help-27

[finite gale]

2

And 5

<@&286206848099549185>

[devout snow]

@finite gale Has your question been resolved?

[latent quarry]

Prove that triangle JMN is similar to triangle MNL

[latent quarry]

If you can prove this

The question can be solved immediately

[finite gale]

how

[latent quarry]

Let Angle MJN = x

Then the angle NMJ = 90-x

[finite gale]

im sorry im not understanding

[latent quarry]

Hmmm

What is the total amount of angle in a triangle

[latent quarry]

Answer this first

Then explain what is similarity in geometry

This is the 2 material we are going to use to solve this question

[finite gale]

180

[finite gale]

Between the two triangles? They both have right angle

[latent quarry]

No

Simmiliarity is when the shape have same angle but one is the enlargement of other

The similarity test in a triangle is if 2 triangles have 2 same angles then they are similar

[latent quarry]

If a triangle has angle 90, x what is the value of the third angle

[finite gale]

90-x?

[latent quarry]

JMN = 90-x right

But JML is 90

So angle NML = ?

[finite gale]

90-x i think?

Or wait

Hold up

Its 90-JMN angle?

[latent quarry]

LMN = X, MNL = 90, we already prove that our triangle is similiar

Do you understand about similarity or should I explain it

@finite gale

[finite gale]

Sorry for the late responses, im tryna do smth rn

[finite gale]

Explain a lil more

[latent quarry]

We know that the smaller triangle have side of 15,8,17

Remember the smaller triangle have base of 8 cm

While the larger triangle have base length of 15

Remember similiar triangle is just a enlargement of the same triangle

So

8*x = 15

Find x and now multiply 15*x and add it by 8 that is the answer

Draw it on paper to help you visualize it better

Make same triangle

One with base 8 and another with base 15

Remember the smaller triangle can be enlarged to the larger triangle by a scale factor

The scale factor is 15/8

[latent quarry]

It is the ratio of the same side length of the 2 triangle

[devout snow]

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[finite gale]

Tysm

[devout snow]

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[vestal bay]

ABC is an inscribed triangle in a circle with point O as the center. From point D on the arc BC, the angles perpendicular to the lines BC, AC, and AB are dropped at E, F, and G respectively.
a. Prove that quadrilateral DEFC is a cyclic quadrilateral.
b. Prove that E, F, and G are on a straight line.

Could someone show me how to draw this figure? (Sorry if the problem sounds a bit off, I translated it from my native language.)

[devout snow]

@vestal bay Has your question been resolved?

[devout snow]

@vestal bay Has your question been resolved?

[devout snow]

@vestal bay Has your question been resolved?

[vestal bay]

<@&286206848099549185> anyone?

[valid silo]

for a. notice the angle DEC and DFC

for b.

3 points are on a line if their angle is 180

try using angle chase, and the fact that DEFC and DEGB are cylic

you are done if you show CEF = BEG

[vestal bay]

I don't know how to draw the figure

[vestal bay]

<@&286206848099549185> (I need help on how to draw the figure)

[valid silo]

draw a circle

and take 4 points on it in order ABDC

make lines

AB

AC

and

BC

then make perpendicular lines from D onto AB, AC, and BC, and that is the diagram

[vestal bay]

I can't make them perpendicular

[valid silo]

you can, if you make lines

not line segments

[strange arch]

not sure why nubudy provided a sketch :c

[vestal bay]

Ohhh

thankkss

[strange arch]

np, odd this was open for 5h :D

are the proof layouts themselves kind of clear?

[vestal bay]

Wdym?

[strange arch]

solving a) and b), that is

[vestal bay]

I'm a little confused on a.

[strange arch]

@valid silo did provide useful tips :]

which property do DEC and DFC share that implies that DEFC is cyclic

[vestal bay]

The opposite angles?

[strange arch]

yoppa

[vestal bay]

One question, why doesn't DEFC & DEGB look like a cylic in the sketch?

[strange arch]

I didn't draw accurately lol

I used the state-of-the-art professional-approved widely in the industry used software ms paint

[vestal bay]

I used like desmos geometry to draw it accurately and it looks similar to yours

[strange arch]

I mean it looks fairly cyclic

[vestal bay]

Oh we are supposed to draw extra circles?

[strange arch]

nono

those are the circles

a cyclic quadrilateral is a quadrilateral with all of its points on a circle

so I drew the circles since you meant DFBG and DEFC don't look cyclic

regardless, for a) note that for any cyclic quadrilateral

DEC + DFC = 180° must hold

and since DEC is right-angled (perpendicular lines), DEC = 90°

and also DFC = 90°

thereby DEC + DFC = 180° is true

thereby DEFC is a cyclic quadrilateral

[vestal bay]

What property is that?

[strange arch]

and for b) I'd use the advice shreyanjha mentioned

[strange arch]

Opposite angles of a cyclic quadrilateral are supplementary

usually the main way to classify a cyclic quadrilateral

[vestal bay]

I don't understand the opposite part

[strange arch]

A+B = 180°

C+D = 180°

[vestal bay]

Ohh okayy

[strange arch]

hm you must have had some way to classify cyclic quadrilaterals in your notes

otherwise how would you know if a quadr. is cyclic

[vestal bay]

I'm a little confused on how DEC and DFC are opposite angles

[strange arch]

DEC and DFC aren't but when the diagonals fulfill DEC + DFC = 180° then the rule holds

[vestal bay]

Could you explain this a bit more? I don't really understand the fulfill part.

[strange arch]

ah sure misunderstood what you were confused by, it implies that the given circle must exist, wait I'll draw it

[strange arch]

you have two right angles at the upper points

which lie on the semicircle defined by the line at the bottom

implying all four points lie on the circle described by the semicircle

it essentially directly tells you DEFC is cyclic

[vestal bay]

Ohh okayy

I still don't really understand where does the DEC + DFC =180⁰ come from

[devout snow]

@vestal bay Has your question been resolved?

[vestal bay]

@strange arch could you help abit more?

[devout snow]

@vestal bay Has your question been resolved?

[devout snow]

@vestal bay Has your question been resolved?

[paper surge]

it comes from this property, its in polish but i think you can understand from the picture

[paper surge]

in this, 2alpha= 180

[vestal bay]

How would I use that to prove its a cylic?

[paper surge]

idk, i'm just anwsering your question

[devout snow]

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[still rover]

No idea how to go about solving this problem

[frosty mantle]

are you familiar with how to calculate the result of a definite integral if you have the function written out

[still rover]

that's the fundamental theorem right?

[frosty mantle]

yes

let me get an image for simplicity here

[frail hull]

hi guys i m new to this server

[frosty mantle]

hello

you can visit #❓how-to-get-help for assistance

[frosty mantle]

so here you see that

[frail hull]

ohh thanks

[frosty mantle]

you’re asked to solve the form on the left

and you’re given values for the form on the right

are you familiar with the form of an indefinite antiderivative

[still rover]

yes

[frosty mantle]

you always add a constant +c

but that constant +c can be anything

so theres multiple forms of the antiderivative

but they only differ in the +c constant

[frosty mantle]

so at x=6 you’re given the values for both F(6) and G(6)

you can calculate the difference in the +C

and from there you can find the values of all F(x) from 1 to 6

and then you can apply the fundamental theorem

[still rover]

ohh ok

so F(5) would be 14?

am i going about this correctly

[sterile geyser]

correct

[still rover]

ok ty

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[topaz scaffold]

A radio tower has a broadcast that can be heard in a circular area with a radius of 40 miles.
Suppose that you are driving along a straight road beginning 45 miles north of the radio tower
and ending 50 miles east of the radio tower.
Consider the radio tower to be at the origin of the Cartesian plane, i.e. the point (0,0).

(a) What is the equation of the circle that the radio broadcast can reach?

$x^2 + y^2 = 1600$

(b) What is the equation of the line the car is driving along?

$y=-\frac{9}{10}+45$

(c) What are the two points where the line intersects the circle?

$x^2+(-\frac{9}{10}+45)^2=1600$

what do I do after this?

[woven radish]

227466

[tall knoll]

the first thing you do is patch up some discrepancies

firstly, where do these two different fractions come from

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[topaz scaffold]

it was supposed to be 9

so then you insert it into x^2 + y^2 = 1600 as the y^2

[devout snow]

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[lime wagon]

how to integrate this 💀

[topaz scaffold]

eh

what happened to my question bro

[lime wagon]

it is not in root over a^2 - x^2

[trail eagle]

Just u sub 16-8x

[lime wagon]

huh 😭

[strange arch]

u = 16 - 8x

if you insert the substitution into your integral it becomes easy to solve

[lime wagon]

damn

sheesh

[strange arch]

just determine the integration values

and sub

[lime wagon]

yes

done

thanks a lot

damn

it wasnt that hard, i was makin it hard in mah head 😭

[lime wagon]

thanks 😭

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[devout snow]

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[dull thunder]

how are we using logistic equations to solve this problem...?

[dull thunder]

i don't understand the 2nd and 3rd one...

[final drift]

so part 2 is asking when the maximum of $\frac{\mathrm{d}M}{\mathrm{d}t}$ is right?

[woven radish]

Arnavutköy

[dull thunder]

yeah?

[final drift]

so the maximum of $\frac{\mathrm{d}M}{\mathrm{d}t}$ will happen when $\frac{\mathrm{d}^2M}{\mathrm{d}t^2}=0$ right?

[woven radish]

Arnavutköy

[dull thunder]

ohh is it like local max min style stuff...? that the derivative should equal 0 and ....

[final drift]

yes

[dull thunder]

so do i take the derivative or?

[final drift]

yes

[dull thunder]

what should i do next...?

[final drift]

set it equal to 0

and solve for M

[dull thunder]

ohh i get the answer 150

what about 3rd one?

[final drift]

okay so you agree that one is correct right?

[dull thunder]

yeah i solved that

[final drift]

moreover, you would have noticed, that when $0<M<300$, that the population of meerkats is increasing right?

[woven radish]

Arnavutköy

[dull thunder]

yes

[devout snow]

@dull thunder Has your question been resolved?

[devout snow]

@dull thunder Has your question been resolved?

[dull thunder]

.close

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[dense scroll]

Maybe a dumb question but I was trying to figure out how to calculate the volume of a solid made by rotating a curve around the x axis (like Gabriel's horn) and what I thought of was first expressing the area under the 2d curve with an integral from 0 to infinite or whatever to infinity, then integrating that times the area of an arc of a circle dθ from 0 to 2pi but it doesn't work,

And I found out a common standard method is integrating the volume of a disk with radius f(x) from 0 to infinity over

It seems like the methods are pretty much equivalent but turns out I'm wrong;

Could someone explain why my method is wrong?

[tall cypress]

what exactly are you trying to do?

find the volume of gabriel's horn by rotating volume of solid?

[violet wind]

"integrating that times the area of an arc of a circle dθ"

what?

[dense scroll]

no I meant like
I tried to derive on my own a method to calculate the volume of a solid constructed by rotating a curve

[dense scroll]

I'll try to explain myself better
Find the area under the 2d curve by integrating from 0 to infinity, then multiply that by slices of a circle

[tall cypress]

you wanna make sure that the integral of that curve from 0 to infinity converges

[dense scroll]

True

[tall cypress]

but have you heard about volume by rotation about x axis

[dense scroll]

Maybe assume the function is just defined between 2 finite bounds
So like from 1 to 5

[dense scroll]

Wdym

Like the standard method for calculating its volume?

[orchid axle]

i feel u

[tall cypress]

the formula is $\pi \int_{a}^{b} (f(x))^2 dx$

[woven radish]

Slender

[dense scroll]

Yeah but I wanted to derive it on my own

I'm trying to figure out what's wrong with what I did because it seems like it's the same thing but in different ways

[tall cypress]

do you know how to find the volume using area of cross section

[dense scroll]

Calculate the area of a disk and integrate over the x axis or calculate the area of a slice and integrate over theta

[dense scroll]

Yeah with the thing you just sent

[tall cypress]

well it's different when you do polar coordinates

[dense scroll]

I'm not looking for practicality I just wanted to do it myself and I made something that in my head should work but doesnt

[tall cypress]

double integral comes when you try to do in terms of polar coordinates

[dense scroll]

I'm not doing polar coordinates though I mean like

Imagine for a second the volume of a slice of Gabriel's horn; not a disk, but a slice as if you were looking at it from negative x but looking to positive x and slicing it like a cake

How would you calculate that

[tall cypress]

sure

so you wanna sum up all slices

to calculate the volume

[dense scroll]

Yeah

I think my error is in the way I calculate the slices

[tall cypress]

Gabriel's horn is generated by rotating 1/x graph

[dense scroll]

Yeah

[tall cypress]

do you know the logic behind this formula

[dense scroll]

Standard way is calculating a cross sectional disk and integrating over the x axis yeah

I wanted to do it another way, calculating a slice instead of a disk

Do you see what I mean or not

[tall cypress]

Something like this?

[dense scroll]

Yeah I guess

[tall cypress]

Yeah so basically

You want to find the area of that slice

Since that slice is "rotated"

It forms a circle

[dense scroll]

Yeag

[tall cypress]

but the radius is variable and changing

[dense scroll]

Yup

[tall cypress]

so let the radius function be R(x)

[dense scroll]

Ok

[tall cypress]

now your area of cross section is pi * R(x)^2

Makes sense?

Each slice is of size pi * R(x)^2

[dense scroll]

Wait but that's just the standard way

[tall cypress]

Yup thats how it is

[dense scroll]

Ok so it's not what I meant

[tall cypress]

You essentially slice

[dense scroll]

Guh

I don't mean slice across the x axis

[tall cypress]

So not like this?

[dense scroll]

Like the standard way cuts cuts that are parallel to the xz plane

Right

[tall cypress]

Okay

You want to look into 2d integration

using polar coordinates

[dense scroll]

I guess that's what I need

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[frozen aurora]

why is f true?

|u(t)| is a scalar function

but the result is a vector

[thin inlet]

is it?

• denotes dot product

so u(t) dot u'(t) is a scalar function

@frozen aurora

[ancient sluice]

|a| = (a•a)/|a|

[ancient sluice]

How did you type that dot ?

[frozen aurora]

ohh

yep i missed that

[thin inlet]

its just because * is in front

[frozen aurora]

thanks

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[viral lynx]

I think this is not the way they want me to solve it, but is there a way to do this by turning it into arctan?

If you put the 3 to the left of the integration symbol it becomes 1/(x^2 - 16)

Which is pretty close to 1/(x^2 + 1)

But im just wondering about that negative sign, cause I cant put that inside a square root

Like if it was 1/(x^2 + 16) I could easily turn it into arctan(x/sqrt(16))/sqrt(16)

Like this^^

[violet pine]

i think this is more of a integration by partial fractions

do you know how that works

[viral lynx]

I think so, but im just wondering if this is possible

[feral agate]

you can write this as $$\frac1{\sqrt{-16}}\arctan\left(\frac{x}{\sqrt{-16}}\right)$$

[viral lynx]

Im just curious hehe

[woven radish]

kheerii

[viral lynx]

Yeah thats what im wondering, is this allowed?

[feral agate]

sure

[viral lynx]

Cause the negatives are inside the square root

[feral agate]

but it's not a very nice form to use

[feral agate]

yes which means you'll have to deal with complex numbers

[violet pine]

if that does not work then it is partial fractions

[winter torrent]

i would not recommend that approach lol

[violet pine]

this problem is mostly suggesting to use partial fractions

[viral lynx]

Yeye I was just wondering if it was possible to do it this way 😝

So I guess the answer is yes but dont

[violet pine]

alr

[winter torrent]

you can, and you can do the opposite (using partial fractions for 1 / (x^2 + 1) ) but it gets very messy

[viral lynx]

Ahh oke 😅

Well that answers my question, thank you everyone!!

❤️

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[lone stirrup]

The pyramid ABCT has a triangular base where AB = 6 4. cm, AC = 5 8. cm
and ∠A = 45°. The side edge AT is 7.4 cm long and forms the angle with 60° with
the ground surface.
a) Find the area of ​​the base.

My first approach was to use the area formula

1/2 * h * b * sin(v)

But that resulted into 1/2 * 5.8 * 6.4 * sin(45) = 13.12cm^2

Why cant I use the area formula here?

[rigid sandal]

anyone who wrote amc8 today?

[devout snow]

@lone stirrup Has your question been resolved?

[devout snow]

@lone stirrup Has your question been resolved?

[devout snow]

@lone stirrup Has your question been resolved?

[devout snow]

@lone stirrup Has your question been resolved?

[marsh osprey]

height would be different

it is the perpendicular distance from the point above

oh you did that

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[dire osprey]

Hi, I am struggling to understand this unit as a whole but especially how to decide whether a value will be less than and equal to, less than, etc than x/y! For example, why is x greater than/equal to -3 in this ss? I sorta forgot to restudy this unit before my final tomorrow and am freaking out a bit hahaha. Thank you!

[wind mason]

look at the graph

[dire osprey]

I'm looking 😭

i have been i promise

i just dont understand

[smoky nimbus]

Observe the function, notice how the function starts at the left then goes right, meaning x values are getting larger, larger than -3

[dire osprey]

Wait this makes sense thank you!! Let me try it out

I think im still lost. i wrote {x|-2>- x >-4

[gleaming socket]

Correct

[gleaming socket]

Do the y axis too

[dire osprey]

sorry the picture i sent is the teacher copy hahaha

[smoky nimbus]

It's an closed circle at x = 4

So it's equal to

[dire osprey]

omg the points decide that??

[smoky nimbus]

Yes

[dire osprey]

i thought they just decided between the brackets when writing coords

i see

'thank you

[smoky nimbus]

That's also another way to write it too

[wind mason]

you can also just think of it as a lower and upper limit

especially for the y-values

[dire osprey]

thanks for the help guys i think im just not cut out to get this lol i dunno

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[gray axle]

Translation for Ex 9
Idk what to do here, idk whhat to do even in "1-"

[strange arch]

start by inserting z

@gray axle

from here try to simplify as much as possible

crunch everything into the ground

[gray axle]

Ty, I will open new ticket If something else is difficult for me

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[devout snow]

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[finite gale]

!status

[devout snow]

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

[finite gale]

2

,rotate

[gleaming socket]

Show work

[finite gale]

tbh i was just putting 1 at each side for the hexagon

i got like nothing done, i got so stuck

[strange arch]

the fraction grayarea/totalarea is saught

[finite gale]

yes

[strange arch]

so you can work with any value for the sidelength or totalarea

starting with totalarea=1 can be fairly intuitive

but you can also use sidelength = 1

if you know the side-length is 1

what's the area of the hexagon?

[dark tundra]

I see there are some lines in your hexagon

Perhaps you can try forming those little triangles all over the hexagon

[finite gale]

i mean i used sidelength = 1, and got 2.6

[finite gale]

ill give it a shot

[dark tundra]

Sometimes the solution is much easier than you think

[strange arch]

notice these triangles occur 6 times in the hexagon

so you can also perceive the hexagon's area as "6 * GreenTriangle"

but how many green triangles is the gray area?

[finite gale]

1 and a bit right?

[strange arch]

how much a bit

[finite gale]

oh yes 1.5

i just saw it in my head

[strange arch]

kk

[finite gale]

wait is that it?

[strange arch]

sure

[finite gale]

what do we do now

[strange arch]

1.5 / 6 green triangles

= 1/4

[finite gale]

oh damn that simple 🤦‍♂️

thx

.close

[devout snow]

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[devout snow]

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Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[latent scaffold]

can anyone help me with this?

[latent scaffold]

<@&286206848099549185>

[formal ingot]

Have you tried all options?

[latent scaffold]

uhh I can't I only have 1 attempt

[formal ingot]

No, as in go through each option without submitting, for example, option 2

Option 2 says a rotation of the triangle by 90 degrees

But has the triangle being rotated?

[latent scaffold]

ohhh

you mean like process of elimination

also sorry for late response

oh

I got the first option

as my answer

@formal ingot

[formal ingot]

Have you tried all the other options

[latent scaffold]

yup

so am I correct?

[formal ingot]

So preimage is a reflection, right?

[latent scaffold]

uhh yeah

pretty sure

yup

[formal ingot]

So it wouldn't be 1

Cause it wasn't mirror in anyway

[latent scaffold]

oh yeee

your right

so it has to be the last one right?

It would be I think

but abc is like

the original

the ones that say A'B'C' were translated 8 units right and 3 units down

[crisp mesa]

I don't know if I am late, however the answer is d) triangle ABC was translated 8 units left and 3 units up if ABC means A1B1C1

[formal ingot]

8 left and 3 up

But I think you get the point

[latent scaffold]

nno like

[latent scaffold]

ohhh

bet bet

thanks guys

[crisp mesa]

No Problem.

[latent scaffold]

.close

[devout snow]

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[devout snow]

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[violet trail]

Hello

[devout snow]

@violet trail Has your question been resolved?

[devout snow]

Channel closed

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[pine bay]

i need to solve B)

i followed the steps yet i got it wrong

[toxic grove]

What did you do?

[pine bay]

let me hsow u

here u go

@toxic grove

brother

[toxic grove]

This absolutely isn't anywhere near my area so no promises

But your approximation of 1.066^(1/12) of 1.0055 isn't accurate at all once you raise it to the power of 240

The error gets quite large

So that could be the problem

It also looks like you copied down the result of 2.7299/0.0055 wrong from your calculator (even though those numbers aren't correct in the first place)

Assuming your formulae are correct then that's what's probably gone wrong

[pine bay]

my formula was wrong

but @toxic grove

general question

when finding interest, we always subtract the Future Value or Present Value from our Total payments to get Interest?

[toxic grove]

Compound interest is always based on how big the debt is at that moment

[pine bay]

okay

so

[toxic grove]

when you say Interest, do you mean how much was charged in total?

Because that's just total payments minus the initial value

I've actually solved it now so I'm not totally clueless anymore

But I'm not an expert

[pine bay]

okay

understood

ok so what i actually mean is

listen i got the right answer

and u get the R value, multiply it by 245280 which gives you the total payments, Subtract 136000 from that and u your interest correct?

@toxic grove

[toxic grove]

By R, do you mean how much he pays each month?

Because total payments would be that multiplied by 240

[pine bay]

R value is what were solving for

[toxic grove]

The answer to part a?

[pine bay]

R= Regular payment each compounding period

[toxic grove]

Okay yes, how much he pays each month

He pays it for 240 months, so the total payment is R * 240

But then yes, subtract 136000 from that and you get the total interest

That's how much extra paul ended up paying

[pine bay]

ok

great

now listen carefully

mb

@toxic grove

u still here

ok now listen carefully

[toxic grove]

Ye

[pine bay]

for 11b

this question right here

it says calculate the future value and INTEREST

correct?

[toxic grove]

Yes

[pine bay]

good

now

[toxic grove]

I assume you know what those mean?

[pine bay]

yes. now listen

when solving for interest. to get the right answer i had to subtract the total payment from the FV

[toxic grove]

Yes

[pine bay]

but in this question we just did, 15b i had to subtract PV from the total payment

[toxic grove]

Okay but in the debt example, it's pretty much the same thing but in reverse

You shouldn't think of interest as a formula like that

In the debt example, interest was how much extra he had to pay on top of the 136000

Because the bank kept charging interest

It is mathematically quite similar, but present value and future value are swapped

[pine bay]

alright so when its FV were subtracting the total payments FROM FV but when its PV were subtracting PV FROM the total payments?

[toxic grove]

Yes

It kind of feels backwards, but that's because in the debt example it was effectively you paying the interest, whereas in the annuity, the bank is paying the interest

It'll be much less confusing if you think of "interest" as an english term rather than a formula to be memorised

[pine bay]

i kind of understand because whenver its FV, The total payments is less than the FV

but when its PV the Total payments is MORE than PV

true?

[toxic grove]

Other things are flipped as well, because in the debt example, the future value is 0, but for the annuity example, the present value is 0

Yes

[pine bay]

understood

fully understood

im about to do this question now, last one on my exam review. tryna solve it too and we see if we got the same shit?

[toxic grove]

That's because in the debt example the bank were charging you interest (so it takes away from your payments), but in the annuity example they're paying you interest on top of your payments

[pine bay]

got it

[toxic grove]

Sure but again no promises, I'm no economist

[pine bay]

its descrete functions 😂

yo

@toxic grove it dont say how much money adams putting in the account per month

only ken

wtf

[toxic grove]

That's what you have to calculate

Ken puts in $100 per month and ends up with some amount of money by age 55

Adam puts in $x per month and ends up with the same amount

Solve for x

[pine bay]

gotchu

this is FV right

[toxic grove]

Yes

[pine bay]

br

im at

FV=R(5.6631)

oh wait

R is 100

im dumb

[toxic grove]

So you think FV = $566?

That doesn't soundn right

[pine bay]

no wait

i thin kits 556

[toxic grove]

If ken puts $100 into the account each month, then he would have at least that amount by month 6

Even without interest

[pine bay]

oh shit ur right

[toxic grove]

He pays into the account for 35 years, it's gotta be a lot more than 500 dollars

[pine bay]

whered i go wrong

i mean

i multiplied 35 by 12

and devided 0.054 by 12

does 556x420 sound right

[toxic grove]

Where did 556 come from?

[pine bay]

here bro

ill send a pic of my work

[toxic grove]

That's a bit more than what I got

[pine bay]

look bro

i couldnt have gone wrong anywhere

[toxic grove]

Okay you're close

[pine bay]

what did i do wrong

[toxic grove]

But i is 0.0044 right?

You've got i = 1.0044

On the bottom of this fraction

[pine bay]

i forgot to subtract the 1 in the denominator

right

oh yeah

cus the its 1+i

so i is without the 1

[toxic grove]

Exactly

[pine bay]

124252.45

right

[toxic grove]

Yep approximately

[pine bay]

thats my FV

[toxic grove]

Yes

[pine bay]

ok so now we plug it in for FV into adams wuqation and solve for R

am i right

or am i right

[toxic grove]

Sounds right

[pine bay]

finally bro

yo

@toxic grove

here we go bro

istg i have to be right

is R = 227.727

[toxic grove]

It might be

Show me your working?

I got 281 but I used a lot more precision than you did

It does feel quite far off so I don't think it's right

[pine bay]

yeah

true

ok well

here

im right

i just typed it in the calculator

this equation equals R447.39

[toxic grove]

Alright a few problems

[pine bay]

wats up

[toxic grove]

Firstly, how many months in 18 years?

[pine bay]

dont even tell me

dont even tell me i reaed the calculator wrong

no fucking shot

its always somthing bro

its always somthing i swear

[toxic grove]

Hahaha it happens to all of us dw

[pine bay]

ok dats all tho right

[toxic grove]

Secondly

The 447 number is wrong because of the wrong number of months

[pine bay]

ok new answer; 282.34

[toxic grove]

But you did the division wrong as well

282 is much more like it

[pine bay]

1 problem bo

oh wait actually

i got it right on the exact decimal

282.34-100=182.34

so he had to pay 182 more dollars than his bro

and would u look at that

got it correct right down to the decimal

[toxic grove]

Yep wd

[pine bay]

ai bro

i wana sleep but i cant

cus yo

@toxic grove my boy

he got a hold of the exam and i was looking at it

1 question i dont get that much

some quadratics shit

pretty easy

i just forget what to do

[toxic grove]

What, the actual one you're practising for?

[pine bay]

nah nah just a old one

[toxic grove]

Okay

[pine bay]

ok so

a)

[toxic grove]

So what are you thinking?

[pine bay]

quick question first

what does the quadratic formula give me again

zeros?

[toxic grove]

Yes

[pine bay]

ok understood

[toxic grove]

If ax^2 + bx + c = 0, then the quadratic formula will give you the possible values for x

[pine bay]

the maximum no ??

that would give u the vertex which includes the max

[toxic grove]

Not necessarily

Wdym?

[pine bay]

making the equation = 0

gives u 2 brackets when u factor it

[toxic grove]

Yes

[pine bay]

and if u fill those in

[toxic grove]

So there are two possible solutions

[pine bay]

those are your zerios

[toxic grove]

Yes

[pine bay]

find the axis of symatry between those 2 numbers

and thats the max

y=max

no?

[toxic grove]

If you're trying to solve for the maximum height of the balloon then yeah that's one way of doing it

[pine bay]

got it

[toxic grove]

Part (a) asks for how long it spends in the air though

[pine bay]

and those 2 numbers that i found the axis of syetry of, 1 is probably gonna be negative and ones probably gonna be positive right?

[toxic grove]

It should be yes

[pine bay]

i use the positive one and thats my x axis

[toxic grove]

Wdym

[pine bay]

ok im confusing my self

forget all that

ok a) how long was the balloon in the air

[toxic grove]

Yeah the positive one is the only "real" solution, the negative one is there just for mathematical reasons

[pine bay]

true

ok so

a) how long was the balloon in the air

what is it trying to ask me here

[toxic grove]

How much time passed between him throwing the balloon, and the balloon hitting the ground

[pine bay]

mm

alr

so

this is a parabola with a maximum cus its facing downwards

[toxic grove]

Yes

[pine bay]

now its getting confusing cus what do u mean in the air

that shit makes no sense

[toxic grove]

The balloon isn't gonna accellerate into the air so that makes sense

[pine bay]

yurrr

so

[toxic grove]

h(t) tells you the height of the balloon at time t right?

If h(t) > 0 then the balloon is in the air

[pine bay]

holdup

h(0)= and solve

am i right

[toxic grove]

No

[pine bay]

fuck

[toxic grove]

h(0) is just the height of the roof

h(t) = 0

solve for t

That's what you want to do

[pine bay]

im saying make t=0

oh

solve for t?

dont u do that by making the equation = 0

or can u use quadratic formula

[toxic grove]

Here's what's actually happening

The red circles are where h(t) = 0

But the solution in the past doesn't really exist, because the ball was thrown to the right

Yes, you make h(t) = 0

But not t = 0

You want to find the value of t

[pine bay]

holdup holdup holdup

so

quadratic formula gives me the zeros, but technically so does making h(t)=0

wat do i do

whic one

[toxic grove]

In theory h(t) has negative values as well but you can see they don't really mean anything

[toxic grove]

Both

They're exactly the same

You set h(t) = 0

[pine bay]

which ones faster

[toxic grove]

Then use the quadratic formula to solve for t

[pine bay]

snm

[toxic grove]

If you know of some other way then you're welcome to use that instead

[pine bay]

nice bro u js got active role

big ups for u

[toxic grove]

Omg my name has color now

[pine bay]

ima use quadratic

[toxic grove]

I don't like it I won't lie lol

It's a bad look for me

Too exciting

[pine bay]

just to be clear h(t)=0 will give me 2 brackets, i need to add a number into the brackets to make them = 0 and those numbers are my 0s right

[toxic grove]

Yes

[pine bay]

but quadratic formula will straight up give my zeroes

[toxic grove]

If you find that h(t) = (x - 5) (x + 3) for example, then the 0s are 5 and 3

[pine bay]

5 -3

are my zeroes

[toxic grove]

Yes sorry that's what I meant

[pine bay]

yeah yeah

but

quadratic

they would just give me 5 and 3

[toxic grove]

So either you can factorise it and then work out the 0s, or you can use the formula to get the 0s (and then factorise it if you wanted)

Yes

Bottom line is it doesn't matter

Do whatever

[pine bay]

say nm

im doing quadratic

brb gang

wait

before i go for a sec

i got a a quick question befores olving this

first: for discrimanent formula: if the x>1 = theres going to be 2 zeroes, if 0<x<1 theres going to be 1 zero, and if 0>x theres no zeroes

correct?

[toxic grove]

No

[pine bay]

wait what

for this formula

[toxic grove]

if x > 0 there are 2 zeroes, if x = 0 there is 1 zero, and if x < 0 then there are no zeroes

Yes

[pine bay]

OH yeah

shit mb

i remmeber now

[toxic grove]

It makes perfect sense if you look at the quadratic formula

You can't sqrt a netagive number

And if it's 0 then +-sqrt(that) are both going to be the same

[pine bay]

oh shit true

because ur adding nothing to both answers

[toxic grove]

So there are still technically two zeroes, just they have the same valune

Exactly

Well

Adding nothing to one of them, and subtracting nothing from the ohter

[pine bay]

basically the answers are going to stay the same and theya re the same

[toxic grove]

Yeah

[pine bay]

yeah yeah

ai ima solve

[toxic grove]

You don't need discriminant for this question though

[pine bay]

yeah ik

i was just wondering

yk what im also wondering

[toxic grove]

Go on

[pine bay]

ok found it

alr so

9y^2-4

I wrote this as (3y-2)^2 why is this wrong

[toxic grove]

Because that's not right

Expand (3y - 2)^2

You don't get 9y^2 - 4

[pine bay]

is itbecause the square root of -4 can be negative or positive so i ned to make it 3y-2 3y+2

[toxic grove]

positive 4, not negative

Negative 4 doesn't have a square root

But yes

2 and -2 are the two solutions to y^2 = 4

[pine bay]

ok

and

theres this other one

i forget what its called

its where theres an a b and c value

and u square root a and c

and use the sign of b

dyk what im talking ab

[toxic grove]

No

[pine bay]

ok for example

9y^2+12y+4

u squareroot 9y^2 and 4

and then u make it (3y+2)^2

u get what im saying

[toxic grove]

Oh are you talking about completing the square?

[pine bay]

no no no

not that bro

[toxic grove]

Well then idk what you're talking about

[pine bay]

wait maybe

but bro

[toxic grove]

Not saying you're wrong

[pine bay]

isnt completing the square

when u factor the first 2

numberes

[toxic grove]

Yeah

[pine bay]

then take b value divide it by 2, square it then add it and subtract it then take the factored number multiply it by the - one and bring it out of the bracket

[toxic grove]

Something like that yes

[pine bay]

ok look

9y^2 + 12y + 4

how would u solve this

like YOU

[toxic grove]

It's just another way of solving quadratics, you don't need to do it if you don't want to

I'd probably divide the whole thing through by 9, then complete the square

[pine bay]

oh?

but what was that thing that the middle number (b) /2 squared or smth has to equal the c value or smth

to use the rule

[toxic grove]

That's what I always used to do when I was at school

Yeah I still don't know what you're talkinga bout

If b^2 = 4ac then the discriminant is 0 and there is only one root?

That's all I can think of

[pine bay]

OH

PERFECT SQUARE TRINOMIAL

ring a bell?

[toxic grove]

No

[pine bay]

No?

perfect square trinomails

this

[toxic grove]

Oh so just (ax - b)^2 for some a?

[toxic grove]

That's literally this

I've not heard it called that before but yeah it's the same thing

[pine bay]

ok so for the middle number

middle number has to equal to the square roots of a and c times 2

[toxic grove]

if $b = 2\cdot \sqrt a \cdot \sqrt c$ then $b^2=4ac$

[woven radish]

depression

[pine bay]

got it bro

love u

ima solve htat question now

the quadratic shit fam

yo

@toxic grove

im at a point where the 2 zeroes are -27-26/-9.8 and -27+26/9.8

do i solve the devision first

even tho the 2 numbers are in the numerator

[toxic grove]

Do whatever the formula tells you to do

Work out the top

Then the bottom

Then divide the top by the bototom

[pine bay]

ok

yeah i fucked up

i defenitely fucked up big time bro

[toxic grove]

Yeah I thought so

Neither of them looked right

That's okay

Try again

[pine bay]

whered i go wrong

[toxic grove]

Missed out a minus sign

/ added one in idk

[pine bay]

wait

its 72.9+47.04 right

[toxic grove]

What is?

[pine bay]

in the bracket

2nd equation i the ss u sent me

729+47.04

is what it should be

no?

[toxic grove]

Yeah thats right

[pine bay]

bet

[toxic grove]

But you did 729 plus 47.04

Sorry no

You did minus but it should be plus

[pine bay]

yea

nah i fucked up again

ok

i got -0.087 and 5.56

as my zeroes

ths shit actually makes no sense to me

how are my zeroes so discombobulated

[toxic grove]

Yeah that looks right

[pine bay]

ok bet

[toxic grove]

Why wouldn't they be?

[pine bay]

so

5.56 seconds is the answer to my first question

[toxic grove]

There's no reason it should have nice numbers

Yep

[pine bay]

how high off the ground was the balloon innitally thrown

now how tf do i know that

[toxic grove]

What do you think?

[pine bay]

i think we plug 5.56 into the equation

[toxic grove]

Try it

[pine bay]

for t

ok

125m?

[toxic grove]

No

[pine bay]

fuck

i was wrong then

[toxic grove]

Plugging in your answer to part a should give you 0

That's what you worked out

In part a

[pine bay]

why the fuck didnt it give me zero then

[toxic grove]

What did it give you? 125?

[pine bay]

yes

[toxic grove]

Then you typed it in wrong

If you got about 1 then it's because 5.56 isn't very accurate

But 125 means you typed it in wrong

[pine bay]

ok look

-4.9(5.56)^2 + 27(5.56) + 2.4

= -27.244+150.12+2.4

= 125.276

[toxic grove]

You didn't square 5.56

[pine bay]

i gnored the square

yup

nice

i always do smth bro

aint gon have an answer sheet on my exam

gon make so many silly mistakes

i got 1.04

[toxic grove]

Yeah

You typed it in right

But 5.56 isn't very accurate

The actual answer is 5.5977

[pine bay]

mmm ok

[toxic grove]

But you probably calculated it correctly

[pine bay]

would the answer be more accuriate if i just plugged 0 in cus that means the height when the ball was at 0m

[toxic grove]

And truncated the numbers on your page is my best guess, I wouldn't consider it a mistkae

Yes

That's the answer

[pine bay]

wait so plugging 0 in and 5.56 in is the same thing?