help-10

[fresh trail]

Almost, you would have y = 0.16667x + b, then plug in (12,3) for (x,y) to get b. so 3 = 0.16667*12 + b, then plug in b

it should be an equation in terms of x, not just a number

[fresh trail]

Not the +3 at the end, it should be +b and you solve it for b

[velvet plover]

i also tried mathway and got b=0.99996

[fresh trail]

Desmos won't help with this part as far as I know (only at the end for graphing everything). The question is basically you have a slope and a point, find the equation of the line

But yes, b=0.99996. Plug that in to y = 0.16667x + b and that equation with x and y is the answer

[fresh trail]

Now it's x instead of 12. The line will happen to pass through the point (12,3) because we made it that way, but we don't need to include it anymore.

When you graph the line with the parabola, the line should just barely touch the parabola at one point

[velvet plover]

i dont got a parabola graphed

[fresh trail]

Oh, the parabola is from the first part of the question with y=sqrt(x-3). If you graph that on the same screen you should see what i mean

(a sideways parabola, not up and down)

And they say to also graph y=sqrt(x-3) at the end of part c now that I look at it again

[fresh trail]

Zoom in, the scale is really big

[fresh trail]

Yeah that seems about right, if you pan to the right a bit you should see more clearly that the line just barely scrapes along the parabola

that's what a tangent line is

[velvet plover]

so y=3 for part c?

this shows that my work is correct?>

[fresh trail]

Your work is correct, but y = 0.16667x + 0.99996, they asked for the equation of the line for part c. And yeah the graph I think is to verify your work

[velvet plover]

or do i put y=.16667(x)+.99996?

[fresh trail]

Yeah that

[velvet plover]

ok thanks man!

[fresh trail]

no problem!

[velvet plover]

it shows that part c is wrong

for y= y=.16667(x)+.99996

y=.16667(x)+.99996

[fresh trail]

Did you put the y= inside the answer box? They already have the y= outside it, inside the answer box should just be .16667(x)+.99996

[velvet plover]

Yeah no y= in the box

just .16667(x)+.99996

[fresh trail]

Oh ok, that's weird, I got .99996 for myself too

Could it be a formatting thing? Like maybe it wants 0.16667x + 0.99996

[velvet plover]

oh brah that was wrong too

i used up all my retry

the answer was .166665*x+1.0002

[fresh trail]

Shoot, sorry about that.

[velvet plover]

its all good

[fresh trail]

I guess there was some slightly wrong rounding that was happening?

[buoyant ingot]

aaaaaa90.

[velvet plover]

i got the other ones correct tho

ill talk to my teacher but thxs again for everything!

[fresh trail]

Yeah that might be one to ask the professor about, your procedure was correct, I think just a rounding thing

No problem!

[velvet plover]

have a good one!

https://tenor.com/view/peace-out-gif-22295199

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[velvet plover]

hi i need help with math 180 calc

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[polar fossil]

(the bot is riding the struggle bus today)

[velvet plover]

whats happening???

.clolse

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[timid silo]

Hi I want to ask if there is a faster way to do question 3 rather than listing all the satisfying values

[obtuse pebble]

@timid silo Has your question been resolved?

[slow maple]

I would recommend looking at it graphically. Specifically:

• When considering our four inequalities, what do the valid input pairs (x, y) look like as a shaded region on a graph?
• Where in the shaded region do we want to look to try and find the "maximum value"?

What may help with that second question is considering the following:

1. Pick some number z, and add the line/curve 2x + y = z (or whatever the function may be)
2. Try a larger or a smaller value of z. Observe how the line/curve placement changes.
3. Extrapolate to see where in the shaded region z could be the largest. Hopefully, you've found either a local maxima or an intersection of lines, and should be able to calculate from there.

[timid silo]

Calculus with inequalities?

I tried that for a question and it worked but this is getting more complicated

[slow maple]

Yeah, that's fair. It shouldn't end up needing any sort of calculus, if I'm not mistaken.

If you'd like, we can walk through a graphical approach to #3

[timid silo]

Ok

That looks a bit messy

[slow maple]

Perhaps, but it gets the point across. I'd like to double check, is it the shaded or the unshaded region that we're interested in?

[timid silo]

Unshaded

Just the small bit in the middle

[slow maple]

Alright, good

I've grabbed some graph paper to help with visual aids, allow me a moment to recreate the graph

Alright

Do you want to look at 2x + y or xy first?

[timid silo]

2x+y

[slow maple]

Alright, what's a value for z that's definitely going to be inside the graph?

[timid silo]

What

[slow maple]

Ah, sorry

We'd like to plot the line 2x + y = z for some z, so we can look at the behavior of the line as z increases or decreases. If we can figure out how the line moves, we can figure out where in the graph it'd have its largest value

[timid silo]

Uh

Ok

[slow maple]

So, for example, I believe the point (12, 6) is in our graph, so we could plot the line
2x + y = (2*12 + 6) = 30

[timid silo]

Yeah

[slow maple]

And we get something like this:

Oh whoops, that's sideways

[slow maple]

,rotate 90

[timid silo]

Yes but how does that help

[slow maple]

Well, if we increased z to say, 40, would we expect the line to move to the left or to the right?

[timid silo]

What

Oh to the right

[slow maple]

Yeah

Also feel free to ask questions if anything isn't completely clear, the goal after all is to learn

[timid silo]

Is the y axis not up to 30?

[slow maple]

The line 3x+2y <= 60 does eventually hit the point (0, 30), yes. I decided it would be clearer to cut that out, since it'd be outside of the region described by x+2y <= 30, (which only goes up to (0, 15)), and by x >= 10. My apologies if that wasn't clear

[slow maple]

Then, since we want the maximum value of 2x + y, and since 2x+y=z moves to the right as z increases, then all we're required to do is find the value z that puts the line as far to the right as possible, while still being within the region.

[timid silo]

Why and how does that work

How do you know that’s the maximum

[slow maple]

We're technically applying the concept of a gradient. In this case, it specifically works because of the following:

• The region we're looking at is closed and continuous (there's no holes, and we're looking at <=, not <)
• The formula (2x+y) is continuous and differentiable at all points within the region.
• The formula doesn't contain any local maxima within the region, except on the edge (we can show that this is true because the partial derivative of x is always the same sign, and so is y's)

If any of these were not true, we could still perform a similar test, but we would need to be way more thorough and careful

[timid silo]

?

[slow maple]

Could you clarify what your question(s) is/are?

[timid silo]

This

[slow maple]

Ah

When we consider the function f(x,y) = 2x + y, we're looking at a plane.

When we set f(x,y) equal to a specific value z, we've defined a line/curve which has the specific property that for every point on that line, we have the same output (by definition).

If we consider the two directions we can go off of the curve (in our case, the 'left' and 'right' sides), the line must either increase or decrease (if it stayed constant, it would just be a wider part of the curve). And, because our equation is continuous and differentiable, we can know that if one point says that going to the 'right' means the value increases, it must also increase if we headed to the right from any of those points (we can demonstrate why if you'd like).

Thus, we know that if we move the line to the right, then the output must increase, because we can show it increases for at least one point.

We would then repeat this until we've found ourselves moving as high up as possible, without falling outside the region we're confined to.

(Since the gradient is constant for a plane, we don't need to re-check that we should move to the right after moving the line, but with more complicated equations, we may want to.)

[timid silo]

Oh I see

Now how are we going to find the ‘high as possible’ point?

[slow maple]

That would be that step of trying to move the line as far right as possible. Since we've decided the function has no local maxima (as it's a plane), we only need to try and move it to the right

And since all we're doing is moving it to the right, and not considering other factors, we should focus on testing points along the border

[obtuse pebble]

@timid silo Has your question been resolved?

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[timid silo]

Do I just multiply the pi to the r's, r squared?

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@timid silo Has your question been resolved?

[supple pumice]

yes

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[reef tapir]

im just confused on how do i do the 3(2x/3) part especially

if thats what im supposed to do

[timid silo]

multiply

if u divide your cookies by three and multiply them by three again

how many cookies do you have

[reef tapir]

same number?

[timid silo]

yep

[reef tapir]

oh so its 2x

?

[timid silo]

yup

dont forget the -7

[reef tapir]

yes

i was just confused on that part

[timid silo]

Yup all good

[reef tapir]

bro godbless thank you

.close.

[timid silo]

lmao no problen

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[reef tapir]

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[azure hatch]

I’m using graphing calculator to try and solve for x by finding point of interception but it is completely different than the answers in the textbook

[royal basin]

hold up

part c says 3cos(x^2) = 2 cos(x) + 1?

what a nasty equation that is.

can you show whata you graphed vs what your textbook says

[azure hatch]

They just want you to put each side = to y and plot on the same graph

[azure hatch]

One sec

[royal basin]

yeah, expected about as much. what's the textbook say?

[azure hatch]

I got x= 0.8 and 2.16
But the answers say 1.91 and 4.37

[royal basin]

try graphing 3 cos^2(x) instead

then skip this question as typoed

[azure hatch]

Sounds good thanks

I have one more question really quick

I got 1.05 and 5.24 but the answers say it is +or- these two solutions… how can that be possible when the domain is greater than zero

<@&286206848099549185>

[raven shadow]

hmmm doesnt make sense, I think x = 1.05 and 5.24 seems like the right answers

[azure hatch]

Alright thanks

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[timid silo]

how did you get 5.24?

[timid silo]

i got x = 1.05 and x = 2.09

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[azure hatch]

@timid silo I just plotted both sides separately and found intercept

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[reef tapir]

i have undertand it is in the 1st quadrant but im confused on what to do after that

[reef tapir]

idk if im supposed to do pi/2 - 11pi/6

[obtuse pebble]

@reef tapir Has your question been resolved?

[reef tapir]

<@&286206848099549185>

[oak shore]

what do you mean by 'reference number'?

[quasi glen]

yeah the question isnt clear

[oak shore]

is it asking for the principle value for some trig function?

[reef tapir]

this is an example problem from the vdieo im supposed to watch

[oak shore]

right

[quasi glen]

it doesnt make it any better

[oak shore]

the principle value

[reef tapir]

im confused

[oak shore]

try pi/6

basically, you want the value modulo pi

so -11pi/6=-2pi+pi/6

so that's just pi/6

[raw egret]

because it's traveling in the same direction clockwise as pi/6 travels counterclockwise.

[oak shore]

(if I am understanding the problem correctly really)

[reef tapir]

okay let me try to understand it for a sec

[oak shore]

here's a slightly bigger breakdown.

Negative angles mean gloing clockwise. Travel 11pi/6 clockwise. Where do you end up? Well, if you travel pi/6 more, you end up at 2pi. Which is back where you started. Thus you initially ended up at 2pi-11pi/6=pi/6

that's your required principle value

[reef tapir]

i thought negatives go counterclockwise

[oak shore]

positives are counterclockwise

see the arrow you drew in your diagram for 7pi/6 (which is positive)

[oak shore]

you drew a counterclockwise arrow

[oak shore]

here

[reef tapir]

omg

your right i got mixed up

[oak shore]

makes sense now?

[reef tapir]

with counter clock wise

i understand it with the picture

[oak shore]

did you get this?

[reef tapir]

i understand that if you travel 11pi/6 clockewise you end up at pi/6

i dont understand what your saying here "Well, if you travel pi/6 more, you end up at 2pi."

no i get it now

[raw egret]

11pi/6 + pi/6 = 2pi

[reef tapir]

yes

i get that now that i think about it

[raw egret]

think of it as a directional change $\fracleft(-11\pi 6\right)$

[warm shale]

Tank_Driver011
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[royal basin]

$-\frac{11\pi}{6}$

[warm shale]

Ann

[raw egret]

yeah, yeah

[royal basin]

or $\left( -\frac{11\pi}{6} \right)$ if you want those parentheses

[warm shale]

Ann

[reef tapir]

so in the video the teacher does 7pi/6 - pi = pi/6 how does that look for the equation im doing

im referring to the 2nd pic i sent of the example question

[oak shore]

so you need to bring the 7pi/6 to the first quadrant. Think of pi also as 0 (as I wrote, we do everything modulo pi)

then 7pi/6=pi+pi/6 and thus your answer is pi/6

[reef tapir]

yeah i get that but im asking how would that look for -11pi/6

[oak shore]

-11pi/6=-2pi+pi/6=pi/6 mod pi

essentially, write your angle as (some integer) times pi plus (some angle in the first quadrant) [if possible]

then just take the (some angle in the first quadrant) as your answer

[reef tapir]

oh okay i get it and the question also asks for a terminal point would that be (sqrt(3)/2,1/2)?

[oak shore]

terminal point?

I am sorry i don't know these terminologies

what's the defn?

[oak shore]

yeah then this is correct

[reef tapir]

okay cool

thats it thank you very much you are great person 😇👍

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[surreal inlet]

How do you solve the equation for x in the last step with a regular calculator? How do you get to 1/8 and 1/2

So just to be clear on a non graphing calculator

[warm shale]

nebula40

[tender stratus]

using the definition of a logarithm

[surreal inlet]

Ohhhhhh

.close

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[proven zephyr]

I showed you how like 10 minutes ago

[surreal inlet]

I did it the wrong way around

My bad lol

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[supple mulch]

how to find the point of intersection between y = 2x+9 and 3x-2y+19=0

[proven zephyr]

Subsitution or elimination is what i’ll do

[supple mulch]

how to do that?

[proven zephyr]

solve it like a system of 2 equation

[supple mulch]

ok i will try

i got x=1

[supple mulch]

.

and then solve for y and then that is the points that they intersect at?

i got 1(,11)

[proven zephyr]

Yeah

Should be correct

Apart from how you wrote the coordinate, everything seems correct

[supple mulch]

thanks

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[wild vigil]

i need to do sin cos tan to solve this right`?

which one of the three for the first one?

[obtuse pebble]

@wild vigil Has your question been resolved?

[amber merlin]

Cos, sin and tan are relative to the position of an angle.

if the angle is close to the line, its adj

if distant, opposite

,align &\cos = \frac{Adj}{Hyp} \quad &\sin = \frac{Opp}{Hyp}

[warm shale]

Thayuno

[amber merlin]

,align \tan = \frac{Opp}{Adj}

[warm shale]

Thayuno

[amber merlin]

so, for the first, you have the r and phi, r is opposite of the angle. t is adjacent

you can use tan to find the value of t

and sin to the hypotenuse

[obtuse pebble]

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[supple mulch]

What formula can I calculate AODM with? AODM is a square

[supple mulch]

??/

[ruby path]

Sure doesn't look like a square

If it is then you can just do a*a where a is the length of the side

[supple mulch]

yeah I know

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[charred shuttle]

Hey hello people

Could someone give me a hand with that calculus?

I'm stuck because I don't really know how I could simply or just separate that ln(x) and 1/x

[spice citrus]

Have you learnt about substitutions?

[charred shuttle]

Yes I did

Though I don't know what to sub

[spice citrus]

What function to sub for comes to mind first?

[charred shuttle]

I would have said ln(x) but It seems to difficult to derivate afterwards

[spice citrus]

What's the derivative of ln(x)?

[charred shuttle]

Oh my Lord

Is it 1/x?

[spice citrus]

Yes

[charred shuttle]

Wow

Dumb of me haha

Thanks for opening my eyes actually It's not as hard as I thought

[spice citrus]

Don't worry

[charred shuttle]

Have a great day thank you!

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[long monolith]

Hello, I was wondering if somebody could help with a method to answer this?

[obtuse pebble]

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[long monolith]

Q. above

[obtuse pebble]

@long monolith Has your question been resolved?

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[pulsar quarry]

if $a_B=\frac{a_A}{2}$, then can we say that $v_B=\frac{v_A}{2}$?

[warm shale]

yamdoot

[pulsar quarry]

a is acceleration and v is velocity

[tender stratus]

nope, this is not true in general.
let's say two cars start from same initial velocity 10m/s at a1 = 1 m/s^2 and a2 = 2 m/s^2.
their velocity at any second after doesnt satisfy this. I guess this works if they both start from rest

[pulsar quarry]

but

if you differenctiate the velocity eq

you get the acc one

[warm shale]

Dyssrupt

[pulsar quarry]

i mean it was pulley system in this case

so they're both initially at rest

[frank monolith]

yeah in that case, it is true

in pulley's case, this is true even for displacement iirc

[fossil crag]

In summary : if $a_B = K×a_A$, then $v_B = K×v_A + constant$

[warm shale]

rafilou2003

[fossil crag]

This literally what happens with integration

If you know furthermore that at some $t_0$, $v_B(t_0) = K×v_A(t_0)$, then that constant is null

[warm shale]

rafilou2003

[pulsar quarry]

yeah

[frank monolith]

ahh NV sir

[pulsar quarry]

wait a sec

let me send my sol

[pulsar quarry]

isnt he good?

first time watching him

hes good tho

[frank monolith]

i havent, but ik he is

[pulsar quarry]

this is what i did?

nvm

its not wrong

until it starts from rest

thNKS

[frank monolith]

it is correct

[pulsar quarry]

thanks

.close

[obtuse pebble]

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[sudden siren]

i am stuck in this logarithm question,
i have attached the question and my approach, but actual ans is 5

[timid silo]

you just need value of ln2
ig

[tender stratus]

no

[timid silo]

then ?

[tender stratus]

everything is right till the last step

when you removed t-5 as a factor

$(t-5)A = (t-5)B \implies (t-5)(A-B) = 0 \implies t = 5 or A = B$

[warm shale]

nebula40

[tender stratus]

basically you've lost a solution when you "cancelled" that term

[timid silo]

yeah thats correct t=5 is a solution

[tender stratus]

solving that last equation will also give you another solution, but I think it's negative(and probably they dont want negative time)

[sudden siren]

ohh got it thanks a lot bro

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[charred shuttle]

Hey it's me again

[teal turret]

Hi me again

[charred shuttle]

If you apply a negative exponent to x
Does it bring only x to the numerator? Or does it also bring the +1?

Hi Stephen

[rain forge]

Why do you mean by that? You want to solve this integral?

[teal turret]

Also the +1

It becomes

[warm shale]

Stephen

[charred shuttle]

Oh I see right

Thank you

[charred shuttle]

Yeah it was to try and solve it

But it doesn't seem to be the solution

Oh no I got it!

[rain forge]

With ln and coeff 3

[charred shuttle]

Ln?

I was thinking about getting the constant 3 out of the integral first

[rain forge]

Yes right

[charred shuttle]

Oh no wait still doesn't work my bad

[rain forge]

And what is the integration of $1/x+1$

[warm shale]

phoestaclies

[charred shuttle]

I wanted to isolate the +1 into it's own integral afterwards

Doesn't work

It's ln (x+1)

?

[rain forge]

Yes

[charred shuttle]

Ooooh right

[rain forge]

So you have all

[charred shuttle]

Hey thank you very much my friend

That really helped me

Have a great day you and Stephen

.close

[obtuse pebble]

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[deft sun]

(over R)

find if W is a linear space (vector space) with + and *

[deft sun]

how do I start?

W is not empty

becuase it has (0,0)
(0,0)

[warm canopy]

Is it closed under addition?

[deft sun]

how do I check that?

I forgot how to add 2 matrixes

😦

[warm canopy]

Take two things in W, and see if their sum of also in W

You don't need to explicitly add them

[deft sun]

what can I understand from A = A^2?

[warm canopy]

Take two matrices A and B that are in W

Check whether or not A+B is in W

[fossil crag]

If A=A² and B=B², does (A+B)² = A+B ?

[deft sun]

I know this one is in W

but I cant find a second one

which is also in W

[fossil crag]

How about Identity ?

[fossil crag]

But right now I'm asking you this

[warm canopy]

Note that in what we are asking you to do, you don't need to explicitly choose/find the matrices A and B, they are just generic matrices that are in W, we don't care what they actually are

[deft sun]

doesnt seem right

[fossil crag]

Yes, but maybe try and expand to see why

[deft sun]

A^2 + 2AB + B^2 != A + B

[fossil crag]

Nuh uh, you fell into the naive binomial expansion

We're dealing with matrices

What bothers me is the 2AB, can you guess why?

[deft sun]

not really

I got a bit confused

[fossil crag]

The matrix product is not commutative

[deft sun]

ohhh right

[fossil crag]

(A+B)(A+B) = A² + AB + BA + B²

[deft sun]

so this is a horrible horrible mistake

ye

[fossil crag]

since A² = A and B² = B

[deft sun]

so AB = -BA?

[fossil crag]

This is equal to (A+B) + AB + BA

[fossil crag]

That is a requirement yes

But is it always true for two matrices in W ?

[deft sun]

not always?

[fossil crag]

Check what happens when A = B

[deft sun]

AB = BA

= A

= B

[fossil crag]

But worse yet

[fossil crag]

From this

A² = - A²

[deft sun]

which means it is 0

?

[fossil crag]

Yes, and this is equal to A

So A = 0

So all we need to do is show that there are non-zero matrices in W

And then this will be false

[deft sun]

!

[fossil crag]

Yep

I² = I

And yet (I+I)² = 4I

Which is not I+I

[deft sun]

nice

thanks for the help! (again)

.close

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[real glade]

is it correct for me to say k>2, k>5 for part (a)?

[valid cradle]

Not quite.

The question asks for how many possible integers k can you put.

So they are looking for a numerical, integer value

And, since you have bounds for what k can be, [-10,10], you can definitely put an integer value.

[real glade]

yes, is it right for me to say can be 2,3,4.....,10?

[timid silo]

they ask how many

[real glade]

hence the number of possible k is 9?

[timid silo]

yes

[valid cradle]

Yep

[real glade]

okiee

[wet moss]

Are there not negative integers to consider?

Also 2,3,4,5 doesn’t satisfy this

[real glade]

oh yup

i did it wrongly

[wet moss]

Try sketching a graph

And then seeing the points where it is above the x axis

[real glade]

yup i have alr done it

thx alot guys

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[wet moss]

This also isn’t correct

[obtuse pebble]

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[spice inlet]

From the top of a cliff, the angle of depression of the top and bottom of a tower are observed to be 45 and 60 respectively. If the height of the tower is 20 m, find : the height of the cliff and the distance between the cliff and the tower

[obtuse pebble]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

@spice inlet Has your question been resolved?

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[queen crow]

me and my friend are literally wracking our brains because we heard two different answers for 4 - ½, is it supposed to be left as is or can someone here explain why it somehow became 4 - ½ = 3 ½

[queen crow]

where n=2

[high lily]

if you had 4 apples and ate half an apple, how many apples would you have left

[queen crow]

oh my god you literally subtract 1/2

i cant believe i spent an hour of time wondering about this

cuz i thought like 3.5 was wrong when it's the same thing as 3 ½

thanks bro

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[forest glade]

how do i solve this??

[runic void]

what have you tried?

[obtuse pebble]

@forest glade Has your question been resolved?

[pulsar quarry]

@forest glade

[obtuse pebble]

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[rose lintel]

need help with
a question I have asked about this morning but couldn't be solved
anyone can help?

[rose lintel]

I will brb
I'm going to take a shower
pls don't claim it

ugh

[versed stratus]

I think expressing h in terms of d may help

Not too sure though

[obtuse pebble]

@rose lintel Has your question been resolved?

[rose lintel]

I had like 23 tries solving that

the correct asnwer should be 202

but couldn't reach that

<@&286206848099549185>

[rose lintel]

if h(X) is equal to cuberoot ( f(x) - x^2 +5 ), and the change of rate of h(X) in the period [1,3] is equal to 4, and ( h(X) + f(x) )^2 = 11, find the change of rate for f(2x+1) in the period [0,1]

[obtuse pebble]

@rose lintel Has your question been resolved?

[rose lintel]

okey gn

[tepid mountain]

Actually I don't think we need the first equation

Let's see the third one
We have
$$h(x) + f(x) = \pm\sqrt{11}$$
Let's get to see the positive first then we would come to the negative
So
$$h(x) + f(x) = \sqrt{11}$$
So
$$h(x) = \sqrt{11} - f(x)$$
Applying this to second equation we get
$$\frac{\sqrt{11}-f(3) - (\sqrt{11} - f(1))}{2} = 4$$
$$f(1) - f(3) = 8$$
So this gives us that
$$f(3) - f(1) = -8$$

But we still have the part of the negative square root
But surprisingly it will give us the same answer which is -8
And you can try it

So what is the purpose of the first equation
Honestly I didn't know but who cares we got an answer:)

[warm shale]

Sherif Player

[tepid mountain]

@rose lintel hmm I think that would help

@rose lintel ?

[obtuse pebble]

@rose lintel Has your question been resolved?

[obtuse pebble]

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[fiery shell]

I've been taught that as long as f(x) is greater than g(x) for all x in [a,b], the area between the curves over that interval is int a->b (f(x) - g(x)) dx

[fiery shell]

however, verifying that f(x) is greater than g(x) sounds like a bit of a pain

would a more general formula be int a->b |f(x) - g(x)| dx?

[grizzled shore]

That doesn’t work

[fiery shell]

no?

[grizzled shore]

It’s because integrals give signed areas

[polar fossil]

um yeah but with the absval it would work

but you basically just rewrote the problem into a more compact but equally useless form

[hexed gull]

it is if you consider "area" the unsigned area

[polar fossil]

since now you have to figure out when |f(x) - g(x)| is equal to f(x) - g(x) and when it's equal to g(x) - f(x)

[fiery shell]

yes cause area can't be negative

[fiery shell]

right I see your point

[hexed gull]

debatable

:3

[fiery shell]

ok then yes, I'm referring to the unsigned area

[grizzled shore]

If you don’t care you can just integrate it normally

[fiery shell]

don't care about what

[grizzled shore]

This is true I just woke up ;-;

[hexed gull]

You usually don't check whether f(x) is greater than g(x) generally, but you can look at the interval, and first check if the first value makes it greater

let's say the interval is [1,2]

first of all f(1) needs to be greater than g(1)

[fiery shell]

and then check the points where they intersect?

[hexed gull]

and now you only need to check that the function f(x)-g(x) is continuous in the interval and isn't equal to zero in the interval

[hexed gull]

or like that

although no

still continuity

but then yes

[fiery shell]

oh right yeah

[grizzled shore]

(We’re also assuming that f and g are continuous on [a,b] which need not be the case to be integrable)

[fiery shell]

I mean, wouldn't it still hold?

if either was discontinuous

[grizzled shore]

It wouldn’t need to intersect to go under the other function

[fiery shell]

although maybe area is less defined if there's a gap in your border

[fiery shell]

right, yes

[hexed gull]

imagine if one function jumps above the upper

[grizzled shore]

Nah thats fine

As long as it’s not too discontinuous, it’s fine

[hexed gull]

you can split it into regions for difficult cases

or use approximations, depending on how you define area

[fiery shell]

alright, I think I understand everything now

thanks everyone!

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[inland thicket]

guys I got this interval and I got all thing here like std.error is unknown

and i try to solve it by making like this set of equations

but it make no sense even tho it theoreticaly should

does anyone know how to get std.error?

[grizzled shore]

Perhaps you should show us the problem itself

[inland thicket]

hmm ok

Using a simple random sample of size n, the following confidence interval was determined
estimates of the proportion of four-member households in the basic set that spend more than 40%
annual income on clothing, footwear and cosmetics:

P(0.36 < p < 0.40) = 0.95. Faction of choice is less than 0.05.

b) What is the standard error of estimating the population proportion?

The normal distribution form is:

[grizzled shore]

Why did you go 0.4 - 1.96SE

[grizzled shore]

In your 2nd line of working

[inland thicket]

basically

0.4 is proportion estimation

the p with cap

[grizzled shore]

Is it?

[inland thicket]

and z _alpha/2 value is 1.96

[grizzled shore]

Oh wait it is

[inland thicket]

yes in task it says

[grizzled shore]

No

It’s not

[inland thicket]

🤔

[grizzled shore]

That 40% has nothing to do with the question

That 40% is talking about how much money the families spend

[inland thicket]

yes but that is p cap value from interval right?

[grizzled shore]

We want the percentage of those families

[inland thicket]

proportion estimated

hmmmmm

OH

w8

i think this whole task is some mistake

because they didn't give sample size or anything

so its imposible

or i have to guess the interval

[grizzled shore]

Even if they were to spent all their money that’s still fine

Well the idea is

Proportion testing is symmetrical

Everything you do here is symmetrical

[inland thicket]

ye

[grizzled shore]

So the point estimate must’ve been the middle of the 2 boundary values

[grizzled shore]

So you must’ve estimated phat = 0.38

[inland thicket]

hmmmm

oh

[grizzled shore]

Does that make sense?

[inland thicket]

but im thinking that the 0.38 value is just result of

0.38 is result of this right

this was my idea

[grizzled shore]

No

0.36 is the result on the left

0.40 is the result on the right

[inland thicket]

isnt this left? 😬

oh

ye 0.36

woah

0.38

OH

OH

!

!

!

[grizzled shore]

Yeah?

[inland thicket]

I just realised what u said

u statistic jesus?

[grizzled shore]

No but I wish I was

;-;

[inland thicket]

⚔️ HMmm

you 100% sure of this or?

like idk

if it was that easy

[grizzled shore]

I’m pretty sure

[inland thicket]

ye but

if it was that easy

hmm

idk

[grizzled shore]

It’s not that easy you still have to divide it through

[inland thicket]

ye

🤔

there is no way for me to know truth of this task

cos its just task no solution

test on 30th of augst

going insane

https://tenor.com/view/dday-omaha-beach-wolverine-origins-saving-private-ryan-sabertooth-gif-21979985

im learning rn day and night with energy drink

🇮🇱⚔️🇮🇱 Spiritual warfare ✅

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[timid silo]

A prime number is a number that is evenly divisible only by 1 and
itself. The prime numbers less than 100 are listed below.
2 3 5 7 11 13 17
19 23 29 31 37 41 43
47 53 59 61 67 73 79
83 89 97
Choose one of these numbers at random. Find the probability that
(a) The number is odd
(b) The sum of the digits is odd
(c) The number is greater than 70

[timid silo]

I solved part a

but i can not understand part b

can someone explain it to me.

[fossil crag]

The sum of the digits of 83, for example, is 8+3 = 11

Here it is odd

[timid silo]

2 3 5 7 11 13 17
19 23 29 31 37 41 43
47 53 59 61 67 73 79
83 89 97

do i need to sum all of these?

[fossil crag]

No, the sum of digits of each individual prime number

So 2, 3, 5, 7, 1+1, 1+3, 1+7, etc...

[timid silo]

okay

so how will i divide all of these numbers by 24 which is total number of possible outcomes.

[fossil crag]

No this is not what you want to do

[fossil crag]

You want to check if every number on this list

Is odd or not

[timid silo]

okay

let me solve it

then tell me if i did it correctly or not

@fossil crag

Is it correct?

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

<@&286206848099549185>

[latent sluice]

no

[timid silo]

A prime number is a number that is evenly divisible only by 1 and
itself. The prime numbers less than 100 are listed below.
2 3 5 7 11 13 17
19 23 29 31 37 41 43
47 53 59 61 67 73 79
83 89 97
Choose one of these numbers at random. Find the probability that
(a) The number is odd
(b) The sum of the digits is odd
(c) The number is greater than 70

this is the question

wait a second

[latent sluice]

you did a correctly

but not b

[timid silo]

Do i need to add values of repitive?

[latent sluice]

yes

[timid silo]

repititive ones?

airbases thank you sir

kind sir

[latent sluice]

and don't forget about 2 3 5 and 7

[timid silo]

why do i need to add those

[latent sluice]

because they're part of the set of prime numbers that are less than 100

[timid silo]

oh okay

@latent sluice

do i need to add 2? cuz it specifically asked for odd numbers

sum of odd

[latent sluice]

you have to consider 2

[timid silo]

it asked for sum of digits that become odd

2 is a single digit

[latent sluice]

0+2

[timid silo]

which disqualify this value

oh okay

[latent sluice]

then 0+3, 0+5, 0+7, 1+1, 1+3 and so on

[timid silo]

it's 12

[latent sluice]

lol

[timid silo]

let me send picture

wait

Thank you

.closed

[obtuse pebble]

@timid silo Has your question been resolved?

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[golden cosmos]

I need some help with latitude and longitude questions?

[timid silo]

show us

[golden cosmos]

Okay, let me grab it!

<@&286206848099549185> Am I perhaps in the wrong server for help?

[bold carbon]

bro what class is this

[golden cosmos]

Well, it is apart of my sciences but I thought because it had to do with calculations that this was the right place. I can leave if this is incorrect, apologies for any confusion.

[gaunt walrus]

I have no idea about ladtitiude and longtitude

[bold carbon]

i think u are supposed to use law of cosines

and maybe vectors

actually nvm

but

wouldnt a be 60 degrees

since u go from 15 N to 45 S

[golden cosmos]

I have no clue, but I appreciate it anyway.

.close

[obtuse pebble]

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[stoic elm]

whats the fundamental principle of counting?

how do i determine how many outfits she has?

[hollow dove]

if you have X ways of doing one thing, and Y ways of doing another, then there are X * Y ways to do both things

[stoic elm]

oookay

[stoic elm]

i still dont get how to actually do the thing tho

[hollow dove]

if I have 3 skirts and 4 shirts, how many outfits do I have?

[stoic elm]

..12?

[hollow dove]

yes

same thing here, we just have more options

[stoic elm]

so she has six blouses 4 skirts 20 pair of socks and 4 shoes

4*20**4?

[hollow dove]

you're missing the blouses

but that's the idea, yes

[stoic elm]

oh yeah

wow

that easy

incredible

im so confused

am i finding the combinations for those 3 balls

or the balls left in the bag

the internet is not giving me a straight answer

[hollow dove]

the 3 balls you pull out

otherwise it doesn't make sense to ask about a "three-ball combination" for 27 remaining balls left in the bag

[obtuse pebble]

@stoic elm Has your question been resolved?

[stoic elm]

so thats like

3*3

every other answer online is soemthing diff

[hollow dove]

not quite

just consider pulling one ball out of the bag

how many outcomes can you have for this

given that there are 30 different balls in the bag

[stoic elm]

wait like

3 ball combos

you only have 1 ball

how

am i getting more balls from the bag?

[hollow dove]

we're considering a simpler problem right now

if there are 30 different balls in the bag, and we pull one ball out, how many possible outcomes are there

[obtuse pebble]

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[gray ember]

Hello I need help on how to get to this answer

[gray ember]

its completing the square

[tardy epoch]

,tex .cts

[warm shale]

riemann

[gray ember]

sorry man im dumb

i know how to complete the square without the number at the front of the variable

but when there is a number, is when idk

for ex. 10v^2 is what i dont understand but without the 10 ex. v^2 I can easily do it

[tardy epoch]

you factor 10 out first

do you know how to factor 10 out of 10v^2 - 20v

[gray ember]

give me a min

[tardy epoch]

you factor out 10 because whatever remains, you can do completing the square

[gray ember]

10(v^2-2v)?

[tardy epoch]

yes

[tardy epoch]

do this on the stuff inside your parentheses

v^2 - 2v

[gray ember]

oh ok

now i get it thank u

[obtuse pebble]

@gray ember Has your question been resolved?

[obtuse pebble]

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[primal perch]

The question asks to show is a rational expression, and the first part of the solution confuses me

does it show that there are rational solutions byt he rational zeroes theorem b/c there are coefficients with the same factor (2sqrt3)?

[restive jungle]

||can you just subsitute and evaluate?||

[polar fossil]

i don't see the word rational anywhere in there

[primal perch]

here's the problem statment

[primal perch]

I think that once I get x^2+2xsqrt3-1-2sqrt3=0, I'm supposed to use rational zeroes thm. to prove

does this prove it ?

how do i show that -(1+2sqrt3) isn't a solution

[obtuse pebble]

@primal perch Has your question been resolved?

[obtuse pebble]

@primal perch Has your question been resolved?

[vast willow]

Hello there

[vast willow]

You can't show that this is not a root because you already showed that it is

Did you show that $\sqrt{4+2\sqrt{3}}-\sqrt{3}$ satisfy the given quadratic equation.

[warm shale]

calculus is fun

[primal perch]

yea i did

im' just not sure how that quadratic proves that the solution is rational

it doesn't seem like i can use rational zero's thm bc there are irrational coefficients

[vast willow]

Is $\sqrt{4+2\sqrt{3}}-\sqrt{3}$ positive or negative

[warm shale]

calculus is fun

[primal perch]

positive

[vast willow]

Ok which of the roots that you got is positive

[primal perch]

1 is positive

makes sense

[vast willow]

The other is negative

And the number of roots of an nth degree polynomial is n (some maybe complex) here we have a quadratic Deg 2 and 2 real roots so no more roots

This implies that $\sqrt{4+2\sqrt{3}}-\sqrt{3}$ is one of the roots

[warm shale]

calculus is fun

[vast willow]

The only positive root is 1 and this is positive to it is equal so 1 which is rational

[primal perch]

ah ok we haven't proved this yet but this makes sense

thank you sm

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[obtuse pebble]

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[vast willow]

Np have a nice day

[obtuse pebble]

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[dapper moss]

Hi guys I'm wondering how to approach this vectors. question using dot product. I drk where to start. Any assistance will be appreciated!

[obtuse pebble]

@dapper moss Has your question been resolved?

[compact shadow]

w=u-v

(u,u)=2(u-v,u-v)

(v,v)=3(u-v,u-v)

Expand these and second equation minus first equation, see what it gives you

[kind hawk]

!nosols

[obtuse pebble]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[kind hawk]

you have given enough

[compact shadow]

Okay

[obtuse pebble]

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[dapper moss]

Thank u sm! We're using the vector squared equals modulus squared property right?

[elder shard]

Cah\

[kind hawk]

(x,x)=|x|^2, yes

although you shouldn't really call the LHS "vector squared"

dot product of the vector with itself

[compact shadow]

Yes you will discover a neat relation between (u,u) and (u,v). I can’t reveal until you calculate it yourself

[obtuse pebble]

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[raw hinge]

Hi, $log_{\frac{1}{3}}(x^2-4x+3) < -1$ how to solve?

[warm shale]

Shachar

[sage geode]

Try exponentiating both sides with base 1/3, don't forget that the sign of inequality ought to be flipped as you do that since 0 < 1/3 < 1

And x^2 - 4x + 3 needs to be positive as well

[raw hinge]

oh so when the base is fraction the sign always flip?

[sage geode]

When the base is between 0 and 1*

[raw hinge]

btw is there any time this wont need to be bigger than 0?

[sage geode]

Logarithms with negative base usually aren't dealt with

[raw hinge]

so lets say I had $log_{-2}(x)$, it would be $x < 0$?

[warm shale]

Shachar

[raw hinge]

or it will be non linear

[candid yarrow]

Think about the definition of log

[raw hinge]

ye i guess x would be - + - + - +...

[sage geode]

The base of a logarithm is defined to be positive (and distinct from 1)

[raw hinge]

so there is no use for logarithm with bases less than 0 right?

[sage geode]

Yeah

[raw hinge]

ok so last question, what are all the checks I need to do to find the final answer?

[candid yarrow]

Sub it back in

To see if it’s extraneous

and then you should have your answer

[raw hinge]

but with inequalities my answer might be more complicated than that

[sage geode]

As long as you do the steps correctly you should be fine

Just don't forget to flip the sign and say that x^2 - 4x + 3 > 0

[raw hinge]

alright ty

.close

[obtuse pebble]

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[rose lintel]

I was asleep
thx for the help but the correct asnwer should be 202 though

[obtuse pebble]

@rose lintel Has your question been resolved?

[obtuse pebble]

@rose lintel Has your question been resolved?

[proven zephyr]

ima copy paste the original question just it becomes clearer

hm... lemme try

hmmm

nvm i failed

[obtuse pebble]

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[hidden stream]

can someone help me with this.
I do not understand how to solve even with solution

[hidden stream]

as i know : this is the solution but i dont understand how they get b and c

[gentle ridge]

the simple method is to substitute the points (0, -1), (pi/4, 1) and (pi/2, 3) that the tangent graph passes through

but here you can utilize the symmetry of the tangent graph to drastically simplify the calculation

youll notice that the two points (0, -1) and (pi/2, 3) are symmetrical about the point (pi/4, 1)

[hidden stream]

im sorry but I have no idea what you are talking about. This is my first time seeing this kind of question

[gentle ridge]

tbh theyre very mean with this problem

i suggest you try a different problem before you approach this one

[hidden stream]

xd ikr but i only have about 1 more month till AS exam

it's a question in a past paper and i have to do it as homework

[gentle ridge]

okok

can you see that the graph passes through the three points (0, -1), (pi/4, 1) and (pi/2, 3)?

[hidden stream]

uhhh yes

[gentle ridge]

do you know the tangent graph is symmetrical about the origin ?

[hidden stream]

not pi/2, 3 tho

[hidden stream]

yes

[gentle ridge]

look again and count from the bottom

[hidden stream]

I hope u have time to explain it to me xd I've been stuck at this for a really long time

[obtuse pebble]

@hidden stream Has your question been resolved?

[obtuse pebble]

@hidden stream Has your question been resolved?

[tepid mountain]

@hidden stream
So do you know what the term c and b mean in an equation like this
$$tan(x-b)+c$$

[warm shale]

Sherif Player

[tepid mountain]

(b,c) is the co-ordinates of the first turning point on the right of the x-axis

Which has the co-ordinates
$$(\frac{\pi}{4}, 1)$$

[warm shale]

Sherif Player

[tepid mountain]

That's how he got the b and c in the first line

You may take an easier way to get them using the obvious points given in the graph as Girl said

Like the points
$$(0, -1)$$
$$(\frac{\pi}{4},1)$$
$$(\frac{\pi}{2},3)$$
To get the answer a bit easier

[warm shale]

Sherif Player

[tepid mountain]

Substitute them into the equation
To get
$$A\tan(-b)+C = -1$$
$$A\tan(\frac{\pi}{4}-b)+C = 1$$
$$A\tan(\frac{\pi}{2}-b)+C = 3$$

[warm shale]

Sherif Player

[tepid mountain]

$$C = - (A\tan(-b))$$
$$A(\tan(\frac{\pi}{4}-b)) - (A\tan(-b)) = 1$$
$$A[\tan(\frac{\pi}{4}-b) - \tan(-b)] = 1$$

[warm shale]

Sherif Player

[tepid mountain]

Then you can use the subtraction of tans identity

$$A[\frac{\tan(\frac{\pi}{4})-\tan(b)}{1+\tan(\frac{\pi}{4})\tan(b)} + tan(b)] = 1$$

[warm shale]

Sherif Player

[tepid mountain]

$$A[\frac{1-\tan(b) + \tan(b)(1+tan(b))}{1+tan(b)}] = 1$$

[warm shale]

Sherif Player

[tepid mountain]

$$A[1- tan(b)] = 1$$

[warm shale]

Sherif Player

[tepid mountain]

So
$$A = \frac{1}{1-tan(b)}$$

[warm shale]

Sherif Player

[tepid mountain]

Actually nevermind
Using the first one is much much easier way than doing that

[obtuse pebble]

@hidden stream Has your question been resolved?

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[fallow rampart]

The context is that I just learned about one sided limits with a hole

Is the professors “yes” to this question only applicable to removable discontinuities? Because the answer doesn’t seem like a “yes” with the example image shown.

[fickle turret]

The example isn't a removable discontinuity

[fallow rampart]

Yes she drew that as an exception but I just want a verification

Because it seems like the answer would just be no

Oh you know what

The first few words slipped my mind

I read the question just about discontinuities as a whole