help-13

[broken mist]

Sort of, but not quite

[forest steeple]

is m the result of -1/3?

[broken mist]

The m, or slope (or gradient if you're not from the united states), can be calculated with given points

It's the result of the change in y (or "rise") divided by the change in x (or "run")

[forest steeple]

oh so X the run is -1 and Y is rise and its plus 3

[broken mist]

So between the point (-1, 3) and (0, 0), x changes by +1 and y changes by -3

[forest steeple]

yes yes

[broken mist]

Yup, and if you divide the rise by the run, you get the slope

So -3/1 (like how I wrote it) or 3/-1 (like you wrote it) both give us -3

[forest steeple]

so 3/-1

yea yea

[broken mist]

Exactly

[forest steeple]

but why is it Y=-3x-, is it always Y?

[broken mist]

So essentially, the equation y = -3x describes every point on that line

If you choose a random number to be x, and then calculate y from that, you get a point on that line

So let's say x = 4

-3(4) = -12

So we'd get the point (4, -12)

Which is on the line there

[forest steeple]

ohhhhhh

[broken mist]

If you're on a computer right now and want to, try going to the website desmos.com/calculator. Write y = mx + b in the box on the left, and then change m and b to random numbers to see what it does

That'll give you some more visuals of other graphs

[forest steeple]

how do you find out what B is

Or in the question it says C

[broken mist]

Ah I didn't notice that

same thing, I just used a different letter

[forest steeple]

yea yea

[broken mist]

Basically, c is the number you get when you set x equal to 0

It tells how high off the x-axis the graph is over the origin (or point (0,0))

I'll get a graph that'll show it better, one sec

[forest steeple]

ok thanks im a bit lost

[broken mist]

Here's the line y = 3x + 4

So the m value there is 3, which means that every time x goes up by 1, y goes up by 3

It's similar to your problem, but the line goes up to the right instead of up to the left

If you look at the y=axis, you'll see that the line crosses it at the point (0, 4)

[forest steeple]

so what is C in my question? 0?

[broken mist]

Exactly

[forest steeple]

ohh its where it crosses?

[broken mist]

Which is why it's not there at all, because adding 0 does nothing

So they didn't write "+ 0"

Yup

Where it crosses the y-axis is the c value

[forest steeple]

and M is -3?

[broken mist]

In your problem, yes

[forest steeple]

and what is X

[broken mist]

X can be anything

Here give me one moment

[forest steeple]

oh so in my equation

ohh

i get it

[broken mist]

So if you choose something random for x, and solve for y, you'll find a point (x, y) that's on the line

[forest steeple]

X can be anything and itll still be on the line

[broken mist]

Exactly

[forest steeple]

yes yes

ahh that makes sense

i got one more question for another problem

[broken mist]

Sure

[forest steeple]

https://gyazo.com/c039785d99b399ff5ae5f47805f2eda6

[broken mist]

And just to note, part a in your original problem is a weird case where you can't write it in y = mx + c form

[forest steeple]

so is it -3 is where you start? then rise 3 up and run 2 positive which means right

then draw a line?

[broken mist]

Exactly

[forest steeple]

ohh i see

https://gyazo.com/accc6c0f6f26d1e5f95fc4c10b59597b

and for A why is it positive? are you meant to work towards the middle?

Because its going left 3? so isnt that negative

[broken mist]

If you think about moving from the left to the right, positive gradient makes you go up

So if you trace the line starting on the left and going right, the m = +1 one goes upwards, and the m = -2 one goes downwards

[forest steeple]

im talking about A

[broken mist]

Yeah you can think of it as working towards teh middle

If you start at the y line and go left, you go down 3 and to the left 3

Which would be -3/-3 or 1

[forest steeple]

ohh they both equal 1 anyway

[broken mist]

Yup

[forest steeple]

cool i get it now thanks

[broken mist]

If this helps, think of it like this

As x gets bigger, y also gets bigger, positive slope

In b there, as x gets bigger, y gets smaller, negative slope

[forest steeple]

yea yea

https://gyazo.com/a1440ff4a631877130673099c73558d3

im a bit confused on this

[broken mist]

Which problem do you want to start with?

[forest steeple]

A and B

or i and ii

[broken mist]

$\frac{x}{9} = \frac{4}{3}$

[wraith dagger]

lexitorius

[broken mist]

There's a couple ways to do this one

There's the way that your solution there shows, and there might be a more simple way if you want that

[forest steeple]

i dont even get where they got the 3 from

in my sdolution

[broken mist]

Essentially they're trying to make the denominators the same, and multiplying 3 by 3 gives you 9

There might be a better way though

[forest steeple]

oh i see

[broken mist]

So we have x/9 on the left side, and we want to get x on its own

[forest steeple]

but in ii the X is at the top

[broken mist]

How do we get rid of "/9"

[forest steeple]

how would you do that

[forest steeple]

with 9?

[broken mist]

Multiply 9 to both sides, yup

So then we'd have $x = 9(\frac{4}{3})$

[wraith dagger]

lexitorius

[forest steeple]

i see

whihc gives 12

[broken mist]

Yup

[forest steeple]

and what aboit ii

[broken mist]

That's the way that I would have done that problem, but we can go through the way they did it too if you want

ii is similar, but the x is in the denominator

$\frac{15}{x}=\frac{3}{4}$

[wraith dagger]

lexitorius

[forest steeple]

so would you have done

15(3/4)

[broken mist]

Not quite

Remember how we had 9 in the denominator, and multiplied it to both sides to move it?

[forest steeple]

wdym both sides

wasnt it still 4/3

[broken mist]

We did $\frac{x}{9} = \frac{4}{3} \implies x = (9)\frac{4}{3}$

[wraith dagger]

lexitorius

[broken mist]

x/9 times 9 is x, and 4/3 times 9 is 9(4/3) or 12

[forest steeple]

im slightly lost

[broken mist]

That's okay

[forest steeple]

so why cant i do 15(3/4)

[broken mist]

You could, but then we would have $\frac{1}{x} = \frac{(\frac{3}{4})}{15}$

Which we could still work with, but it might be harder

[wraith dagger]

lexitorius

[broken mist]

Since 15 is on top we'd actually have to do that

Divide both sides by it

[forest steeple]

oh so whats the other way of doing it

[broken mist]

So what I would do is multiply both sides by x

$\frac{15}{x} = \frac{3}{4} \implies 15 = x(\frac{3}{4})$

[forest steeple]

and what is X? 15

[wraith dagger]

lexitorius

[broken mist]

x is just x, we can multiply it to both sides to move it just like we did with 9

We don't actually know what x is yet, but it is a number so we can do the same things to it as we can numbers like 9 and 15

Another way to write what I got there is $15 = \frac{3x}{4}$

[wraith dagger]

lexitorius

[broken mist]

Which gets x on the top, so we can solve it similarly to how we did part i

[forest steeple]

im lost man sorry

[broken mist]

That's alright lol

[forest steeple]

maybe i didnt understand i

[broken mist]

$\frac{x}{9} = \frac{4}{3}$

[wraith dagger]

lexitorius

[broken mist]

Another way we can do i is by making the denominators teh same

Let's try it that way, if you'd like

[forest steeple]

yea sure

youd do x3

oh and then 4x3

is 12

ok i got that now

[broken mist]

Yup, to the right side which we can do because 3/3 is equal to 1

You can always multiply something by "one" without technically changing it

So even though we're changing the numbers themselves, they still mean the same thing

Namely, 4/3 and 12/9 are equal

In the same way that 1/2 and 2/4 are

If that makes sense

[forest steeple]

yea yea

[broken mist]

And we can work with 12/9 much better

$\frac{x}{9} = \frac{12}{9}$

[wraith dagger]

lexitorius

[broken mist]

Now if we multiply 9 to both sides, we just get x = 12

[forest steeple]

but all you need to do is find x no?

so after ive done 4x3 im done

[broken mist]

Exactly

[forest steeple]

ok cool cool

now for ii

[broken mist]

$\frac{15}{x} = \frac{3}{4}$

[wraith dagger]

lexitorius

[forest steeple]

can you make the top the same?

like 3 x 5 is 15

and 4x5 is 20

[broken mist]

Yeah that's actually a shortcut

Not sure if it'll always work but

Here's a trick

$\frac{15}{x} = \frac{3}{4}$ is the same as $\frac{x}{15} = \frac{4}{3}$

[wraith dagger]

lexitorius

[broken mist]

If you flip both sides like that, they will still be true

[forest steeple]

oh is it the same?

[broken mist]

And now you can do that problem just like we did i

[forest steeple]

ohh i see

cool thanks

[broken mist]

Be careful about when you do that though because if there's more stuff on one side it won't work

$\frac{15}{x} + 2 = \frac{3}{4} \text{is not the same as} \frac{x}{15} + 2 = \frac{4}{3}$

[wraith dagger]

lexitorius

[broken mist]

Holy beans that's finnicky

[forest steeple]

LMAOO

[sacred grail]

try doing something like

[wraith dagger]

$\frac {15} x + 2 = \frac34$ is not the same as $\frac {x} {15} + 2 = \frac43$

[broken mist]

Oh I did that before but I had "not" underlined and it messed it up

[wraith dagger]

$\frac {15} x + 2 = \frac34$ is \underline{not} the same as $\frac {x} {15} + 2 = \frac43$

[broken mist]

$\frac{15}{x} + 2 = \frac34$ is NOT the same as $\frac{x}{15} + 2 = \frac43$

Valid

[wraith dagger]

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

[forest steeple]

https://gyazo.com/02314730e8bec4bf520270380a2f5c75

what about this one

[broken mist]

So in that one, we want to start by isolating the term with x in it

We want to get x/2 onto its own side of the equation

[forest steeple]

oh ok

[broken mist]

$\frac{x}{2} - 2 = 1$

[wraith dagger]

lexitorius

[broken mist]

But that -2 is in our way, so we need to find a way to move it to the other side of the equation

[forest steeple]

so you do -2=1?

[broken mist]

What's a way that we can make x/2 - 2 change to x/2

[forest steeple]

ohh +2?

[broken mist]

yup

But if we do that to the left side of the equation, we need to do the same to the right side

[forest steeple]

ohh

so youd get 3?

[broken mist]

Yup

So now we have $\frac{x}{2} = 3$

[wraith dagger]

lexitorius

[forest steeple]

then you put 3 over 1?

[broken mist]

You can if you want, or we can also find a way to get rid of the 2 in teh denominator

Similar to how we got rid of the 9 earlier

But yeah 3 is equal to 3/1

[forest steeple]

i think i rather do 3/1

[broken mist]

Go for it

[forest steeple]

so then x/2 = 3/1?

[broken mist]

yup

Now how can we make the denominators the same?

[forest steeple]

not sure

[broken mist]

When we had $\frac{x}{9} = \frac43$, how did we make the right side be $\frac{12}{9}$ instead?

[wraith dagger]

lexitorius

[forest steeple]

we made the denominators the same

x2?

so 3 x 2 is 6

[broken mist]

correct

[forest steeple]

so then we got x/2 = 6/2

[broken mist]

Mhm

[forest steeple]

now im kinda lost tho

[broken mist]

$\frac{x}{2} = \frac62$

What is 6/2 equal to?

[forest steeple]

3?

[broken mist]

So we now know that x/2 is equal to 3

[forest steeple]

is x/2 equal to 2x?

[broken mist]

nope x/2 is x divided by 2

Or half of x, if that helps

[forest steeple]

yea i thought so

but idk where to go from there

[broken mist]

So if half of x is 3, what is a whole x?

[forest steeple]

6?

[broken mist]

Yup

To do it symbolically, we would want to get rid of the denominator 2

Which we can do by multiplying everything by 2

[forest steeple]

ohh so x = 6

makes senes

[broken mist]

x/2 times 2 is x, and 3 times 2 is 6

[forest steeple]

ohh ok

[cedar kiln]

@forest steeple Has your question been resolved?

[forest steeple]

https://gyazo.com/82d598cdbbbb484784ad258a275838ee

@broken mistbit lost on this one

[broken mist]

So this is called the distributive property

[forest steeple]

i got the first line of the pink writing but not sure about second

[broken mist]

Okay cool so you got the multiplication part down?

[forest steeple]

yup

[broken mist]

Remember before when we had a -2 on one side of an equation, and we wanted to get rid of it?

[forest steeple]

yea

[broken mist]

We can do something similar here

What we'll want to do is get all of the terms with an x in them onto one side, and all of the terms without an x to the other sidew

[forest steeple]

yea

[broken mist]

So we have $2x + 2 = 3x - 6$

Let's start with the +2, we want to get it to the other side

[forest steeple]

so 2 - 6

and 2x + 3x

[broken mist]

not quite

TO get the +2 to the other side, we have to subtract 2

[forest steeple]

oh are we removing them from the one side?

so -2 to get rid off it?

[broken mist]

Yup

[forest steeple]

and -2x to get rid of the x?

so -1x and -8?

[broken mist]

we could, but we want to keep the x terms on one side

So we would instead move the 3x from the right side to the left side

[forest steeple]

oh so u saying we just get x?

[cedar kiln]

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[forest steeple]

.reopen

[cedar kiln]

✅

[forest steeple]

bcs 3x -2x?

.close

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[cedar kiln]

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[waxen kestrel]

Hello, I need some help with this one: Find all the asymptotes for the function

[bold vine]

Try considering what happens as x -> ±∞

One option is (now I wouldn't do this immediately), but you can check your answer by graphing this on Desmos

[waxen kestrel]

Alright, so, when x -> +infinity, f(x) -> x

when x -> -infinity, f(x) -> 2

does that mean I have a horisontal asymptote at y = 2=

?*

[cedar kiln]

@waxen kestrel Has your question been resolved?

[dire geode]

No

First divide all terms by the largest one

[waxen kestrel]

u mean something like this?

[dire geode]

That's not the largest term

x^3 e^(3x) is

[waxen kestrel]

mb, thought the biggest term had to be in the denominator

[cedar kiln]

@waxen kestrel Has your question been resolved?

[dire geode]

Oh you're right. It should have been x^2 e^(3x)

[waxen kestrel]

okay, so it's something like this?

[cedar kiln]

@waxen kestrel Has your question been resolved?

[cedar kiln]

@waxen kestrel Has your question been resolved?

[dire geode]

Yes

[cedar kiln]

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[inland totem]

how is this wrong

$arccos\left(\frac{\left(11\right)^2+\left(14\right)^2-\left(8\right)^2}{2\cdot 11\cdot 14}\right)$

[wraith dagger]

Kold

[inland totem]

= 34.8

[haughty wraith]

whats that formula

?

just use heron's formula?

[crystal raptor]

That looks like you've solved for an angle using cosine rule

[inland totem]

cos A = (b)^2 + (c)^2 - (a)^2 / 2 * b * c

[crystal raptor]

It wants area

[inland totem]

omfg im so stupid

i thought since

the opposite of the sides

are capital

it wanted side length A

[crystal raptor]

Ah I see it tbf

[haughty wraith]

ohhh

lol

[inland totem]

so A = 1/2 * b * h

[haughty wraith]

no

herons formula

[inland totem]

i have no clue what that is

😭

[haughty wraith]

you dont know base or height

you could solve for angle

and use

like sin(A) thing

[inland totem]

wtf why did she never teach us this

[crystal raptor]

There's also a formula for when you have an angle so your work might not be for nothing

[haughty wraith]

yea ^

Area = (ab sin(C))/2

[cedar kiln]

@inland totem Has your question been resolved?

[cedar kiln]

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[pure gale]

I would like to know if this is correct

<@&286206848099549185>

[haughty wraith]

where did the 8.7 come from

[pure gale]

8.7 is the velocity

I used v = u + at to try and find the acceleration

[weak lake]

a = change in velocity / change in time

ya that formula should work

34/3.9

[pure gale]

is the acceleration correct

2.2m/s

[weak lake]

what?

34/3.9 doesnt give 2.2

its gotta give some much larger number

for example something around 7

its 8.7

[pure gale]

so the acceleration and velocity are the same?

[weak lake]

?

no

[pure gale]

The velocity is 8.7m/s^2

[weak lake]

no

acceleration is 8.7 m/s^2

velocity goes from 0 to 34

[pure gale]

34 is not the velocity though

[weak lake]

then what is it?

[pure gale]

34 is the displacement

[weak lake]

if it was displacement it would be given in meters instead of m/s

[pure gale]

Ohh

That’s what I got wrong alright thanks

.close

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[outer agate]

Im looking to show that the set of points with foci A,B {P|(d(A,P)^2)+(d(B,P)^2)=a} where a is constant

[outer agate]

I think the center would be something like d(A,B)/2, and know appollonius proved this using ratios of the distance

would a be the curvature?

[cedar kiln]

@outer agate Has your question been resolved?

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@outer agate Has your question been resolved?

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@outer agate Has your question been resolved?

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@outer agate Has your question been resolved?

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@outer agate Has your question been resolved?

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[barren parcel]

I assume it's safe to ignore this case because it's a false statement?

[barren parcel]

what if it was a true statement like 3 > 1?

would that have any meaning?

[rancid perch]

What are you trying to do? the first inequality is equivalent to x<1

[barren parcel]

This is the exercise. I need to specify the set of all numbers in R for which this is true, in the simplest way possible.

[wraith dagger]

madmike

[barren parcel]

basically my question is if I am doing things correct

I don't know what to do if I get gibberish from a case, like 1 > 3

I am just ignoring it now

On a different exercise I also have a case that just results in a true statement

What should I do with this?

Am I supposed to ignore it?

<@&286206848099549185> 🙃

.close

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[undone ice]

Hello can you simplify this question

[rugged palm]

yes

Try factorising the numerator

[undone ice]

idk how

[cosmic steppe]

What's the greatest common factor between $15a^2b^2$ and $3ab$

[wraith dagger]

Umbraleviathan

[undone ice]

there is only (A) and (B)

or it is that there is a ^2

[cosmic steppe]

The greatest common factor

If you're doing stuff like this you should be proficient in factoring

You always yield to the factor(s) with the lowest exponents (for variables)

Which term has the lowest exponents

[undone ice]

can u simplify the answer

my english not good

[cosmic steppe]

You can simplify it

[undone ice]

ok help me with this then

or this what is N?

ik that the answer is not 3

[dusk finch]

if 3+3+3+3=4*3 then your equation is equal to...

2+2+2=3x2

5+5+5+5=4*5

Try to do the same with your equation

[undone ice]

5*5?

[dusk finch]

almost. You cant forget the 5^n

[undone ice]

so n+n+n+n+n = 5*n

?

[dusk finch]

Yeah! this is correct

[undone ice]

oh nice

[rancid perch]

not quite

[undone ice]

so what is the answer?

[crimson sedge]

5 times 5^n equals 5^1 * 5^n

[rancid perch]

5ⁿ+5ⁿ+5ⁿ+5ⁿ+5ⁿ=5x5ⁿ

[undone ice]

oh

[crimson sedge]

equals 5^1+n

[undone ice]

is that the answer?

[crimson sedge]

the answer is n = 14

[rancid perch]

no, you need to find n

[dusk finch]

Yep

[dusk finch]

This one is right

[undone ice]

but how is it 14?

[crimson sedge]

bro thats a diff question

[undone ice]

my question is

[crimson sedge]

Listen

[dusk finch]

This is solution

[undone ice]

can u tell me the answer and how that is the answer

[crimson sedge]

How many times do you count 5^n

5 times right

5 * 5^n = 5^1+n

5^1+n = 5^15

1+n = 15

n = 14

[dusk finch]

You gotta use parenthesis

[crimson sedge]

where?

[dusk finch]

Here are his steps written in the way you may be more familiar with

[crimson sedge]

This

[dusk finch]

You can't just say 5^1+n because it would be 5+n. You have to say 5^(1+n)

[crimson sedge]

Got it

[dusk finch]

And this one is (Broken to the most basics steps)

[dusk finch]

You should be able to do this one by yourself if you look to my solution of the previous equation

[cedar kiln]

@undone ice Has your question been resolved?

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[grand forge]

https://gyazo.com/58bdd3451734eb42b22fd98af8a92e61

[grand forge]

bottom doesnt factorise

so im a bit lost

have to use that sub

[gaunt hamlet]

Complete the square

[grand forge]

omg

ur a life saver

.close

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[crimson sedge]

How did the a & (1-a) end up in front of the ln ?

[obtuse gyro]

ln(xy) = ln(x+y)

and ln(x^n) = n ln(x)

[grand forge]

no

ln(x) + ln(y) = ln(xy)

[obtuse gyro]

my bad

yea

[tame granite]

it's a special proof of logarithms
whenever the parenthesis is raised to a power
it's equal to the logarithm being multiplied by that number

so if you have log((x + 3)³) it's equal to 3log(x + 3)

[crimson sedge]

Its like a math law ?

Like for logarithms?

[long swan]

Not quite

log((x+3)³) = 3log(x+3)

Pay special attention to parantheses

[tame granite]

ope

ok there

[tame granite]

Yes

[cedar kiln]

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[crimson sedge]

differentiation

[crimson sedge]

Hi

Okay

So first thing, we need to use chain rule

We define $u = x(x-5)^4$ so that by the chain rule $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

[wraith dagger]

♡LexQa♡

[crimson sedge]

mhm

i already went wrong there

well not really but erm still

So first, $\frac{dy}{du} = \frac{1}{u}$ and $\frac{du}{dx} = (x(x-5) ^4)'$

yeah

[wraith dagger]

♡LexQa♡

[crimson sedge]

i think i got a strange answer for du/dx

[eager sparrow]

You can use the product rule for that

[crimson sedge]

So now we have to do apply product rule for du/dx and chain rule

So we define point v = x-5 and say $\frac{du}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx}$ we apply product rule to get
$\frac{du}{dx} = x'(x-5)^4 + ((x-5)^4)'x$

[wraith dagger]

♡LexQa♡

[crimson sedge]

hm

So: $\frac{du}{dx} = (x-5)^4 +4(x-5)^3 \cdot x$

is it not defining $v = (x-5)^4 ?$

[wraith dagger]

♡LexQa♡

🍌Hannah🍌

[crimson sedge]

I kind of skipped the chain rule for the second one, but if u r confused I can write it in detail

[crimson sedge]

i got exactly this

[crimson sedge]

We want to get it into the form of $u^n$ where $(u^n)' \to n\cdot u^{n-1}$

[wraith dagger]

♡LexQa♡

[crimson sedge]

oh right

well anyways

i did get what you got

Anyways, adding it all together gives us $\frac{dy}{dx}= \frac{(x-5)^4 + 4x(x-5)^3}{x(x-5)^4}$

We can simplify

But first thing,is everything clear?

yeah

[wraith dagger]

♡LexQa♡

[crimson sedge]

i did the stupid thing of trying to simply this

We're going to do exactly that now

okay

oh lol it is the right answer

i complicated it too much before...

thank you so much

So we factor $(x-5)^3$ from the numerator to get $\frac{(x-5)^3(x+4x-5)}{x(x-5)^4}$ which means $\frac{5x-5}{x(x-5)}$

[wraith dagger]

♡LexQa♡

[crimson sedge]

yeah i did that

Aye

tyy

Good jobb

😄

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[cedar kiln]

@glad pagoda Has your question been resolved?

[glad pagoda]

<@&286206848099549185>

[glad pagoda]

<@&286206848099549185>

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[desert solstice]

This is how I've interpreted the question

I dont know if this is correct though since the way the question is worded is a little confusing

I would advise finding dx/dt and substituting dx/dt and x into y

then differentiating using either the product rule and/or chain rule

[cedar kiln]

@glad pagoda Has your question been resolved?

[glad pagoda]

How about the 20 tents that are built ? Where will it apply ??

[desert solstice]

yea then the equation for y is wrong

you would only start adding 2(dx/dt)t when the tent number (x) reaches 20

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[wet swallow]

hi, i’m confused about this question. it’s graph theory, so if anyone knows it it would be really helpful! thanks

[cedar kiln]

@wet swallow Has your question been resolved?

[wet swallow]

<@&286206848099549185>

😦

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@wet swallow Has your question been resolved?

[wet swallow]

damnansdljfa;sjd

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[thin roost]

Hello

[thin roost]

Find the equation of a pair of line passing through the point ( 2,3) and perpendicular to the Line Pair 2x²- xy - 6y² + 4x + 6y = 0

Well I've separated the equation
(x-2y+2)(2x+3y)=0

But idk what to do now

[spice tundra]

instead of factorising like that, try make y the subject in the second equation

you want it in the form of y = mx + b

oh sorry

hang on

[thin roost]

Huh

[spice tundra]

didn't read the second half

[thin roost]

Ok

[clever lintel]

" equation of a pair of line passing"

what's a pair of line?

do you just want a line equation?

or two lines, or what. I don't get it

[spice tundra]

i've never heard of this "pair of lines"

[thin roost]

Ye two lines

[thin roost]

By pair they mean the combine the line and make it in homogenous form

[clever lintel]

so yeah the second part I get, it's two lines, so we can call it a pair

[thin roost]

Ye ye

So how am I suppose to do it??

[clever lintel]

ok ok I gochu. it's just that the grammar of the first part makes no sense

maybe you misspelled something

[thin roost]

Ok ok

[clever lintel]

you should get the equation for perpendicular line

*formula

you take that second pair, and isolate these two lines

so you do it twice

then, when you have the A coefficient, I don't recall what's it's called in english, you know, the one that tells you the slope of the line

you plug in the point they want you to cross, so you raise or lower your newly found lines to make them work

[thin roost]

So you mean smth line

2x+3y=0
y=(-2/3)x

Then you find the slope

Then

[clever lintel]

yeah, -2/3 is the slope

[thin roost]

(y-3)=-2/3(x-2)

Like this??

Then simplify

[clever lintel]

no no, wait

[thin roost]

The bot is offline

[clever lintel]

solve this for y

get the two possible lines from it

bam, you have two line equations

(you should get 2 x solutions)

[thin roost]

Well I've separated the equation
(x-2y+2)(2x+3y)=0

This??

[clever lintel]

no get in the standard form

like y=ax+b

[spice tundra]

you can just use this

[thin roost]

Ohh

[clever lintel]

then use the thing

[thin roost]

Slope intercept

[clever lintel]

to find perpendiculars

[spice tundra]

bro

you already have the two equations factored

$x - 2y + 2$

$2x + 3y$

just make y the subject

find the gradient

[thin roost]

Ok ok

So

[spice tundra]

find the negative reciprocal gradient for perpendicular

then slap it into the formula

y - y1 = m(x - x1)

and you'll get two equations

[thin roost]

Kk

[spice tundra]

and the final res will be

(eq1)(eq2)

[thin roost]

Ye I did the Same thing

I get quit diff ans

Let me show ya

[spice tundra]

yes

x - 2y + 2

x + 2 = 2y

y = (x + 2) / 2

he already has it all factored

don't see what the issue is

[spice tundra]

what was the answer?

[thin roost]

This is the equation right??

[clever lintel]

well he did only half of it

[thin roost]

Huh

[clever lintel]

(x-2y+2)(2x+3y)=0 this is just the pair in the original question

now he has to get the perpendicular pair right?

[spice tundra]

you didn't do negative reciprocal

if a line is perpendicular to another line, they have negative reciprocal gradients

for example

if we have y = 2x

[thin roost]

m1*m2=-1

This thing?

[spice tundra]

yes

[thin roost]

Ok

So

[clever lintel]

doesn't look perpendicular to me :p

[spice tundra]

tf have you done you spoon

hm

okay

if we have

2y = x + 2

y = (x + 2) / 2

y = 1/2x + 1

negative reciprocal is -2

[thin roost]

Finally I got it

Thanks man

[spice tundra]

np

[clever lintel]

tell us your result

[clever lintel]

we can plot and check

[thin roost]

I've chevked

Seems fair

Thanks

[clever lintel]

looks okay, but did you make sure they cross at point (2,3)?

[thin roost]

Yes

[clever lintel]

aight awesome. good luck on the next one

[thin roost]

Thanks

.close

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[summer gazelle]

i don’t know where i’m going wrong for this proof

[modern compass]

,rccw

what are you trying to do?

[summer gazelle]

i did the basis, inductive step etc and now i’m just trying to prove the rhs=lhs

i can show that part if you’d like

but i’m getting 1/4 and -1/4

[dusty hazel]

The first line in LHS, 5^(k-1) is wrong. It's 5^k - 1

[summer gazelle]

omg my handwriting fucked me at some point THANK U

wait nevermind i get it wrong still

[modern compass]

what did you try

[modern compass]

you can't cancel like that

write down what you're trying to get to and keep that in mind while you simplify the fraction. You're basically there, but forgot where you were going

[summer gazelle]

like i can’t cancel the 5^k/5^k and the 5/20?

[modern compass]

no

for the same reason you can't cancel in (2-1)/2
since that is obviously 1/2
but if you cancel the way you did you get (1-1)/2 = 0

[summer gazelle]

oh i’m stupid i see now bless!!!

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[covert jacinth]

Did I do number 4 right? Practice sheets for that extra learning

[broken mist]

💯

You did

Looks great

[covert jacinth]

haha Ws only 😎

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[lunar crystal]

how the heck do i work this out?

[mortal merlin]

gaussian elimination

[frosty ocean]

o_O

Do you not see they have asked to use substitution?

[lunar crystal]

And how do i do that?

[mortal merlin]

oh

I dind't see

[frosty ocean]

Substitute the y in the first equation in the second. Solve for x. Plug in x in either of the given equations to get y.

[lunar crystal]

Wait what, how do i substitute the y?

[frosty ocean]

2x+1=x+3

Solve for x.

[lunar crystal]

oh okay

x = 2

so y = 5

?

oh

[frosty ocean]

You're right.

[lunar crystal]

oh lol

thanks alot :D

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[wind crest]

Find three integers m such that 13≡7(mod m).

[south tundra]

Recall the definition of congruency in mod m

[wind crest]

hmm, congruency means they're the same. so eg 5mod 2 = 3, and 10 mod 7 = 3

excusing my use of =

[south tundra]

a ≡ b (mod m) <-> m divides (a - b)

So in this case you need to find integers m which divide 6

[wind crest]

ohhh

[south tundra]

I am not sure if mod 1 is acceptable so the answers are 2, 3 and 6

[wind crest]

ohh, is it like

in the world of modulo 2, even if we pass a number that is 0 mod 2, we can still go on

to get 7 mod m that is congruent to 13

um actually, idk what i was going on about

.close

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[lapis storm]

isnt f'(x) always gonna be 0 ?

[elfin hemlock]

Why would it be that?

[lapis storm]

derivative of a whole number is 0 no?

uncropped^

[elfin hemlock]

But its not a whole number?

[lapis storm]

wait im fucking stupid

(-inf, -3)

and (2,inf)

right

[elfin hemlock]

Yes

[lapis storm]

ok thanks

myb

@elfin hemlock

sorry one more question

would i just enter -9999

like i dontunderstnad the answer formatting

ok wait nvm

got it

$close

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[amber epoch]

I don't get how it went from the first step to the second

[south tundra]

cos^2 * (1 - sin^2) - sin^2 * (1 - cos^2) = cos^2 * cos^2 - sin^2 * sin^2 = cos^4 - sin^4 = (cos^2 - sin^2)(cos^2 + sin^2) = cos^2 - sin^2

[amber epoch]

how'd you get cos^2 * cos^2 - sin^2 * sin^2

oh

nvm

but you can't get cos^4 because one is with alpha and the other is with beta

[south tundra]

Oh

Have you tried expanding?

[amber epoch]

yeah but it's the same issue with one being alpha and the other being beta

So you can't cancel them out

<@&286206848099549185>

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[empty topaz]

hello can someone help me with this A 5.0 cm tall object is laced 36.0 cm away from a convex mirror whose focal length is 90.0 cm. What is its magnification??

[bright agate]

@empty topaz you may consider physics server #old-network

[cedar kiln]

@empty topaz Has your question been resolved?

[hushed spoke]

Mirror formula to find v

You should be able to find magnification then

[empty topaz]

i tried it

but the answer is always 0.6

and it's not in the choices

[empty topaz]

these are my solutions can u tell me if there's an error?

[cedar kiln]

@empty topaz Has your question been resolved?

[cedar kiln]

@empty topaz Has your question been resolved?

[eager sparrow]

Hey, I got 5/3 or 1.666666

[eager sparrow]

,rotate

[eager sparrow]

Can you please send another image that isn't rotated

[empty topaz]

me too, but 1.6 isn't in the choices

[eager sparrow]

I can't really see the other one

Maybe it's asking for the height of the virtual image

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[wanton lake]

Hello could I get some help on what to do equation wise for this word problem plz?

"The average person drinks about 3⅕ cups of coffee per day. If a person works 5 days a week for 50 weeks every year, estimate how many cups of coffee that person will drink in a working lifetime of 51¾ years."

[wanton lake]

It's once again my final question so help is needed <@&286206848099549185>

My first impression of this server was awesome, but having to wait the 15 mins, tagging helpers and seeing others getting help long after I've asked, and my name going further and further down the occupied list begs to differ. Like it's been nearly 20 mins now and no help.

<@&286206848099549185>

[balmy apex]

you're acting like you're entitled to help, which you're not. no one is being paid to help you. don't expect people to put more time into helping you than you put into asking your question. read the message the bot sent.

[wanton lake]

I pinged after the 15 mins and ita now 30 and all I'm seeing is my name going down the list.

This was NOT the impression I was given a few days ago when I first joined the server. I never said I was entitled, just wasn't given this impression the first time asking and joining.

So either help, or get off the chat so someone else who is more kinder to someone with anxiety and PTSD can.

[balmy apex]

you're certainly acting like it lmao

read the bot message

also don't try gaslighting, that's just an asshole move.

[wanton lake]

Not gaslighting at all. I have anxiety and PTSD and this is only bringing back memories of when I've asked for help before and had to wait weeks to months on end only to need the question answered long before then, as I was badly falling behind then, just like I am now.

[balmy apex]

no one is being paid to help you buddy

don't expect people to put more time into helping you than you put into asking your question.

[wanton lake]

I'm not ur buddy. Never will be with ur disrespectful behaviour towards me.

[balmy apex]

if you actually put time and effort into your question, maybe someone will come and help you.

[balmy apex]

you're the one being disrespectful and rude.

[wanton lake]

I did the same thing last time and someone was right there ready to help.

[balmy apex]

okay and

[wanton lake]

So don't tell me what is right and wrong to ask for help with.

[balmy apex]

lol okay buddy

[wanton lake]

U do u and I'll do me.

Ok PAL?!

Cuz all ur doing is making my anxiety go up.

[cedar kiln]

@wanton lake Has your question been resolved?

[acoustic pivot]

@wanton lake what have you tried

When you ask a question on this server, it is easiest for someone to help you if you explain where you got stuck.

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@wanton lake Has your question been resolved?

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[crimson sedge]

q

[crimson sedge]

hey

guy

s

how i do

tan3pi/5-tan2x>=0

[glossy wave]

you could say that both sides of the division

need to be >=0

tan 3*pi is 0

[crimson sedge]

@glossy wavewait

tan3pi/5 is 0 or

tan3pi

[glossy wave]

tan3pi

is zero

[crimson sedge]

where is the 5

[glossy wave]

but the 5 is in 5-tan2x

[crimson sedge]

no

is not

the is is the denominator

of tan3pi

of 3pi

how i do nunu

[glossy wave]

oh

ok

then its

(tan3pi/5)-tan2x

[crimson sedge]

yes

[glossy wave]

this is what you wrote

[crimson sedge]

no

[glossy wave]

oh shit ur right

my bad

xD

[crimson sedge]

this is

[glossy wave]

oh that tan(3pi/5) - tan(2x)

[crimson sedge]

yes

[glossy wave]

are you supposed to use angles?

[crimson sedge]

wdym

i'm supposed to find solution

[glossy wave]

do you know what the value of tan 3pi/5 is

[crimson sedge]

i dont know

[glossy wave]

ok, ok

[crimson sedge]

but i can use tanx-tany orn o ?

factorization formula

[glossy wave]

mmh

[crimson sedge]

this

or what you would do

[glossy wave]

you could try that

but does it help u any way?

[crimson sedge]

i dont think it hel pme

[crimson sedge]

what should i do for this

[glossy wave]

lemme think for a sec what you could do

[crimson sedge]

okey

maybe put the 0 in pi/4 then do 3pi/5-2x>=pi/4 ?

[glossy wave]

what did you do with the tan?

[crimson sedge]

what i can do with the tan

🤔

[glossy wave]

mmmh

who the heck put this inequality to you xD?

[crimson sedge]

my teacher

but wait where are you in school

what is your grade

[glossy wave]

im 18, first year of career

[crimson sedge]

Ok

im 17

[glossy wave]

you can

move tan(3pi/5)

to the right side

being negative

multiply by -1

[crimson sedge]

tan2x<=tan(3pi/5)

[glossy wave]

so you reverse the inequality and make both positive

yes

you cant do them numerically right

[crimson sedge]

wdym by numerically

[glossy wave]

nah nothing

okay

you can use this

lemme draw it

[crimson sedge]

im not good at english

im from belgium and i speak french

[glossy wave]

yeah, np

i'll do a drawing

[crimson sedge]

Okey

[glossy wave]

there aint no way

ur supposed to know this

but copy this

xD

[crimson sedge]

no its not for a homework

[glossy wave]

<@&286206848099549185> do u know any other solution?

[crimson sedge]

its because tomorrow i have a test equation/inequation and she often put exercice like this

[glossy wave]

initial problem is tan(3pi/5) - tan 2x >= 0

[hearty basin]

what is the range of x

[crimson sedge]

@glossy wavei can do arctan3pi/5 ?N

[glossy wave]

nah, that isnt exact

[hearty basin]

can you tell me the range of x because it can have infinite values

[glossy wave]

thats why i did it with n

i suppose he doesn't know it

[crimson sedge]

0,2pi

[hearty basin]

ok

so there will be 4 inteervals

and tan 3pi/5

is negative

so you to satisfy the equation you need the value of tan of 2x that can have a more negative value

[hearty basin]

this is a conceptual method and i asuume you know less of equations of trigonometry

so first interval is [pi/4,3pi/10]

second interval will be [3pi/4,7pi/10]

similarly find the next two interval

[crimson sedge]

but

i understood nothing

[hearty basin]

do you know the graph of tanx

take the help of it

[crimson sedge]

yes

[hearty basin]

plot a grapf of tan a

where a =2x

[hearty basin]

do you understand this

[crimson sedge]

i need to get a value inferior to tan2x

[hearty basin]

yes value of tan2x will be btween (-inf, arctan3pi/5]

[crimson sedge]

Ok

and how i find these value

[hearty basin]

take the help of graph

you will find it easily

[crimson sedge]

i dont know the value of arctan3pi/5

[hearty basin]

sorry, something came up. I need to go now

[hearty basin]

you can find that it is negative with info

[crimson sedge]

i hate inequation

i dont understand