help-28

[dense edge]

And therefor f(x) would be output

So these would be the same values for input and output .. x and f(x)

Do I have that correct?

That’s why displacement and time look similar

And = slope

[nova basin]

you're literally just asked for (s(3) - s(1))/(3-1)

[dense edge]

Yep

So if it runs through s() argument that’s the output

The numerator

The denominator is just the delta of time so the inputs only

(Delta of output) / (Delta of input)

Slope formula

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[torn jolt]

Hoi

[torn jolt]

Formules i learned till now:

Sin^2 a + cos^2 a = 1

Tan^2 a + 1 = 1/cos^2 a

Cot^2 a + 1= 1/sin^2 a

Thats it i need help with d only

<@&286206848099549185>

[gritty rose]

can you also show the original paper without the translation

[torn jolt]

Why ? Its in dutch

[gritty rose]

some math got translated and shouldn't have

[gritty rose]

so don't know what variables are which

[torn jolt]

Calculate sin and cos when tan is given….

[gritty rose]

is the triangle supposed to have hypotenuse 1?

[torn jolt]

Its not with trizngles

Its just calculating

Wait

[gritty rose]

there are infinite triangles with tan(alpha) = 3/4

[torn jolt]

Look at a) calculate cos when sin is given

[gritty rose]

are they the same alpha?

[torn jolt]

Change the first formule out my list a bit it becomes cos= 1-/sin

[torn jolt]

Ye

The given is supposed to go replaced with the alpha

So to solve a) cos= 1-15/17

Get it?

[gritty rose]

plug that into tan(alpha)

$\tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)}$

[glossy valve]

riemann

[gritty rose]

you used pythagorean theorem wrong here

[gritty rose]

the squares are important

[torn jolt]

Hi so i dont yave neither sin ir cos

[torn jolt]

And no those formules are given by the teacher not part of pythagoras

[gritty rose]

where?

take another picture

[full forum]

@torn jolt Has your question been resolved?

[torn jolt]

I wrote it on a paper u couldnt read my handwriting but look u think u know how to do it with those formules or get one similar or?

[gritty rose]

the formulas you wrote are correct

[gritty rose]

you're just not using them correctly

$\cos^2(a) \neq \cos(a)$

[glossy valve]

riemann

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@torn jolt Has your question been resolved?

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[brazen girder]

Let's say I have a certain n by n matrix A written in the canonical basis over a given field K. This matrix represents an endomorphism from K^n to K^n, as it can act as a linear transformation. Now let's say I compute its characteristic polynomial p=|A-xI_n|. How would I go about finding a basis of K^n such that the endomorphism would be represented by the companion matrix of p?

[brazen girder]

Is this always possible? If not, what criteria determine it?

I don't have any good resources on the topic and I'm kind of lost

my work up until now has basically been resolving isolated cases, which leads me to believe it's not always possible. I'm pretty sure I've found some impossible ones.

But idk what criteria determine it

[full forum]

@brazen girder Has your question been resolved?

[brazen girder]

<@&286206848099549185>

[gritty rose]

maybe try #groups-rings-fields

[brazen girder]

oh okies

[brazen girder]

do I just ask there?

[gritty rose]

copy and paste 2 messages: 1 with the original question and 2 with your progress and questions about the question

[brazen girder]

okok

I'll check in linear algebra once it's free

thanks

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[spiral grail]

how do you prove the diagonals in a kite are perpendicular

[spiral grail]

i dont even know how to start this

i think you use the converse of perpendicular bisector thereom

?

[robust slate]

Hint: Consider congruent triangles

[spiral grail]

yeah i did that

there was another question that was like prove the opposite angles in a kite are congruent

but i dont get how i use that here

[robust slate]

It's actually very similar in the fact that you use angles to prove it - consider ||the angles that form a straight line||

[spiral grail]

OK

so

it would be perpendicular

but how do i write that in a 2 column proof

im given a kite abcd and the diagonals are drawn in

[robust slate]

Did you try anything on your own? I think I gave you a good starting point.

[spiral grail]

yeah

so i have

the kite and the congruent sides as given

but i dont know how to write that in a proof

do i just write straight lines are perpendicular

[robust slate]

You could establish equality between two angles, and then say that since they add to $180^{\circ}$, they're both right angles, hence the diagonals are perpendicular.

[glossy valve]

Civil Service Pigeon

[spiral grail]

OHHH

ok

cuz they form linear pairs they add up to 180?

[robust slate]

mhm

that's what I was talking about earlier with the angles on a straight line

[spiral grail]

ok ty

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[raw aurora]

I don’t know where to go from this

[meager dew]

,rccw

[meager dew]

draw a triangle

$\sin(\beta)$ represents a ratio of a right triangle's sides

[glossy valve]

a disappointing son

[raw aurora]

Ok

[meager dew]

it may also help to see that the ratio you're given has been rationalized

[raw aurora]

Bro

[meager dew]

you can get a correct answer without realizing that

[raw aurora]

How do I draw the triangle

Wait ok

[meager dew]

grab a pencil and press it down on paper in the shape of a triangle

[raw aurora]

I haven’t learned how to do it with the a triangle

I have only been taught to do it by equations

[meager dew]

label one of the non-right angles beta

then label the sides of your triangle

you know sin(beta) represents opposite/hypotenuse

so the numerator of your given ratio is the side opposite your angle, the denominator is the hypotenuse

[meager dew]

you're given $\sin(\beta)=\frac{4\sqrt{29}}{29}$

[glossy valve]

a disappointing son

[meager dew]

so what can you say x=?

[raw aurora]

Um idk

Wait

1?

[meager dew]

not quite

$\sin(\beta)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{4\sqrt{29}}{29}$

[glossy valve]

a disappointing son

[raw aurora]

Ohhh

[meager dew]

exactly

now you can find y

[raw aurora]

Ok give me a min

[meager dew]

not quite the method you'd want to take

you have two sides of a right triangle and you're looking for the third

what theorem can you use?

[raw aurora]

Cos?

[meager dew]

i'm thinking of a more... pythagorean method

[raw aurora]

Ohhhh

I’m sorry today has been rough

Give me a min

I got stuck here

What do I do with the square root 841

Nvm

[meager dew]

,w calc sqrt(29^2-(4sqrt(29))^2)

[glossy valve]

Results provided by WolframAlpha

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

Requested by a disappointing son#0540

[meager dew]

looks good

[raw aurora]

Yes!

[meager dew]

so you know what y equals

now you can get your other ratios

[raw aurora]

What do I round the 19 too

Like whole number, tens, hundredths

[meager dew]

normally two decimals is appropriate

[raw aurora]

19.42?

[meager dew]

yeah

[raw aurora]

Does this right/correct?

[meager dew]

yep

actually

you might want to leave it as sqrt(377)

for an exact answer

[raw aurora]

Oh

Ok

Mathaway says it correct

Thank you for helping and being so patient

I’ll let you help the other people, good night!

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[flat anvil]

.close

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[mortal viper]

how do i solve integers by the order of operation? i literally have watched videos and my friends tried to help me and i still dont understand at all😭😭😭😭😭

[mortal viper]

i dont get anything only bedmas but its still confusing

im probably overthinking it

but i dont get it

[clear lily]

solve integers

[mortal viper]

IDK HOW TO DO IT

[clear lily]

show an example

[mortal viper]

okay

[clear lily]

we can work through it

[mortal viper]

this is an example our teacher made us write in class today

im trying to do the homework and my friend tried helping me

and it was no use

[clear lily]

,rotate

[mortal viper]

i dont get it

at all

[clear lily]

you have the right answer

where do you get stuck?

$7 + 3 (4^2 - 10)$

[mortal viper]

i know but that was just an example our teacher wrote for us

[glossy valve]

heavy

[mortal viper]

i didnt do it

LMAO

[clear lily]

oh lol

[mortal viper]

hold on

heres a question im stuck on

the one circled

i dont understand how to solve it

[clear lily]

,rotate

[clear lily]

alright

[mortal viper]

like i get bedmas to an extent

[clear lily]

so we have $[5 + 3(8 - 2\times 3)]^2$

[glossy valve]

heavy

[clear lily]

dont think about the []^2 yet

first look at

[mortal viper]

okay

[clear lily]

$5 + 3(8 - 2\times 3)$

[glossy valve]

heavy

[clear lily]

do you know how to do this?

[mortal viper]

yea isnt it always do the brackets first

the thinga inside

things*

[clear lily]

yep

[mortal viper]

so 8-2x3?

or do i only do 2 numbers

[clear lily]

yes we look at 8 - 2x3

[mortal viper]

oky

[clear lily]

so thats m in bedmas

[mortal viper]

okay

yes

[clear lily]

so we do 8 - 2x3 = 8 - 6

because 2x3 = 6

[mortal viper]

okay

[clear lily]

so we now have whats inside the bracket

$5 + 3(8 - 2\times 3) = 5 + 3(8-6) = 5 + 3(2)$

[glossy valve]

heavy

[mortal viper]

the thing that i get stuck at is the formatting of it

like where i start the next line

and such

if u understand what i mean

[clear lily]

oh alright

[mortal viper]

like how do ik when to switch lines and do “=“

[clear lily]

well usually you want a string of = signs

[mortal viper]

yes

[clear lily]

like you would probably write this as

[mortal viper]

oh

wtf

[clear lily]

$5 + 3(8 - 2\times 3) \newline = 5 + 3(8-6) \newline = 5 + 3(2)$

[glossy valve]

heavy

[clear lily]

like that

[mortal viper]

oh

OH

oh

and then

after the last line

i do

3x2 right

or am i wrong

[clear lily]

exactly

3x2

so we get

[mortal viper]

6

so 5-6= 1?

the answer is 1?

wait

am i soow

slow

LMAO

[clear lily]

hahah its okay

weve all been there

$5 + 3(8 - 2\times 3) \newline = 5 + 3(8-6) \newline = 5 + 3(2) \newline = 5 + 3 \times 2 \newline = 5+ 6 \newline = 11$

[glossy valve]

heavy

[mortal viper]

so is this right

WTF

ok see

i get lost

and then i mess up

😭

[clear lily]

where did you get minus?

[mortal viper]

BECAUSE

3x2

=6

5-6

=1

LMAO

[clear lily]

but it says 5 + 3(8-6)

[mortal viper]

oh

YEAH

I GET STUCK

THERE

ALWAYS THERE

[clear lily]

alright so 8 - 6 = 2

[mortal viper]

and then

5+3 right

wait

[clear lily]

so you get $5 + 3(8-6) = 5 + 3(2)$

[mortal viper]

wtf

[glossy valve]

heavy

[clear lily]

$5 + 3(8 - 2\times 3) \newline = 5 + 3(8-6) \newline = 5 + 3(2) \newline = 5 + 3 \times 2 \newline = 5+ 6 \newline = 11$

[glossy valve]

heavy

[mortal viper]

OH

I GET IT

[clear lily]

very good

[mortal viper]

THANK U SO MUCH

😭😭

[clear lily]

im just happy to help

but we cant forget the last step

we had $[5 + 3(8 - 2\times 3)]^2$

[glossy valve]

heavy

[clear lily]

and now we know that $5 + 3(8 - 2\times 3) = 11$

[glossy valve]

heavy

[clear lily]

therefore

$[5 + 3(8 - 2\times 3)]^2$ = [11]^2$

[glossy valve]

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

[clear lily]

and that [] is the same as ()

its a bit confusing

[mortal viper]

ohh

[clear lily]

but its just normal brackets

[mortal viper]

oh i see

i get it kinda now

[clear lily]

$[5 + 3(8 - 2\times 3)]^2$ = (11)^2$

[glossy valve]

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

[clear lily]

and 11^2 = 11 x 11 = 121

[mortal viper]

OH

IHHH

okay i get it

[clear lily]

thats good to hear

[mortal viper]

thank u

[clear lily]

np

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[steep herald]

can someone help and explain the steps please 🙏🏿

[full forum]

@steep herald Has your question been resolved?

[sinful python]

try to think about what C being the midpoint of AG means for AC & CG, and hopefully that should help

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[astral sinew]

is there a good formula for the nth derivative of quotient?

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@astral sinew Has your question been resolved?

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[deft minnow]

The answer says, correct option is (c)....i don't know how

Questions 17*

Opps sorry ...i guess i figured it out .

. close

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[empty grotto]

hey can anyone help me with this

[empty grotto]

idk how my teacher got to where they got

[sharp vine]

which one exactly

[empty grotto]

a to start of

[sharp vine]

can you find vertex of quadratic?

[empty grotto]

no

ik how to make vertex tandard

but idk how to do it vice versa

[sharp vine]

we need the other way around here

[empty grotto]

oh vertex to standard?

[sharp vine]

no, standard is given and we want vertex

[empty grotto]

yea

[sharp vine]

so it's standrad to vertex

[empty grotto]

can u help me with doing that

idk how to reverse it back

[sharp vine]

idk if you are familiar with formulas for h, k? or you prefer completing the square

in y = a(x-h)^2 + k

[empty grotto]

we were not taught anything bout hk or k yet

so i assume were not doing it in this test

[swift bluff]

hold on

[empty grotto]

k

[swift bluff]

ok so the vertex of a parabola is at x=-b/2a

[empty grotto]

ohhh ik that one

we were taught that today

theres a b and c

[swift bluff]

yeah

so what do you get from -b/2a

[empty grotto]

i did not do taht yet

thx for the reminder 😄

[swift bluff]

k

[empty grotto]

doing it rn ill ping u when finished

[swift bluff]

k

[empty grotto]

x=30/2 times 4.9

right?

[swift bluff]

its "-30"

because a quadratic equation is in the form of ax^2+bx+c

you have the form of (-4.9)t^2+(-30)x+2

so a=-4.9, b=-30, c=2

[empty grotto]

ohh yeah i forgot u gotta reverse it

[swift bluff]

yep

[empty grotto]

so than i just answer it

[swift bluff]

yes

[empty grotto]

and sub in the rest

[swift bluff]

yep

[empty grotto]

lemme solve that rq

-73.5=x

[swift bluff]

nope

when you do -b/2a on this, you should get a value of about 3.06

[empty grotto]

so-30/2 times 4.9 is wrong?

for x

[swift bluff]

its -4.9

[empty grotto]

...

these lil mistakes gonna fuck me up

aight lemme fix it

[swift bluff]

the signs are important

[empty grotto]

oh yeah

i just put in my calcuklater

so now we put in vertex since we got x right

[swift bluff]

yes

[empty grotto]

do we put it in as p? or just as x

[swift bluff]

so plug in 3.06 into the equation and you will get your maximum height

[swift bluff]

i didnt get you

[empty grotto]

oh nvm

ye ima just plug in x

[swift bluff]

yep

[empty grotto]

47.91=h(t)

right? @swift bluff

[swift bluff]

yep

you got it

[empty grotto]

so would that be the max height?

[swift bluff]

yes

[empty grotto]

YESSSSSSSS

thank u we have been stuck on this for a while me and my friend

[swift bluff]

remember the vertex is the x value for the maximum or minimum y value of a function

[swift bluff]

good luck

[empty grotto]

:))

[swift bluff]

@empty grotto if you're done, please type in ".close"

[empty grotto]

i still got a bite more in me

[swift bluff]

alright

[empty grotto]

im just explaining what u told me to my friend

[swift bluff]

sure

[empty grotto]

he getting another answer in his calculater

he forgot to square smth

aight now my next question would be

for 8 b

were just solving the time taken for it to reach the flair

but idk what letter would represent taht or what were solving by like x p or q or a

OHH

its x

[swift bluff]

this is the graph of the path travelled by the flare. the x axis represents the time. What do you think is the time the flare was in the air?

first off, where was the flare launched?

[empty grotto]

like -2.8 smth

[swift bluff]

hmmm

how would you get that value from the equation?

[empty grotto]

i just guessed from the line

[swift bluff]

no. look at the graph. where was the flare launched

[empty grotto]

0/

[swift bluff]

at what value of y was the flare launched

[empty grotto]

• smth??

the line is under the bar

[swift bluff]

no. the flare was launched from a y value of 0

from the ground

in the graph, it would be the intersection point of the graph and the x axis

[empty grotto]

sp thats where it launched high in the air?

[swift bluff]

yes

and it lands somewhere around 2.55

[empty grotto]

okk

[swift bluff]

how would you get these values from the equation

[empty grotto]

i would look at my calculater table

if it wasnt in a neat line

[swift bluff]

this is what the graph represents

what you would do is equate the equation to 0

so -4.9t^2+30t+2=0

and then solve it

im assuming you know how to solve a quadratic equation?

[empty grotto]

yes

[swift bluff]

ok so you would get two values for t

find the difference between both and that should be your answer

[empty grotto]

isint t the t from earlier

that was 3.06?

[swift bluff]

no

i dont think you understand the concepts of a quadratic graph. just a sec

,rotate

[empty grotto]

3.1

[swift bluff]

the x value of the vertex represents the time from when the flare was launched to when it is at the maximum height

[empty grotto]

is the answer rounded

ye

so i fgured it would just be t

[swift bluff]

no one second

[empty grotto]

so 1?

[swift bluff]

its these two values

find the difference and its the answer

[empty grotto]

6.12205

[swift bluff]

yes

sorry i went to eat

its more about 6.24

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[full forum]

Please don't occupy multiple help channels.

[sharp flame]

Please stick to your old channel

[full forum]

@autumn cipher Has your question been resolved?

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[high vale]

Four math books, three ecology books, two music books and three economics books are to be arranged on a bookshelf. None of their books are identical. How many different permutations of these books are there?

[high vale]

<@&286206848099549185>

[full forum]

@high vale Has your question been resolved?

[thorny horizon]

this is very similar to ur previous problem

of arranging the letters of the word love

treat each of the book as different letters

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[thorn fjord]

Stuck here

Any advice?

[devout valley]

Were you given the value of R?

[thorn fjord]

Nope

(polar equation to cartesian)

[devout valley]

Ohhh I see, fair fair!

[thorn fjord]

If it helps, this is the answer

$x+\sqrt{3y}=8$

[glossy valve]

rainy

[rose rain]

oh they just want ya to convert it

[thorn fjord]

Yes

[devout valley]

Think you should have that an r here too btw

[rose rain]

my first thought was to try thinking of it as an equation involving complex numbers

[thorn fjord]

Still cant really make sense of it

[devout valley]

Wait is it 3/pi or pi/3?

[devout valley]

And is it $x + y\sqrt{3}$ or $x + \sqrt{3y}$?

[glossy valve]

chartbit

[thorn fjord]

First one, mb!

[devout valley]

Oh and also your LaTeX game is improving

[thorn fjord]

🧐

Are you spying on my latex

[devout valley]

Maybe

I'm always lurking, even in my sleep

Anyways, if it was $\frac{\pi}{3}$ that they meant, then it would be better as you would get to
$$
x\cos\left(-\frac{\pi}{3} \right) - y\sin\left(-\frac{\pi}{3} \right) = 4
$$

[glossy valve]

chartbit

[devout valley]

Both cos(-pi/3) and sin(-pi/3) being easy to work out

[full forum]

@thorn fjord Has your question been resolved?

[thorn fjord]

Yep. I just realised pi and 3 are flipped lmao

Let me quickly redo it

[thorn fjord]

Yeah with the CORRECT numbers this wasnt too hard 🤣 , Thanks

Guess im blind

[devout valley]

No worries it happens at times!

I was like "damn they must have been mean to set that for you!"

[thorn fjord]

Same lol

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[atomic vortex]

so ik the point of intersection

x = 6 and y = 12 right

but then

when integrating

do I do 0 to 12 thats all?

and then if I do on y axis the x and y switch right

so x = 12 and y = 6?

for intersection

[hexed solar]

Is it necessary to use integration here

Like is it expected in course

[atomic vortex]

yes because im doin calc II

so its basically all

integration

[hexed solar]

Ok sorry for the interruption

[red cypress]

That would only be the intersection of the two first. It's also bounded by x=0. If you graph these it would maybe help you

[atomic vortex]

the 0,18 point?

[atomic vortex]

so I would integrate from 0 to 12 right and then subtract

6 to 12

ahhhhh idk

[red cypress]

Well it's being rotated around the y-axis

Think about how you slice up the solid to integrate, which direction do you slice?

[atomic vortex]

oh since this is y

it will be on the left side

right?

like it wont go down

Im still confused

about what my bounds will be

[torn jolt]

Whats the intersection point?

[atomic vortex]

x = 6?

we only want the area of the shaded side right?

$pi \int_{-6}^{12} (\frac{y+6}{3}) ^2 , dy - pi \int_{12}^{18} (18 - y)^2 , dy$

[glossy valve]

Calc II Victim

[atomic vortex]

or wait since

its y

0 to 12

12 to 18?

$\pi \int_{-6}^{12} (\frac{y+6}{3}) ^2 , dy - \pi \int_{12}^{18} (18 - y)^2 , dy$

[glossy valve]

Calc II Victim

[atomic vortex]

brah why isnt it working

[gaunt nest]

Why don’t u just find area of two cones

That formed

Instead of integrating

[atomic vortex]

omg

im stupid

I wrote a negative

it was +

[gaunt nest]

Silly mistakes I feel ur pain

[atomic vortex]

I understand now

finally

[torn jolt]

Integral 0 to 6, 2pi(x(24-4x) dx I think would work

[atomic vortex]

why 24-4x

[torn jolt]

18 - x, 3x - 6

[atomic vortex]

also idk but like is it kind of unfair that im graphin these on desmos?

after finding the point of itnersection

[torn jolt]

Subtract them

[atomic vortex]

I see but they wanted wrt to y

not x

[torn jolt]

That's for the rotation

[atomic vortex]

just wanna make sure

but

like when they say interval

they r talking abt on the y axis?

or x axis?

[0, 1]

[torn jolt]

This is different

[atomic vortex]

oh i know im done the prev question

I just wanna know if they are referrin to the points on y axis or x axis for this

before I start

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[lyric cargo]

order matters for like

word problems or word choices

is 2,4 different from 4,2?

say for a

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[olive tendon]

is the following statement true?

[olive tendon]

2<4>3

[river mulch]

yes

[olive tendon]

thanks

.close

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[plain dune]

how do i rearrnage this

[plain dune]

to get

ignore that star

I have spent a long time on this, and its really hard i dont know why

<@&286206848099549185>

[stable heart]

do you want help to get there or just show that it is indeed true?

[plain dune]

get there

[stable heart]

have you tried anything yourself?

if so then please show your work and i can guide you in the right direction

[plain dune]

this is the question

i got to here after differentiating using product rule

but i cant reach that answer

no matter what

[stable heart]

i did the problem myself so this is right

[plain dune]

i dont even understand how they went from the thing at the top to the bottom?

i feel like im missing something

[stable heart]

\begin{align*} t & =\frac{25-\ln(t+1)}{1+\ln(t+1)} \ &= \frac{25-\ln(t+1)+1-1}{1+\ln(t+1)}\ &=\frac{25+1}{1+\ln(t+1)}-\frac{\ln(t+1)+1}{1+\ln(t+1)}\ &=\frac{26}{1\ln(t+1)}-1\end{align*}

[glossy valve]

Duh Hello

[stable heart]

its just a trick where you add one and subtract one from the numerator to get the ln(t+1)+1 term to be in the numerator so you can reduce it to 1

[plain dune]

thank you @stable heart

i have not seen this before

[stable heart]

but either way, to get there first you have this right

multiply both sides by (t+1)

then distribute and collect the terms with t (but not the one inside the ln)

if you instead rewrite the equation to $$\frac{10-0.4t}{t+1}=0.4\ln(s)$$where $s\equiv t+1$ then its just about solving for $t$ now

[glossy valve]

Duh Hello

[stable heart]

maybe that becomes a little bit easier for you?

[plain dune]

a bit easier

im still struggling with it

i do udnerstandthis tho:
\begin{align} t & =\frac{25-\ln(t+1)}{1+\ln(t+1)} \ &= \frac{25-\ln(t+1)+1-1}{1+\ln(t+1)}\ &=\frac{25+1}{1+\ln(t+1)}-\frac{\ln(t+1)+1}{1+\ln(t+1)}\ &=\frac{26}{1\ln(t+1)}-1\end{align}

[glossy valve]

xboxking_55

[plain dune]

but not how you get there

[stable heart]

yeah thats what im trying to help with. so we now have $$\frac{10-0.4t}{t+1}=0.4\ln(s)$$what do you propose is the first thing we should do here to solve for $t$?

[glossy valve]

Duh Hello

[plain dune]

i would times 0.4ln by t+1

[stable heart]

so just this?$$\frac{10-0.4t}{t+1}=(t+1)0.4\ln(s)$$

[glossy valve]

Duh Hello

[plain dune]

i mean make t+1 replaced by s too

yes but it would also remove that fraction

[stable heart]

ok good

just making sure you are on the right track

but in this particular case we do NOT want to replace t+1 with s

[plain dune]

why

[stable heart]

we are pretending that only the t inside the ln is irrelevant

since we can see in the final answer they leave it on one side

[plain dune]

ok

[stable heart]

but we need to get rid of all the other t's

so we have this now right?$$10-0.4t=(t+1)0.4\ln(s)$$

[glossy valve]

Duh Hello

[plain dune]

yes

[stable heart]

what do you think we should do here?

[plain dune]

we can group the ts in one side

[stable heart]

yep, how would we go about doing that?

[plain dune]

so move -0.4t to the same side as 0.4ln(s) and expanding that

so we get 1-0.4ln(s)=0.4tln(s)+0.4t

then factorise t out

then divite

divide

[stable heart]

so we have $$10-0.4\ln(s)=0.4(\ln(s)+1)t$$and you say we divide so now we have $$t=\frac{10-0.4\ln(s)}{0.4(\ln(s)+1)}$$

[glossy valve]

Duh Hello

[stable heart]

now this is starting to look pretty familiar to me dont you think?

[plain dune]

i got 1 instead of 10

[stable heart]

ah i thought it was a typo

[plain dune]

oh wait sorry nvm

[stable heart]

so how did you get 1?

[plain dune]

type

typo

ye

[stable heart]

👍

[plain dune]

ok yea i have what you sent but now im stuck

on that

[stable heart]

well now we seem to be very close to getting there, now we can look that they want us to have $\ln(s)+1$ in the denominator

[glossy valve]

Duh Hello

[stable heart]

so what can we do to make that be the case?

also we can switch back to using t+1 instead of s now

[plain dune]

times by 0.4 but then we will also have that on the t, on the other side

[stable heart]

that is true. but what we can do instead is to instead of multiplying both sides by 0.4 we can multiply AND divide the right side to not need to change the t

[plain dune]

you cant just divide the right side?

[stable heart]

you can however multiply the right hand side by $\frac{0.4}{0.4}$ since this is just 1

[glossy valve]

Duh Hello

[stable heart]

$$t=\frac{0.4}{0.4}\cdot\frac{10-0.4\ln(t+1)}{0.4(\ln(t+1)+1)}$$

[plain dune]

0.4 * 0.4 != 0

[glossy valve]

Duh Hello

[stable heart]

good thing we are multiplying it and not adding it

[plain dune]

i mean multiplying

[stable heart]

multiplying by 0 would indeed change both sides

[plain dune]

you will not get rid of the 0.4

[stable heart]

but multiplying by 1 doesnt change anything

[plain dune]

you will have another number

whats the next step

[stable heart]

might be easier to see what this does if i instead write it like this $$t=\frac{\frac{0.4}{0.4}(10-0.4\ln(t+1))}{0.4(\ln(t+1)+1)}$$

[glossy valve]

Duh Hello

[plain dune]

ok, whats the next step

[stable heart]

well now we can cancel the 0.4 on top and the 0.4 in the denominator

$$t=\frac{\frac{1}{0.4}(10-0.4\ln(t+1))}{\ln(t+1)+1}$$

[plain dune]

oh ok

[glossy valve]

Duh Hello

[plain dune]

now what?

[stable heart]

well now you just distribute it

and you get here

[plain dune]

thank you

you are very helpfull

thank you alot for your time, i was stressing over this alot 😃

[stable heart]

no worries

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[torn jolt]

hello

how can i do part i)

[jade radish]

do u know how to do z scores?

[torn jolt]

is that inverse normal

[jade radish]

what?

wym

[torn jolt]

where did you apply 491 times @jade radish

just curious

[jade radish]

Do u know how to calculate a z score

[torn jolt]

um yes

[jade radish]

I have a long list of most of the places in another discord I can screenshot to u

[jade radish]

Do that first

[torn jolt]

go ahead

i got z = 1/2

is that right

and what do i do after that

[jade radish]

One sec

Lemme do it

,w (240-247)/14

[glossy valve]

Results provided by WolframAlpha

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

Requested by Jukelyn#9848

[torn jolt]

isn't it 247-240/14

what should i do after i get z

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[heady fulcrum]

could someone please help with part iii of this q. Sadly I dont have a MS, but I have an answer - just dk if its right or not

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[split heath]

for example if i have 5x/3 + 2(x-2)/3 what do i do

[split heath]

cause i know ijust add numerator

5x+2(x-2)

[split heath]

but do i add it like this^

or like this: 5x + 2x -4 or
5x+ (2x-4)

[ebon spoke]

Yea

Or u doing fractions

[split heath]

?

which way?

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[calm field]

hello, im trying to convert this 14 bit binary number to decimal: 11 1111 1100 1000.
what are the steps in doing this? im confused if this would be positive or negative.

[safe trench]

You begin with 2^0 from right

2^0 * 0 + 2^1 * 0 +.... 2^13 * 1

Actually, it depends if the binary number is unsigned, signed or 2s complement

I assumed it's unsigned

[full forum]

@calm field Has your question been resolved?

[calm field]

its a 14 bit 2s complement

[safe trench]

So it's negative since the MSB is 1

Now do you know how to convert it to it's positive compliment?

[calm field]

hmm i think u first take its complement

then add one to it

[safe trench]

Yeah you basically invert all the bits and add 1

For example if it's 110 you do 001 + 1 = 010 which is 2 in decimal so the answer is -2

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[torn jolt]

i got this as false

is this right

because cot is cos/sin

[maiden lodge]

👍🏼

[torn jolt]

and if u mult 1 by sin/cos thats not equal

ok ty openglobe

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[torn jolt]

.reopen

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✅

[torn jolt]

i did this by doing inverse of cos

and i got pi/6

then i subtracted 2pi

and got -11pi/6

but thats not an option

[maiden lodge]

Take the absolute value

[upbeat comet]

pi/6 and 11pi/6 is the result

[torn jolt]

y do i take absolute value

[upbeat comet]

Personally I remembered a table for exact values of sin/cos/tan

Looks like this basically

[torn jolt]

did u sit and memorize it like daily until u knew it

[upbeat comet]

It's actually really easy to remember once you learn how it works

[upbeat comet]

But to answer this question

You did the subtraction backwards

[torn jolt]

if i add 2pi

i get 13pi/6 tho

[upbeat comet]

Which has a relative acute of pi/6

[torn jolt]

in my head rn

i think pi/6 is the only angle that is in 0 and 2pi

[upbeat comet]

11pi/6 is also between 0 and 2pi

because it's 11/6 * pi, where 12/6 * pi = 2pi

[upbeat comet]

Here's how to read/remember this diagram

You write the numbers 0 through 4

And beneath it write them again but backwards

Like this

0 1 2 3 4
4 3 2 1 0

Easy to do

Now wrap the whole thing in one big fat square root

For the top row of numbers, you have sin, the bottom row you have cos

[torn jolt]

yea but how would i get positive 11pi/6

if when i subtract 2pi i get negative

im thinking it becomes outside 0 and 2pi range

[upbeat comet]

2pi - pi/6 not pi/6 - 2pi

[torn jolt]

oh

big mistake by me

[upbeat comet]

Anyways do you want me to explain the diagram I don't have to if you don't think you'll use it

But I personally found it very helpful for those exact angles it's one of the few things I just decide to memorize in math lol

Because in my school they tried to get us to memorize it like this

Which I found to be way harder since my memory is poop

[torn jolt]

i see

then u put 2 under

[upbeat comet]

And on the top you have your angles 0 to 90, and you can also write them in radians

[torn jolt]

did someone tell u this or u made it up this way

[upbeat comet]

so when someone says what's sin(60) I look at the first row of numbers (because sin), and look under 60, and see 3
So I say it's sqrt(3)/2

[upbeat comet]

No I found this many years ago lol

I definitely didn't make it

So cos(0) = sqrt(4)/2 = 2/2 = 1

Because that's what the diagram shows

And then if you want to know tan of an angle, like tan(45), you just use the table to get sin(45)'s exact value, cos(45)'s exact value, and then divide them

since tan(x) = sin(x)/cos(x)

[torn jolt]

will i use trig in calc 1

[upbeat comet]

Yep

Trig never leaves you lol

[torn jolt]

wow ok

ty for helping me

this one idk either

i got pi/6

by taking inverse of tan

but i added and subtracted 2pi

and dont get 7pi/6

[upbeat comet]

pi/6 is 30 degrees, and is also positive for tan

Which means, the related angle must also be positive for tan

So looking at the CAST diagram

We see that in order for the angle to be positive for tan, it must be in the bottom left quadrant

So, given x, where

$0 < x^{\circ} < 90$

[torn jolt]

7pi/6 is in q3

[glossy valve]

Kappa Mikey

[torn jolt]

oh ok

[upbeat comet]

We see we need add 180 degrees, or pi to our answer

[torn jolt]

so how would i get 7pi/6 from pi/6

[upbeat comet]

Let me explain how I'd do the question from the beginning

We have

$\tan t = \frac{\sqrt{3}}{3}$

[glossy valve]

Kappa Mikey

[upbeat comet]

Find t on the interval [0, 2n)

Okay, cool

First, I use whatever method I like to find the value of t

Again, personally, I use that diagram I sent before

You can use your method

And I determine

$t = \frac{\pi}{6}$

[glossy valve]

Kappa Mikey

[upbeat comet]

Now, given that this is an acute angle, I know where it resides

In this cast diagram, we see anything less than pi/2 is in "A"

Meaning "All"

This diagram represents the values of x in which a trig function is positive or negative

So if my angle is in A, all the trig functions are positive,
If it's in S, only sin is positive

T, only tan is positive, and C, only cos is positive

And in the problem

[upbeat comet]

Right here, we see our number is positive

So we need a positive value for tan

Well, we already found one in the "All" quadrant, it's pi/6

[upbeat comet]

In order to find the other one, according to this CAST diagram

I need to add 180 degrees to my angle

But we're using radians, so that's just pi

So that means I need to say

$\pi + \frac{\pi}{6}$

[glossy valve]

Kappa Mikey

[torn jolt]

oh ok i see

[upbeat comet]

$= \frac{6\pi}{6} + \frac{\pi}{6} \
&= \frac{7\pi}{6}$

[torn jolt]

i tried this method u just taught me on another problem and i messed up

[glossy valve]

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

[upbeat comet]

Oof let's take a look

[torn jolt]

this one on the graph

says 2pi - angle

https://media.discordapp.net/attachments/903486480985489478/1073058710294909038/image.png

i got pi/4

oh i get 7pi over 4

[upbeat comet]

Yep lol

$2\pi - \frac{\pi}{4} = \frac{7\pi}{4}$

[glossy valve]

Kappa Mikey

[torn jolt]

so if the denominators are the same i can just subtract the numerators

[upbeat comet]

Yep

[torn jolt]

ok ty

[upbeat comet]

$\frac{8\pi}{4} - \frac{\pi}{4}$

np

[glossy valve]

Kappa Mikey

[torn jolt]

what if its a sqrt root

like this one

[upbeat comet]

Wdym they're all square roots lol

You mean negative?

[torn jolt]

yea

[upbeat comet]

Same rules apply just don't forget the negative

It's still 2pi - angle

Since your angle is negative, that should work out to

$2\pi + |x|$

[glossy valve]

Kappa Mikey

[upbeat comet]

Where x is your angle in radians

Actually wait

My bad

[torn jolt]

i would get 4pi/2 + sqrt2/2

[upbeat comet]

Cos is an even function

[torn jolt]

but how do i add 4pi to sqrt2

also this one

[upbeat comet]

Here we see sin(90) = sqrt(4)/2 = 2/2 = 1

So the answer is 90

Rip why'd you delete it

[torn jolt]

cuz i put it wrong

this one tho with -sqrt

is confusing tho

[upbeat comet]

Negatives don't change anything

Not really

If you wanna simplify it

Multiply both sides by -1

So your new problem is

$-\cos t = \frac{\sqrt{3}}{2}}$

[glossy valve]

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

[torn jolt]

i mean this is it right