
second thing heres follows by differentiating with respect to x

problem is better after u correctly rewrite the integrand

[odd edge]

@novel venture Has your question been resolved?

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[mystic saffron]

I am asked to derive the formula for the volume of this figure, is my proof correct?

And can someone explain why the first line applies to begin with? (It is simply given by the task)

<@&286206848099549185>

<@&286206848099549185> 🙏

<@&286206848099549185> 😔

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[digital jacinth]

can someone help me why we have different answer

[digital jacinth]

this topic is about vector and scalar

the problem is our teacher's example still added sin and cos to a vector that only has one direction

while the first one followed the rule where we dont use the trigonometric when we only have one direction and the second followed our teacher

I am quite confused which one is right

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@digital jacinth Has your question been resolved?

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[trail merlin]

This is physics, but does someone know how to find the total tension on the horizontal section here?

[quartz trellis]

F=ma

4*9.8(or 10)

you can simplify it to 40N

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[waxen hornet]

Average value is 0 right?

since its a sinusoidal function

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[silent plaza]

Im literally struggling in a simple question but could somebody please help me

[pastel orbit]

apply log rules

[limber glade]

what is log a - log b?

[silent plaza]

loga/logb

[pastel orbit]

not quite

[sterile blaze]

log(a/b)

[silent plaza]

yh thats what i ment

so once its x+1/x-4 = 3 what do i do

[pastel orbit]

hi esthesia

[sterile blaze]

hi gher

[pastel orbit]

this form never arrives

you have log_2(x+1/x-4) = 3

now you need to get rid of the log_2...

you can do that by remembering that the inverse of log_2(x) is 2^x

[silent plaza]

oh

oh wait yeah ik what to do now

ty

[pastel orbit]

[silent plaza]

.clow3

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[somber gulch]

I have no issues with part (a)

[honest turtle]

Make x and y into perfect squares.

[mystic saffron]

and then you will be left out with constants,
get it to the right side
now every thing on the right should be positive since it is equal to r^2
and r^2 is always positive

[somber gulch]

But for part (b) how can we convert this into an inequality like what would be the conceptual significance of circle inequality like the range of values which make the circle

so the inequality is always (x-a)^2+(y-a)^2 >r^2

[honest turtle]

Think of a circle, what is the domain of possible perimeter?

[somber gulch]

no idea not taught this but thinking

r

greater than r

2 pi r should be greater than r

otherwise it will not be a circle

[honest turtle]

2 pi r is the formula for the length of the circle's edge.

[somber gulch]

yes circle perimeter

[honest turtle]

... my bad, I meant radius.

[somber gulch]

I don't understand what you mean do you mean the circle equation equaling the radius

squared

so the perimeter should be greater than the radius

[honest turtle]

Think of a circle, if you we're to define a "logically" and "visually" possible circle, are there any values that the radius can't be?

[somber gulch]

thinking

negative values negative square roots

but isn't negative radius radius in another direction

[honest turtle]

So a radius with minus sign can't be true.

Think of a football, how could a ball with r = -1 visualized? It doesn't mean that the radius is in "the other direction".

[somber gulch]

but on a coordinate plane negative value radius can be possible

[honest turtle]

This is not the situation. If you're at the center, and the distance from the center to the edge is always r, not -r.

[somber gulch]

so always absolute value is r and that is what is the domain of the circle function

so even in the negative scenario the equation would take the absolute value inpurt

[honest turtle]

Not absolute value of r.

If you are standing on the ground, you can jump 1m above the ground, but you can't jump -1m above the ground, you'll just stop at 0m (hard ground).

You can't just "jump" into the ground.

[somber gulch]

So the conceptual barrier of a circle is not going into the negative because then the shape would not be true

[honest turtle]

Yes. So don't think of the "direction", there's no direction.

You'll most like get something like (x-a)^2+(y-b)^2 =k+a2+b2, where a2 and b2 are constants.

[somber gulch]

Got it so (x-a)^2+(y-a)^2<r^2 and r^2>0 so these are axioms of a circle

[honest turtle]

(x-a)^2+(y-b)^2=r^2, r>0

[somber gulch]

So an equation which I had memorized became an inequality which I had to conceptually think and apply

[honest turtle]

The x- and y-coordinates don't necessarly needs to be same.

[somber gulch]

so I can input x coordinate of (a,b) and y coordinate of (c,d) so I can input x-a and y-d?

[honest turtle]

The center is (a,b).

[somber gulch]

oh sorry i mean x coordinate of (c,d) and y coordinate of (e,f) so I can input c-a and y-f?

in the circle equation like if there is a tricky question where I have find coordinates

[honest turtle]

Where did c-f came from?

[somber gulch]

sorry that was a mistake

I mean x-f

sorry I mean c-a

[honest turtle]

It's not always that easy to explain some graphical with pure words. Watch a video might give a better understanding https://www.youtube.com/watch?v=u_39J-syjB0

[somber gulch]

c, f is coordinates on the circle

[somber gulch]

Watching the video now Ty pls wait

[somber gulch]

Watched it but he didn't talk about domain range or that r^2>0

Is this inequality true

[honest turtle]

r>0

[somber gulch]

(x-a)^2+(y-b)^2<r^2?

[honest turtle]

This is how a circle functionn looks like $$(x-a)^2+(y-b)^2=r^2$$

[clever fjord]

Good

[honest turtle]

a is the x-shift, b is the y-shift.

r is the radius.

radius can't be negative, so r>0

[somber gulch]

Can't see the image

[honest turtle]

(x-a)^2 + (y-b)^2 = r^2

[somber gulch]

so this inequality is false right?

[honest turtle]

This inequality doesn't tell anything

[somber gulch]

wdym

radius square is always greater than the other part of the circle equation

[honest turtle]

The center is less than the radius.

[somber gulch]

so the inequality is the other way

?

[honest turtle]

Read the chapter in your book again, or watch other videos. You're confusing yourself with something.

[somber gulch]

it'

s

there are tricky questions related to the circle

many times they ask for range of possible values where I need to convert an equation to an inequality

so maybe there is some trick in this equation

with the inequality being the other way I don't agree unless someone esle

[honest turtle]

I'm out of time, so you should find a other way.

[somber gulch]

says yes as you have more experience with more tricky questions

Got it Ty so much for your help

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[haughty sage]

i am pretty confused on how to solve this, I tried making the mapping point and then doing the opposite operations of the g(x) points to try to get the f(x) points but its still wrong

[haughty sage]

I made the mapping notation before this is what I got (x, y) -> (-x+3, 2y-4)

here is what i have so far

@wanton bison is it correct?

<@&286206848099549185>

[wanton bison]

Almost, it should be (-x-3, 2y-4)

[haughty sage]

Are u sure

Dont u have to factor the negative out

[wanton bison]

You can

[haughty sage]

On the graph its on positive 3

[wanton bison]

but then -(x-3) implies that you you reflected it on the y-axis and then shifted it by 3 units to the right

So then you should have -(x+3) instead

not -x+3

sqrt(x)
-> sqrt(-x) reflection at the y-axis
-> sqrt(-(x+3)) shift by 3 units to the left
So now your origin moved from (0,0) to (-3,0) = (-(0+3), 0)

With -x+3 the origin would have been moved to (3,0) instead but it went to the left not right

[haughty sage]

Hmm okay

So (-x-3, 2y-4)

Once we have that what do we do next to get to the original function

undo ?

[wanton bison]

You would do
-x-3=-6
2y-4=2

[haughty sage]

ohhh

then solve for y and x?

x = 3 and y = 3

[wanton bison]

wait

[haughty sage]

Ok

[wanton bison]

yea you would equate them but for x it would be -x+3=-6

[haughty sage]

So my original mapping was right?

wait im confused ngl

[wanton bison]

no

[haughty sage]

hmm ok but why did u switch it to -x+3=-6

[wanton bison]

let me think for a second i got myself confused

[haughty sage]

ight

[wanton bison]

-6
-> -6-3=-9 shift to the left g(x+3) coordinate wise we did x-3
-> -(-9) reflection at the y-axis g(-x) coordinate wise we did -(x-3) = -x+3

yea

i think you were right i am so confused right now i'll stop

[odd edge]

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[haughty sage]

ight thanks for u rhelp

u helped me on the last part cuz i had no idea what to do

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[wise zodiac]

did i write this correclty?

[wise zodiac]

by writing (-infinity, 1] twice?

the prompt is to find the intervals of the polynomial

[odd edge]

@wise zodiac Has your question been resolved?

[sterile blaze]

the question is when is the polynomial negative or 0

when x is 8 the polynomial is positive

[sterile blaze]

unless theres a different question

[wise zodiac]

i didnt notice that

[wise zodiac]

why

[sterile blaze]

each term is positive

multiplying positive terms gives u a positive term

[wise zodiac]

did u mean 7

if not, idk where the 8 came from

[sterile blaze]

8

but 7 works too

[wise zodiac]

why 8

[sterile blaze]

but u get a 0

[wise zodiac]

a 0 grade?

[sterile blaze]

im trying to tell u your interval would result in a positive number

p(7)=0

[wise zodiac]

so

is the interval [7, 0)

?

[sterile blaze]

how did you get that

[wise zodiac]

here i had [7, inf), and i replaced the infinity with 0

since 7 is positive and

p(7)=0

[sterile blaze]

when is the graph=0

[wise zodiac]

x=1, and x=7

[sterile blaze]

so the graphs sign cant change outside of those regions right

[wise zodiac]

ya

[sterile blaze]

draw a number line and mark 1 and 7

[wise zodiac]

ok

[wise zodiac]

idek what a sign is i gotta look that up rq

[sterile blaze]

when x is in the first region (below 1) what is the sign of the polynomial

[sterile blaze]

if its positive or negative

[wise zodiac]

negative

[sterile blaze]

inbetween 1 and 7?

and above 7?

fill it in in the number line

and that would help u answer the question

[wise zodiac]

would that be positive? i havent done this before so idk

[sterile blaze]

pick a random value in that range

so maybe use 2

[wise zodiac]

would the interval [1, 7] work?

[sterile blaze]

yes

[wise zodiac]

so i can write p(x)_<0 at [1,7] and its correct?

[sterile blaze]

wait sorry

it should be <=7

0 works

[sterile blaze]

because its negative below 1

[wise zodiac]

how do i write this answer

[sterile blaze]

when x<=7

or when x in (-inf,7]

either works

[wise zodiac]

ty

.close

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[sterile blaze]

np

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[deep turret]

How to do this?

Here’s my attempts

[remote abyss]

those are some good attempts

ill help you one second

[remote abyss]

the i on the bottom is a variable but it is invisible.

so it would be written like i^1

if u break it down a lil more

it will look like this

1/i^1 + 4/i^3

now there is exponents on both sides

you gotta multiplying the second term to eliminate the imaginary number from the denominator

[deep turret]

I did

[remote abyss]

what was your answer

you should get 1/i + 4i

as your answer

is that all?

[deep turret]

By what? Also what’s on the other side?

[remote abyss]

when you want to remove the imaginary unit 𝑖 i from the denominator of a fraction, you multiply both the numerator and the denominator by the conjugate of the denominator. The goal is to turn the denominator into a real number.

[deep turret]

That doesn’t work when there’s only an imaginary number in the demoninator

[forest sky]

multiplying by the conjugate still works when the denominator is imaginary

[remote abyss]

ur right when there's only an imaginary number like i (or a power of 𝑖 i) in the denominator, you can eliminate the imaginary unit by multiplying by i, without needing to use the conjugate (which is used for complex numbers with both real and imaginary parts)

[deep turret]

No it doesn’t

Wait…oh

Yea right

I forgot what i^2 was

Uhh I see what I did wrong now

I simply needed to take a break

and reread

Let me do that thank you

.close

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[deep turret]

.reopen

[odd edge]

✅

[deep turret]

I should post the right answer before I go

[remote abyss]

sure!

[deep turret]

-4 - i

That’s my answer

I was close to finding it with my previous work

[remote abyss]

so close

[lost karma]

.reopen

yo can somone help me

[remote abyss]

you would take out the negative

[lost karma]

with this

[remote abyss]

go over here

[deep turret]

The book said 3i…

[lost karma]

where

[remote abyss]

#help-14

[lost karma]

its not working

[remote abyss]

oh.

[deep turret]

…what?

[remote abyss]

i rechecked my work and it says 4i is the answer

[forest sky]

the book is correct

[remote abyss]

wait how tho?

there is no 3 in the equation besides the exponent?

[forest sky]

what do you get after making the denominators real?

[deep turret]

-1 + -4i

[remote abyss]

yea same

[forest sky]

4/i^3 != -4i

[remote abyss]

it would be postive tho

[forest sky]

and 1/i != -1

[remote abyss]

cause over here theres 2 -'s

[forest sky]

remember you have to multiply both the numerator and denominator by i, otherwise that is not multiplying by 1

[deep turret]

idk

[remote abyss]

im postive that 4i is the right answer

[forest sky]

the trick we are trying to employ here is "multiplying by 1", which is where we multiply something by a fraction which is the same in both the numerator and denominator (anything divided by itself is 1)

because anything multiplied by 1 is itself

[deep turret]

-i + 4/i^3

[forest sky]

we can do the same thing to 4/i^3

[deep turret]

-4i

[forest sky]

not quite

[deep turret]

4i

-i + 4i = 3i

[forest sky]

well there's your answer then

[deep turret]

.close

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[lost karma]

Hi

uh im not a helper im a person that needs help-

.close

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[livid dust]

hi just curious if the difference is meaningful between these two in a calculus course

the difference i mean is between putting a negative before or instead writing ytan(x)sec(x)-9tan(y)

[golden marten]

there shouldn't be a difference

I would argue that the one on the left is slightly simpler in this case

since the negative is an extra symbol

[livid dust]

in a previous question i found the second through implicit differentation though my online assignment only accepted the first version

[livid dust]

this is what i thought

but then in the previous question only the second expression was considered correct
guess im just curious if there was anything i was missing but maybe the software is just bad

[golden marten]

I reckon it's notational but both are correct

it depends on preference in certain cases

[livid dust]

Alright thanks, guess i should just deal with it lol. probably bad programming

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[chilly isle]

Im very quite not sure what I have to prove

If someone could give me hints or clarifications on the question that might help

Is what I wrote sufficient to prove it?

[odd edge]

@chilly isle Has your question been resolved?

[odd edge]

@chilly isle Has your question been resolved?

[odd edge]

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[lucid shadow]

Can anyone help with question 3?

[odd edge]

Please don't occupy multiple help channels.

[lean raptor]

Please keep this question contained to one channel

[wanton bison]

!1c

[odd edge]

Please stick to your channel.

[wanton bison]

They have already one

[wanton bison]

.

[lucid shadow]

I understand it now

Someone can take the channel

[ocean torrent]

You need to close it

Type .close on both this one and your other one lol

[lean raptor]

.close

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[lean raptor]

I did it for you

[sterile blaze]

goat

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[odd edge]

@ivory vine Has your question been resolved?

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[supple frigate]

hi i just need someone to check my answers

[supple frigate]

Anyone??

[odd edge]

@supple frigate Has your question been resolved?

[supple frigate]

.close

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[tidal ingot]

. Bits are transmitted over a communication link. When a 1 is sent, it is received as a 0 with probability 0.15. To reduce the probability of error, each bit is transmitted 5 times. If at most one bit differs from the others, then the bit that was received the most often is assumed to have been transmitted. What is the probability that a 1 that is transmitted 5 times is received correctly as a 1?

[brittle notch]

So basically

U need

Either 3 or 4 or 5 1s

Right?

[mystic saffron]

YO

VND

[brittle notch]

Yes

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[feral igloo]

so uh would the answer be 7/x^2 ?

[feral igloo]

I got a feeling thats wrong but

7

wait...

thats just divisible by 7

[simple surge]

343 = 7^3. The root can't "destroy" the exponent (3)

[feral igloo]

hm

[simple surge]

it would turn into a fractional power

(7^3/2) / (x^2)

[feral igloo]

Oh so not like this?

[clever fjord]

faiyrose

[feral igloo]

yes

[clever fjord]

faiyrose

[feral igloo]

sure

7

[clever fjord]

faiyrose

[feral igloo]

erm

probably 1?

Might be a decimal

lets see, comparing how he solved his problem, for the ...

this cant be the right answer, can it?

[clever fjord]

faiyrose

[feral igloo]

oh

[feral igloo]

@icy vault so um is that the answer?

huh.

[feral igloo]

now i geget it

thanks.

.close

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[feral igloo]

would like to know if im supposed to cancel out the sqrt first

[feral igloo]

or just +1 to both sides so its sqrt6x-2 = x + 1

then cancel out the sqrt

[lean raptor]

Plus one first

But you will also have to square the x + 1

[feral igloo]

yeah ok then

thank you!

[feral igloo]

(x+1)(x+1)

[lean raptor]

Yep

[feral igloo]

i think i probably was supposed to leave that 6x alone? ...

[lean raptor]

Well you are correct to say that x^2 + 2x + 1 = 6x - 2

And now you’re supposed to take the x’s to the other side and solve for x

[feral igloo]

hm

Im pretty sure 4x/4 isnt the way to go, but uh

not sure, do i just divide x^2 by x?

[lean raptor]

You got rid of a 1

You have x^2 = 4x - 3

[feral igloo]

ohh

mb with the -3

[lean raptor]

So do you know how to solve a quadratic by factoring?

[feral igloo]

i think,

[feral igloo]

Basically, this for example,

x+1=0

[lean raptor]

Ok so you have x^2 - 4x + 3 = 0

[feral igloo]

factor,

[lean raptor]

Would you be able to factor that?

[feral igloo]

yeah

[feral igloo]

i gtg for a bit, but will be back with an answer (class is starting)

@feral igloo

.close

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[jade flame]

Did i do 2a correctly?

Also how would 2b be any different?

[hollow plaza]

help

[long bridge]

please get your own channel

[hollow plaza]

how

[quasi sparrow]

#❓how-to-get-help

[quasi sparrow]

doesn't look like you found x_0 yet

[hollow plaza]

help

[jade flame]

Doesnt x-x_0 just become t?

[long bridge]

there really shouldn't be a t in your answer

[long bridge]

there are instructions on how to get help in #❓how-to-get-help

[jade flame]

What should my answer even look like?

[long bridge]

something like y = -6x+8
but different numbers

[forest sky]

it asks for a parameterization

but your answer should be in the form
x = f(t)
y = g(t)
where the functions are only in terms of t, all the constants like x_0 should be replaced by the actual numbers

[jade flame]

Ok now i get it

But how is the answer in a supposed to be different from b?

[forest sky]

for the line t should be any number, whereas for the line segment you have to restrict its domain

[jade flame]

To what exactly?

[forest sky]

whatever values of t would restrict it to only going between the two points

[jade flame]

So are the parametric equations still the same?

Just the domain is different?

[forest sky]

yes

[jade flame]

Ok thanks .

.close

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[jade flame]

Also do i figure it out the same way for the parabola and circles?

[forest sky]

the process will be similar

[odd edge]

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[lyric lintel]

help

[lyric lintel]

been stuck on this for almost 40 minutes

number 8. your help would be greatly appreciated. plz @ me when replying

[runic swift]

just put x = -4 into f(x-2)

[lyric lintel]

y

[runic swift]

?

[lyric lintel]

like i wna be able to understand why we are doing this step

and if u put -4 as x shouldnt u get 1 out as the y value @runic swift

[runic swift]

yep

[lyric lintel]

then y would i plug in x as -4

?

[runic swift]

yes

then you'll get f(-6)

which is the original function f(x)

[lyric lintel]

so f of -6 is also 1 in y value?

[runic swift]

yes

[lyric lintel]

oh ok cause y=f(x) is no transformations

yet

[runic swift]

yes

[lyric lintel]

so then i have 1, -6 in original graph