help-39

[midnight haven]

this is the solution for a) but i do not understand

[light helm]

which part in their work don't you understand

[buoyant trench]

some1 help me on the simplest thing in this server

pls

this is so ez

[sweet bone]

!occupied

[pearl pond]

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).

[buoyant trench]

@light helm '

[light helm]

what have i done to deserve this ping notification?

[pearl pond]

@midnight haven Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[glass smelt]

This is what i have solved for the (a) part

<@&286206848099549185>

[pearl pond]

@glass smelt Has your question been resolved?

[glass smelt]

.close

[pearl pond]

Channel closed

Closed by @glass smelt

Use .reopen if this was a mistake.

[glass smelt]

.close

[pearl pond]

Available help channel!

Send your question here to claim the channel.

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

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[gleaming flame]

im doing a) is there only AQ or are there more

[gleaming flame]

im kinda blind

[sweet bone]

this might just be different notation, but usually angle-ABC refers to the angle at B

if your course uses it to mean the angle at a, then just AQ is fine

[tender tartan]

i think it's not angle abc but triangle abc

[gleaming flame]

yea it is triangle

[sweet bone]

then there are more

[gleaming flame]

wait wha

idk if a angle bisector of a triangle requires to touch the vertex

[sweet bone]

ah wait, I misread

yeah it needs to touch a vertex

[gleaming flame]

thanks

how do i close?

[sweet bone]

.close

[gleaming flame]

thx

.close

[pearl pond]

Channel closed

Closed by @gleaming flame

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

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

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[covert lotus]

b.?

[covert lotus]

no clue how

[solemn rock]

Do you know how to sketch the set on the left side of the intersection?

[pearl pond]

@covert lotus Has your question been resolved?

[cursive sphinx]

Please send me solution

[solemn rock]

nobody will give you a soln, and please use another channel

[cursive sphinx]

Ok

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[vital cradle]

How would you find the modulus and argument in question c.i ? That's where I'm stuck.

[pearl pond]

@vital cradle Has your question been resolved?

[vital cradle]

<@&286206848099549185>

[pearl pond]

@vital cradle Has your question been resolved?

[pearl pond]

@vital cradle Has your question been resolved?

[vast berry]

for modulus

wouldnt it be sqrt(sin^2(4x) + (1-cos(4x))^2)

and by expanding the square i got sqrt(1-2cos(4x))

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

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

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[slim canyon]

I'm a bit confused on how to answer the second part of the question.

From my understanding I should be able to create two equations.
0.0500N=k*((Q^2)/r^2)
0.160N=k*((q1-dQ)(q2+dQ)/r^2

This however leads me to a solution of dq=1.79, which is surely incorrect.

[pearl pond]

@slim canyon Has your question been resolved?

Channel closed

Closed by @slim canyon

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[unreal pike]

hello

[austere pike]

That is my task

and i dont know how they got what they got

317 * p^25 =346 25th root of 346/317 = 1,0035

Thats the solution and i dont understnad it

[pearl pond]

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

Available help channel!

Send your question here to claim the channel.

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

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[frosty badge]

I got a triangle the 2 sides are 6cm and 8cm the medians on thees sides are mutuallyperpendicular need the triangles 3rd side

[frosty badge]

<@&286206848099549185>

[pearl pond]

@frosty badge Has your question been resolved?

[frosty badge]

<@&286206848099549185>

[open ivy]

Here is an image that helps maybe too much

[frosty badge]

so do i use pyth here or what?

[open ivy]

Ye, mostly.

[frosty badge]

mhhh how so

[open ivy]

You want BC, normally it would be given by $x^2+y^2=BC^2$ as in the image.

[jolly parrot]

Crystopher

[open ivy]

So you have two unknowns, $x$ and $y$, find those.

[jolly parrot]

Crystopher

[frosty badge]

so i need to find x and y

do i need to make a system?

with pyth

?

@open ivy

[open ivy]

yes, you have 3 right triangles in the figure to use at your disposal.

one of them is 'useless' for now, so use the other two

[frosty badge]

This is correct yes

got it thanks a lot

mind helping me with 1 more problem

@open ivy

[open ivy]

sure, let's see.

[frosty badge]

ok so i have a isosceles triangle the base is 12 and legs are 18 and 18 what lenght line do i need from the triangles apex to the legs when i connect them to make a 40cm perimeter trapezoid

@open ivy

you here man?

[open ivy]

yes

I'm trying to parse the formulation

Today I learned what an apex of a triangle is

[frosty badge]

do you get the info i gave?

if not ask

[open ivy]

Let me see if I understand. In this, image, do they want a sum of CE, DE and CD that is 40?

[frosty badge]

you see the trap right

that trapezoids perimeter is 40

[open ivy]

I haven't reached a solution yet, but I think that to solve the problem you need to think about similarities in the figure

[frosty badge]

wait

the drawing is a bit wrong

[open ivy]

how so

[frosty badge]

the line has to come from the base apex to the legs

[open ivy]

I don't get it. Isn't the apex the point C in this triangle?

[frosty badge]

yes your correct

but the problem says i has to connect to the leg

so saying apex was mb

from A to the leg

[open ivy]

How could you form a trapezoid by drawing a line from A to the leg BC?

[frosty badge]

wait i think i solved it

your drawing is correct

so look

w know that the bas is 12 right

base

[open ivy]

yes

[frosty badge]

de is a middle line in the triangle

so it has to be half of 12

so 6 is the correct answer

[open ivy]

but then 12 + 6 + 9 + 9 = 36 != 40

[frosty badge]

mhhh

your right

dose d e cut the legs in half?

[open ivy]

That is just a concept figure, it doesn't have to be correct, I was trying to see if I understood the problem correctly. The measures of the sides are 1:1, but the figure is possibly not the solution.

[frosty badge]

is there a way we could find the legs somehow

[open ivy]

Anyways, we can let DC = x, since ABC and DEC are similar, then $\frac{12}{18}=\frac{DE}{x}$. This would mostly solve the problem.

[jolly parrot]

Crystopher

[frosty badge]

ok

what next tho

[open ivy]

Well, what is the perimeter of the trapezoid expressed in terms of $x$ and $DE$?

[jolly parrot]

Crystopher

[frosty badge]

wym?

[open ivy]

Find first the sides AD and BE

[frosty badge]

how did you get 5

?

[open ivy]

That's a question mark, not a 5.

[frosty badge]

ok

[open ivy]

I notice, I drew it the wrong way.

[frosty badge]

40-x-12

?

[open ivy]

I mean the question marks

[frosty badge]

/2

40-12-x/2

[open ivy]

What is that?

[frosty badge]

the ? marks

in the terms of the perimeter no?

[open ivy]

No, find the side AD in terms of x.

[frosty badge]

ADx wym?

[open ivy]

sry, typed wrong.

I fixed it

[frosty badge]

18-x

[open ivy]

So, how do you calculate the perimeter of a trapezoid?

[frosty badge]

add all the sides

36-2x+12 equals 40

[open ivy]

You forgot DE

[frosty badge]

+x?

[open ivy]

No, remember:
Anyways, we can let DC = x, since ABC and DEC are similar, then $\frac{12}{18}=\frac{DE}{x}$. This would mostly solve the problem.

[jolly parrot]

Crystopher

[frosty badge]

12x equals 18DE ?

[open ivy]

yes, or $DE = \frac{2x}{3}$, if we were to simplify it.

[jolly parrot]

Crystopher

[frosty badge]

now we add that too

[open ivy]

yes, and solve for $x$.

[jolly parrot]

Crystopher

[frosty badge]

would it not be 3/2??

oh we did De

your correct

x is 6

[open ivy]

Checks out.

[frosty badge]

yup

thanks a lot

how old are you

?

[open ivy]

you are welcome, 21.

[frosty badge]

how lon have you been doing maths for?

[open ivy]

I'd say I started taking maths seriously since the first year of high school, although nowadays I'm too busy with Computer Science to dabble in math.

[frosty badge]

i am planning to study cp too

[open ivy]

Nice to hear, it may not be my favorite cup of tea, but it is quite enjoyable anyways.

[frosty badge]

sry to ask mate but can you help me with 1 more?

[open ivy]

sure.

[neat harbor]

nice

[frosty badge]

i got a right triangle with a circle inscribed in it the curcle splits the hypotnus 2x and 3x and the inscribed circles center is distanced from the right angle by sqrt8 need to find all sides of the triangle

you here?

@open ivy

[open ivy]

yes, I'm here, I'm drawing an image.

[frosty badge]

ok thanks a lot

[open ivy]

Maybe this is somewhat accurate.

I notice angle ADB does not have to be a right angle.

[frosty badge]

MD make a right angle?? why

[open ivy]

That's what I say, I noticed that does not have to be the case. Perhaps I was fooled by the figure.

[frosty badge]

ok ill ignore that angle

[open ivy]

ye, sry about that.

[frosty badge]

its ok

so what do we do?

[open ivy]

Well, if we draw a radius perpendicular to AB, we get a right triangle, and we can find the value of $r$. Maybe that is a good starting point.

[jolly parrot]

Crystopher

[frosty badge]

ok i see the triangle

the hypot would be sqrt8

what about the 2 other sides

[open ivy]

Well, we know that the circle is the incircle, as such its center is the intersection point of the triangle's angle bisectors. This means that angle MAB = CAM = 45 degrees.

[frosty badge]

so AD is a bisector?

[open ivy]

Not necessarily, you may see AM and MD as not being part of the same segment.

[frosty badge]

so AM splits the angle in half

[open ivy]

But the line through A and M is a bisector.

[frosty badge]

that makes the triangle isosceles

the right one

so the sides are sqrt4

right?

[open ivy]

yes

[frosty badge]

so r is sqrt4

[open ivy]

or 2.

[frosty badge]

yea

[open ivy]

Now I wonder what we can do with this information, I actually don't know yet.

I'll try to draw a better image.

[frosty badge]

ok

[open ivy]

Maybe this is better, this time MD forms a right angle since BC is a tangent to the circle.

[frosty badge]

ok so that makes 2 more isosceles right triangles right?

[open ivy]

where?

[frosty badge]

ADB and ADC no?

thats wrong

[open ivy]

But they are not isosceles?

[frosty badge]

there not?

[open ivy]

I don't see how they must be isosceles.

[frosty badge]

A is 45 right D is 90 and B must be 45 no?

[open ivy]

That is under the assumption that the angle ADC is 90 degrees, but it is not. It is the angle MDC that is 90 degrees.

A, M and D are not (necessarily) colinear.

[frosty badge]

oh

so what do we do?

[open ivy]

We notice that two tangents through the same point to the same circle have equal lengths, like in the following image

[frosty badge]

yea thats a rule

[open ivy]

We get the following

,rotate

[open ivy]

CA and CB are both tangents to the same circle, through the same point C, so EC = CD = 2x

[frosty badge]

yea ok

[open ivy]

Similar reasoning can be done about AF, EA, BF and BD.

Now you just use Pythagora's to solve for x

[frosty badge]

(2x+2)^2+

like that?

[open ivy]

ye

[frosty badge]

ok

ill solve that wait

x is 6

2

i mean

[open ivy]

yes, x=2

Well, I gotta go now. I hope most of it was clear. Bye.

[frosty badge]

thanks a lot

[frosty badge]

.close

[pearl pond]

Channel closed

Closed by @frosty badge

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

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

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[fallen osprey]

i missed lesson on function now i’m confused on how this work

[fresh night]

By direct substitution, f(a+2)= 5-(a+2)

Then simplify

[fallen osprey]

so it is 2-a?

-a + 3

[fresh night]

Yes

If no more information is given

[fallen osprey]

yeah

thanks

.close

[pearl pond]

Channel closed

Closed by @fallen osprey

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

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

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[warm sorrel]

Trying to do an inverse Laplace transform, not sure where I went wrong

[warm sorrel]

But this is the expected result

[pearl pond]

@warm sorrel Has your question been resolved?

[pearl pond]

@warm sorrel Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[covert lake]

hi

[covert lake]

Need help with this

[quick yacht]

What have you done so far?

[charred ocean]

!status

[pearl pond]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[covert lake]

is this the correct setup?

[quick yacht]

Binding is wrong

[covert lake]

?

wdym

[quick yacht]

Using y with dx

[covert lake]

its dy?

[quick yacht]

Yep

Also, you got it backwards

e^y-(y^2-5)

[covert lake]

ok bet

I got this

and it was marked wrong

@quick yacht

dont think thats the right seup

setup

[quick yacht]

Im not sure how you got that

[covert lake]

nevermind

I added wrong

ty

now I need help with this

how do I figure out the interval

like -1 to the blank

do I set equal to 0?

[quick yacht]

When does y^4=2-y^2

Normally quartics are a nightmare to solve, but this one is easily factorizable

[covert lake]

y = -1 and 1

right?

[quick yacht]

right

[covert lake]

so it would be -1 to 1

[quick yacht]

yeah

[covert lake]

and inside would be y^4 - 2-y^2

right

[quick yacht]

what?

[covert lake]

inside the ()

like this

[quick yacht]

think about their order

[covert lake]

well like this

[quick yacht]

pick some x from -1 to 1

say, 0

[covert lake]

yea

oh

2-y^2 would be bigger

[quick yacht]

0^4 = 0, 2-0^2=2

yeah

[covert lake]

so that comes first

right

thats how it works

the bigger comes first?

[quick yacht]

typically

the only difference is the sign in the area

• or -

everything else will remain the same

[covert lake]

wdym the sign

when do u do + and -

[quick yacht]

thats what i mean by sign

you generally want positive area in these problems

unless you are given some specific orientation

[covert lake]

oh ok

[quick yacht]

but, if you are never given an orientation, it should be positive
i.e: y1>y2, counterclockwise, are orientations

[covert lake]

yea that makes sense

for this

the last graph would be the correct choice fight

right*

[quick yacht]

its -x

,w graph y=x^2+7, y=-x-1

[jolly parrot]

Results provided by WolframAlpha

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

Requested by Cycadellic#0025

[covert lake]

oh what

so my first step in these type of questions would be to graph both equations right?

to figure the line

[quick yacht]

youd only graph it if youre asked or you think its easier than finding the inequality

[covert lake]

well I need to graph for this question right?

[quick yacht]

right, because it asks for the graph

[covert lake]

because I need to choose the right graph

[quick yacht]

yeah

[covert lake]

ok

so the 2nd graph is the right one

and this would be the correct setup right

[quick yacht]

yeah

thought the lower bound was -1

that was the last problem

[covert lake]

yep

,w graph y=5/x, y=5/x^2

[jolly parrot]

Results provided by WolframAlpha

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

Requested by abindagoat

[quick yacht]

as for this, they are continuous, so we are only concerned about when the lines are equal, then one sample point in our interval will tell us which each function is greater than the other

[covert lake]

ok

so here the first graph would be the right choice right

[quick yacht]

this is where you just get a graphing calculator, like desmos

or, like, a ti84

[quick yacht]

this is the intermediate value theorem, by the way

[covert lake]

this is my setup

lets goo i got it

,w graph y=4cos(x), y=4e^x

[jolly parrot]

Results provided by WolframAlpha

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

Requested by abindagoat

[covert lake]

,w graph y=15-x^2, y=x^2-3

[jolly parrot]

Results provided by WolframAlpha

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

Requested by abindagoat

[covert lake]

need help with this one

I got -1944 as area

but ik thats wrong

this was my setup

@quick yacht can u help me with tihs

this

I was on a roll and now I got stuck on this

[quick yacht]

15?

upper bound should be 3

[covert lake]

so the graph goes up to 15

I solved for x and got -3 and 3

but im not sure y its 3 and not 15

[quick yacht]

when does 15-x^2=x^2-3

since they are continuous, they will only be enclosed by their solutions to this equation

so

,w 15-x^2=x^2-3

[jolly parrot]

Results provided by WolframAlpha

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

Requested by Cycadellic#0025

[quick yacht]

$$x=\pm 3$$

[jolly parrot]

Cycadellic

[covert lake]

yea i got that

I just got 15 from the graph

since it goes up to 15 on the y axis

but its supposed to be from x=-3 to x=3

,w graph x=6y^2, x=20+y^2

[jolly parrot]

Results provided by WolframAlpha

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

Requested by abindagoat

[quick yacht]

youre thinking about it wrong

$$\big\int_{x=-3}^{x=3}(15-x^2-(x^2-3))dx$$

[jolly parrot]

Cycadellic
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[quick yacht]

x=... because the integral is wrt x

[covert lake]

yea i got it

[quick yacht]

now, we can do
$$\big\int_{y=-3}^{y=15}(15-x^2-(x^2-3))dx$$

[jolly parrot]

Cycadellic
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[covert lake]

for this why is the 3rd graph not right

[quick yacht]

but this requires u sub and inverting y(x)

,w graph x=6y^2, x=20+y^2

[jolly parrot]

Results provided by WolframAlpha

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

Requested by Cycadellic#0025

[covert lake]

yea I graphed it

that looks like the 3rd one

but it said I was wrong

when I clicked on it

[quick yacht]

because the axis are wrong

[covert lake]

oh

so it would be the last one

[quick yacht]

looks like it

[covert lake]

because its going from y = -2 to y=2

right

not x= -2 to x =2

,w graph y=3/x, y=12x, y=1/3x

[jolly parrot]

Results provided by WolframAlpha

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

Requested by abindagoat

[covert lake]

,w graph y=3/x, y=12x, y=1/3x, x>0

[jolly parrot]

Results provided by WolframAlpha

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

Requested by abindagoat

[covert lake]

im confused about this

cant figure which graph it is

and what equations would I subtract to find the area of the region

@quick yacht

<@&286206848099549185>

[spark parcel]

well

which one

,work

ugh

!work

WHAT WAS IT I FORGOT

anyways show ur work

[covert lake]

@spark parcel

I looked at that

,w graph yx+1, y=0

[jolly parrot]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by abindagoat

[covert lake]

,w graph y=x+1, y=0

[jolly parrot]

Results provided by WolframAlpha

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

Requested by abindagoat

[spark parcel]

the graph is wrong

,w graph (1/3)x

[jolly parrot]

Results provided by WolframAlpha

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

Requested by waitara

[covert lake]

wdym

[spark parcel]

did you atleast see which graph was the right one?

[covert lake]

wdym which graph

[spark parcel]

out of these

[covert lake]

thats what im trying to figure out

😭

[spark parcel]

okay

so

which graphs do u think u can rule out?

[covert lake]

the last two

[spark parcel]

alright, perfect

notice

y=12x and y=(1/3)x

[covert lake]

yea

[spark parcel]

when you multiply x with a larger number

would the graph come closer to or farther from the y axis?

[covert lake]

wait what happens to 3/x

[spark parcel]

just take y=12x for now

[covert lake]

it would be further

cause the number would keep getting bigger

[spark parcel]

you sure?

take y=x

if y=5, x=5, correct?

if you take y=6x

substitute y=5, x= 5/6

,w graph y=x

[jolly parrot]

Results provided by WolframAlpha

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

Requested by waitara

[spark parcel]

,w graph y=6x

,w graph y=(6)x

[jolly parrot]

Results provided by WolframAlpha

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

Requested by waitara

[spark parcel]

nvm

[covert lake]

so which graph is right

😭

im so confused

[spark parcel]

the black line is y=6x, purple is y=x

if you multiply x with a number, it comes closer to the y axis

its a rule of graphing

so if you multiply it now by 12, it will be quite close to the y axis

[covert lake]

oh oj

so the first 1 is the graph

now how do I find the area

because there are 3 lines

[spark parcel]

that uh... idk.

haven't done integrals yet

I AM GOING TO THIS YEAR THO

if u can stick around for a few more months i can help then

:))

[covert lake]

need @quick yacht help again

he knows what he is doing

[spark parcel]

HAHAHAHAH

SHHH

I HELPED U FIGURE THE GRAPH OUT

[covert lake]

yea ik

ty for the help

[spark parcel]

ur welcome

[quick yacht]

Whats the problem?

[spark parcel]

giggling

[covert lake]

im just saying he knows more about it

lol

[spark parcel]

kicking my feet

[covert lake]

this one

how would I go about figuring out the area

[quick yacht]

We do piecewise regions

When does 12x=1/3*x

Then, 3/x=1/3x and 3/x=12x

[covert lake]

1/6

so why do I do 3/x equal to both

thats y I am confused

[quick yacht]

We are trying to find the three points where the graphs intersect

This way, its easy to break the regions into pieces

[covert lake]

so I got 1/6

1/2

and 3

• and - for all

[quick yacht]

Which solutions are what?

[covert lake]

1/6 for

and 3 for 3/x=1/3x

and 1/2 for 3/x=12/x

yea one solution for each one

[quick yacht]

12/x isnt a term we have

So the sols should be x=1/2, 3, 0

[covert lake]

3/x=12x

is 1/2

[quick yacht]

Right

[covert lake]

i got 1/6 for this

[quick yacht]

Its $12x=\frac{x}{3}$

[jolly parrot]

Cycadellic

[quick yacht]

So x=0

[covert lake]

oh yea

so now what do I do with the solutions

[quick yacht]

Then we have the regions
0<x<1/2 with 1/3 x < y < 12x and
1/2<x<3 with 1/3 x <y<3/x

And then it should be pretty much the same as the others from here

[covert lake]

yea but what equations would I substract

and is the interval from 0 to 3

[quick yacht]

We are doing pieces now

[covert lake]

so am I adding

[quick yacht]

So simply consider 0<x<1/2 and 1/3 x<y<12x

Write the integral

Do the same with 1/2<x<3 with 1/3 x<y<3/x

Then, add them together

[quick yacht]

Then, you have the area of the region

[covert lake]

is this one of them

[quick yacht]

Bounds are wrong

[quick yacht]

This is from 1/2 to 3

[covert lake]

so theres 4 total

right

[quick yacht]

4 total?

[covert lake]

4 total integrals

[quick yacht]

2

One for each region

[covert lake]

so this is one

[quick yacht]

Right

Then what is the other one

[covert lake]

this but idk the bounds

[quick yacht]

Bounds are wrong

[covert lake]

wait

[quick yacht]

0<x<1/2 with 1/3 x<y<12x

[covert lake]

0 to 3?

or 0 to 1/2

[quick yacht]

Yes

[quick yacht]

Because that is the x interval of this region

[covert lake]

this right

and then add them to together

[quick yacht]

Not 3/x

1/3 x

[covert lake]

right

i got this as the answer

but doesnt look right

[quick yacht]

For this?

Looks like you still did 3/x if so

[covert lake]

no when added together

[quick yacht]

Yeah, its a little off

[covert lake]

not sure what is wrong

that exponent looks wrong

[quick yacht]

probably a computational mistake

$$\big\int_{0}^{.5}(12x-\frac{x}{3})dx+\big\int_{.5}^{3}(\frac{3}{x}-\frac{x}{3})dx$$

[covert lake]

i used mathway for it

what should the answer be

[jolly parrot]

Cycadellic
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[quick yacht]

about 5.375

[quick yacht]

this is approx 5.04

[covert lake]

ln(216)

right

yep it was that

now doing washer problems

idk which washer it is

[quick yacht]

think of the x,y cross-section first

which graph has a cross-section of y=x+1 and y=0

[covert lake]

x=-1

[quick yacht]

no, which of these graphs has those cross-sections

[covert lake]

the 2nd one

so is that the answer?

wait how do I get cross section

@quick yacht

[quick yacht]

yeah

[quick yacht]

thinnk about where the solid intersects the x,y plane

that is, z=0

[pearl pond]

@covert lake Has your question been resolved?

[covert lake]

,w graph y=e^x

[jolly parrot]

Results provided by WolframAlpha

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

Requested by abindagoat

[covert lake]

@quick yacht sorry I keep tagging u but I need help with the last part of this question

how do I find what the sketch looks like

,w graph x=2squareroot7y

[jolly parrot]

Results provided by WolframAlpha

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

Requested by abindagoat

[quick yacht]

well, there is an issue

[quick yacht]

there is nothing on the last two graphs

and one of them has to be the answer

[covert lake]

oh wait

let me refresh

[quick yacht]

but you got the cross section right

just imagine the cross section revolving about y=0

[covert lake]

these are the last two

[quick yacht]

yeah, those are blank

[covert lake]

thats all it gives me

the shadow thing

let me back out and back in

[quick yacht]

you should email your teacher

it is impossible to know

[covert lake]

yea

[quick yacht]

maybe try the description

that may provide useful info

[covert lake]

i got it on my phone

Those are the graphs

@quick yacht

[quick yacht]

so, which graph has this cross section

[covert lake]

3rd one?

[quick yacht]

yes

[covert lake]

yep ty

for this its asking about the y-axis

[quick yacht]

i need to go to bed after this one

[covert lake]

alright np

[quick yacht]

so, how do we find volumes of solids of revolutions?

do you remember the formula?

[covert lake]

its pi times the function squared

but idk what the bounds are

0 to 5?

[quick yacht]

y goes to 5

[covert lake]

yea

[quick yacht]

so, examine the cross section first

[covert lake]

is this it?

[quick yacht]

i think so

[covert lake]

so I got this wrong

y would it not be the one I selected

oh its the first one

wait idek now

[quick yacht]

,w graph x=2 sqrt(7y)

[jolly parrot]

Results provided by WolframAlpha

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

Requested by Cycadellic#0025

[quick yacht]

oh i see, you picked the wrong one

[covert lake]

so is it the 1st one?

[quick yacht]

this

it looks like you picked the graph, but you actually picked the one to its right

excuse my terrible highlighting

[covert lake]

oh shit yea lmao

u going to sleep?

,w graph y=x^2, y=4x

[jolly parrot]

Results provided by WolframAlpha

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

Requested by abindagoat

[covert lake]

,w graph y=x^2, x=y^2, about y=1

[jolly parrot]

Results provided by WolframAlpha

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

Requested by abindagoat

[pearl pond]

@covert lake Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[civic cliff]

I need assistance on question 5

[civic cliff]

On how to find the domain part

[lavish portal]

what is the domain of log(x)

@civic cliff

[civic cliff]

(3, infinity)

(3, infinity)

And (-5, infinity)

[lavish portal]

what??

[civic cliff]

For domain of log3

[lavish portal]

no

what

what is $log_3(1/3)$

[civic cliff]

Are you asking for the standed domain of a log function?

[jolly parrot]

MoonCaik

[lavish portal]

the domain for log functions are all the same

[civic cliff]

you mean (0, infinity)

is what I'm assuming then

[lavish portal]

yes so you have to find when (x^2+2x-15) >0

that's the domain

[civic cliff]

what about the 3^2

x^2 + 2x - 24 = 3^2

and then you get 0 = (x + 6) and (x -4)

and x = -6 cannot happen and x = 4

can happen and you use that number plus the potential domain of (3, infinity) and make a weird plot line

I'm confused on how we solve it from there

?

/help

<@&286206848099549185>

[lavish portal]

why can't they happen

that's just when the graph is 0 right

oh well you forgot the >

[civic cliff]

Wait, but i thought log functions the argument can nto be anything less than zero

[lavish portal]

(x+6)(x-4)<0

[lavish portal]

yes?

[civic cliff]

that's why I'm assumign x = -6 cannot happen

[lavish portal]

it can't happen because it makes the inequality untrue

[civic cliff]

right

[lavish portal]

but you also have to find when (x^2+2x-15) >0 because that's the domain of the log function

[civic cliff]

what about the 2 at the end?

[lavish portal]

doesn't matter

we're talking about the domain of the log function not the inequality itself

so you're partially correct

[civic cliff]

Well for (x^2+2x-15) >0 I got x = -5, and x = 3

[lavish portal]

since the graph is pointing upwards then (-5,3) is invalid also

hopefully that makes sense

[civic cliff]

yea that makes sense

so how does that help us find that the solution is (3, 4)?

[lavish portal]

the other domain is 0>(x+6)(x-4) correct

you keep removing the inequality signs

[civic cliff]

so now what?

[lavish portal]

you find the valid solutions so... it is

[civic cliff]

the valid solution is 4?

[lavish portal]

(-6,4)

[civic cliff]

(-6, 4)

so is that the answer?

[lavish portal]

no

INVALID domain is (-5,3)

valid domain is (-6, 4)

[civic cliff]

(-6, 4) is also an invalid domain

[lavish portal]

no they are valid

[civic cliff]

so what then?

wait -6 is valid?

[lavish portal]

so subtract invalid from valid

no it isn't because it's (

[lavish portal]

this is finding when theparabola is LESS then 0

[civic cliff]

its parenthesis i guess, which means it's not -6

[lavish portal]

(-6, 4) is VALID
so subtract invalid from valid

the answer is [3,4) I think

[civic cliff]

waitwhen you say subtract invalid from valid

you saying like -6 - (-5)?

which woud be -1

[lavish portal]

no

[civic cliff]

How we doing the subtract?

[lavish portal]

(-6, 4) is VALID for 0>(x+6)(x-4)

correct

so these are teh only valid solutions for x

BUT

the original equation has extra domain restrictions

which are (-5,3)

[civic cliff]

right

[lavish portal]

so those are invalid x solutions

if we highlighted those solutions you can see they overlap

so some of the valid x values are actually invalid because of the logarithm domain

[civic cliff]

ok right

So explain how we subtracted and get (3,4)

What I'm lost in is that we did a whole other method, when on the worksheet my teachers uses a completely different method utilizing the number line

I'm just wanting to know how to use that numberline to get the solutions

[lavish portal]

I didn't ACTUALLY subtract like 1-5

[lavish portal]

that number line is a sign diagram

to help with finding the domain

[civic cliff]

I see

do you know why my teacher used it?

and how she used it to find (3,4)?

[lavish portal]

so she wanted to find when y was less than 0

to find the domain of the simplified equation

[civic cliff]

alright

[lavish portal]

on the number line, your teacher deduced that y was negative between (-6,4)

here is what I meant by subtraction

the blue region is the INVALID domain

the green is the VALID

so I subtracted the blue region from the green region to get the final domain

which is [3,4)

[civic cliff]

ok

makes snese

so how do we get there without using a calculator

[lavish portal]

I didn't really use a calculator

this is just what i imagine in my mind

[civic cliff]

alright

[lavish portal]

(-6,3) are invalid x values
(-5,4) are valid

so i imagine in my head the valid and invalid

idk lol

[civic cliff]

ok go on with the sign diagram

[lavish portal]

??

[civic cliff]

With how my teacher go the answer through the sign diagram

I understnad how she plotted 3, 4, 5, 6

[lavish portal]

what do you not understand

[civic cliff]

I'm sorry I'm an idiot, but I just do not understand how you did all that in your head

I just wanna know how my teacher got (3,4) as the solution

through the line diagram she used

[lavish portal]

the dotted line on 3 is the excluded domain from the logarithm

[civic cliff]

right

[lavish portal]

well actually

[civic cliff]

from the log3 (x-3)

[lavish portal]

your teacher is a little suspicious

and then the line on 4 is the INCLUDED domain

[civic cliff]

Cause it was from the equation

[lavish portal]

the - and + sign show when the graph is below or above 0 right?

[civic cliff]

x^2 + 2x - 24

yes

[lavish portal]

so the valid part is the -

[civic cliff]

So how do we know whats minus or +

or should I say where to we plug in these x's to determine if it is plus or minus

like say for explain 3.9

[lavish portal]

you plug in x on either side of x intercept

so if there is an x int at 3

then you plug in 2 and 4 because one of those values will be negative and the other is positive

i hope that makes sense

[civic cliff]

alright

[lavish portal]

the sign diagram is helpful but for me since I have more experience I can just imagine the graph

[civic cliff]

but why plug in 2 and 4?

[lavish portal]

since the degree of the quadratic is x^2

uhhh

[civic cliff]

don't we plug in one and see

[lavish portal]

i just chose numbers that were on either side of the x intercept

[civic cliff]

what about a number like 3.5?

its between 3 and 4

[lavish portal]

yes

it works

wait

no

you need to choose 2 numbers for a sign diagram

actually 1 number sorry

choose 1 number and determine whether the output is positive or not

then the other side of the x intercept is the opposite

[civic cliff]

well with 3.5 it would ve positive

[lavish portal]

u sure?

[civic cliff]

Would it?

[lavish portal]

sign diagram says it would be negative

[civic cliff]

how?

[lavish portal]

cuz the -

we can just calculate it

[civic cliff]

3.5 - 3 would be positive

[lavish portal]

noooooo

[civic cliff]

and 3.5 Plus 5 is positive

I thought we plugin those x?

[lavish portal]

the sign diagram is an indicator of the sign of the y value

the graph is (x+6)(x-4)

you plug in 3.5

(9.5)(-.5)

[civic cliff]

so 3.5 plus 6 and 3.5 -4

so it would be negative

[lavish portal]

yes

[civic cliff]

what about 2?

[lavish portal]

it's also negative

we know that the x intercepts at -6 and 4 right??

[civic cliff]

then how come the domain is (3,4)?

meaning its between 3 and 4 are solutions

[lavish portal]

because the logarithm domain

the other domain is (x^2+2x-15) >0

[civic cliff]

with (3, infinity( and (-5, infinity)

which would be x > -5

and x > 3

[lavish portal]

...

[lavish portal]

these are VALID solutions

for the log domain

you need to recognize that there are TWO domains where you need both to be true

[civic cliff]

So that's why 2 is not a solution

[lavish portal]

yes

[civic cliff]

those two domains being the (3 infinty) and the logs being less than 2?

would be the two domains that need to be true I'm hoping?

right?

[lavish portal]

idk what you are talking about

but there is the LOG domain and the INEQUALITY domain

[civic cliff]

that

[civic cliff]

that I meant

.close

[pearl pond]

Channel closed

Closed by @civic cliff

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

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

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[lean mason]

my question is: Find and write down in all three forms the equation of quadratic whose graph has symmetry axis x = -2, one of x - intercepts = —4 and y - intercept = 5

[lean mason]

like i basically dont get it

i tried to draw it out but i cant connect it

<@&286206848099549185> help

is it this

[pearl pond]

@lean mason Has your question been resolved?

[lean mason]

NO

[high quarry]

are you sure the question is correct 😭

[high quarry]

a quadratic has to be symmetrical which means the other root has to be at x=0 as you rightly put on this

a y-intercept of 5 wouldn't make sense considering the axis of symmetry

[pearl pond]

@lean mason Has your question been resolved?

[lean mason]

yes it was literally on my school test and im pretty sure its incorrect but idek 💀

[lean mason]

thats what i thought too right?

okay well then i guess its wrong

thanks for ur help anyways

[pearl pond]

Channel closed

Closed by @lean mason

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

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

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[opaque iris]

I think I got part a down, it’s pretty intuitive to me, but part b is confusing me

[opaque iris]

It’s uploading rn hang on

It’s AP Calculus btw, unit 4 stuff

Why is the double prime of W (the volume) equal to the prime of the rate that water is coming in, minus the rate that water is going out

[pearl pond]

@opaque iris Has your question been resolved?

[opaque iris]

<@&286206848099549185>

[opaque iris]

This my main question

[leaden wadi]

May I assume that you have begun learning about integrals?

[opaque iris]

Is that the tall squiggly line? I have not :p

[opaque iris]

I’m hella behind in my course

[leaden wadi]

Anti-derivatives?

[opaque iris]

No

Are those needed here?

[leaden wadi]

To properly understand the answer to your question, yes.

[opaque iris]

Can you give a basic explanation on what I would need to understand that the answer is the way that it is? Because I saw that anti derivatives are what I’m learning after this test that this problem is a study problem for

[leaden wadi]

An anti-derivative is just reversing the operation of finding a derivative.

eg.

f(x) = x^2
f'(x) = 2x

It would be going from 2x to x^2.

[opaque iris]

why would I do that here?

[opaque iris]

Makes sense

[leaden wadi]

There is a relation between the anti-derivative of a function and the area under the curve of that same function.

[opaque iris]

Does that have to do with concavity?

[leaden wadi]

No, give me a moment to make a graph that will make it more clear.

[opaque iris]

I stg my online course feels so out of order some times. This isn’t the first time I’ve not been able to understand a problem because the content was in a future lesson

[leaden wadi]

https://www.geogebra.org/classic/ghbkczsb

Pay particular attention to the y-value of B'. Note that it has the same value as the blue-shaded area.

[opaque iris]

Ok

[leaden wadi]

The grey line is the anti-derivative of the blue line, f(x).

[opaque iris]

Wait what?

[leaden wadi]

f(x) = x^2, g(x) = (x^3)/3

[opaque iris]

Oh

Yeah

[leaden wadi]

g'(x) = x^2 = f(x)

[opaque iris]

Makes sense

[leaden wadi]

In your problem, F(t) gives the rate of change at a given time, t.

[opaque iris]

so because there is a relationship between the anti derivative of a function’s point, and the area from that , does that mean I can conclude the answer in my problem somehow?

[leaden wadi]

We'll get to that.

If you were to sum up all of those little rate of changes on an interval from a to b, you could calculate the volume in the container.

[opaque iris]

I misclicked the friend request btw, idk how to take it back on mobile

[leaden wadi]

This is related to the average rate of change of a function that you should have learned about

[opaque iris]

I think I’m getting what you’re saying

[opaque iris]

This is if you are given a value t right

[leaden wadi]

Yes.

[opaque iris]

Ye ok

[leaden wadi]

So the function F(t) that you were given that shows the volume of water flowing into a small tub is the rate of change.

If you calculated the anti-derivative of F(t), you can calculate the volume of water in the tub.

The problem actually gives you the initial value that you would need to solve for the funciton W(t).

At t = 3, W(t) = 2.5. The reason for this is that with any anti-derivative, there is always some constant c tacked on to the end.

[opaque iris]

Yea

[leaden wadi]

This problem is arguably difficult if you don't know how to find the anti-derivative of arctan().

Kind of odd that you would have this type of problem this early.

[opaque iris]

The basic collegeboard explanation doesn’t really say anything about anti derivatives so maybe it was added by accident to the review for the test? Idk

[leaden wadi]

No, you should learn about anti-derivatives early on in Calculus. Calculating the anti-derivative of an inverse trig function usually comes in Calculus 2 though.

[opaque iris]

Hmm

Either way at my point in the course I haven’t even done normal anti derivs yet so yeah

[leaden wadi]

,w antiderivative arctan(pi/2 - t/10)

[jolly parrot]

Failed to get a response from Wolfram Alpha.
If the problem persists, please contact support.

Jan 29, 2024, 9:22 PM