can someone help me with general maths revision. I'm 14 so I'm doing basic relationships stuff

[sterile elk]

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[tardy heart]

Well, what do you want to clarify?

[coral rover]

oh hi

I was just gonna post questions that on the homework I got wrong

this is the first one

I went from 4p - 11 < p - 2 to 3p < 13 to p greater than or equal to 4

do you know where I went wrong?

[tardy heart]

13 is incorrect.

Note that we have p - 2 on the right, not p + 2.

[coral rover]

is it -13?

[tardy heart]

No.

[coral rover]

what do you actually do to find the right number

[tardy heart]

Well, we have:
4p - 11 < p - 2
First, we subtract p from both sides.
3p - 11 < -2
Then, we add 11 to both sides. What do we get?

[coral rover]

3p < 9

I understand

the goal is to get 1 letter on the left and one number on the right

[tardy heart]

That's correct.

[coral rover]

so you change the subject until you have that

and then obv its p <

3

[tardy heart]

Better to call them variables and constants, not letters and numbers, but yeah.

[coral rover]

we don't get taught names like that or anything

is a constant a number

cause they always mean the same thing

and a variable, letters because it can be changed

[tardy heart]

Really? Huh, that's odd... Oh well.

[tardy heart]

Not all letters are variables. There can also be parameters.

Better to use the correct terms, anyway.

[coral rover]

yeah that's why I don't come to this server often cause people alays come at me with these names and I just don't know what's going on lol

[coral rover]

okay cool

next question

I missed a couple marks on this

but I got a half tick half x at the end

I missed one mark

I'll send you what I did

[tardy heart]

Alright, so what did you do?

[coral rover]

was it just because I didn't put the brackets in?

[tardy heart]

Ah, I see. 4(a + 5) = 4a + 20, not 4a + 5.
Also, x^2 = y is equivalent to x = ±√(y), not just √(y). So, the correct answer is b = ±√(4a + 20). Or b = ±2√(a + 5), whatever you prefer.

[coral rover]

so do you actually have to expand the bracket and do the equations on the left of the equals sign

[tardy heart]

What do you mean?

[coral rover]

like instead of me just putting a + 5 x 4 I had to multiply the 5 and 4 and then put a in wherever

[tardy heart]

Not sure what you mean.
The distributive law is x(y + z) = xy + xz, not xy + z.

[coral rover]

oh right yeah

that makes sense

okay

onto the next question

I didn't get either marks for this

I put a = 2 and b = 3

[tardy heart]

Can you explain why?

I mean, that is wrong. Just want to know your thought process.

[coral rover]

because that's where the parabola intersects with the corners of the squares

idrk why I thought that was it

the only thing I could think of lol

oh wait b is 2 right?

and a is 1

is it or am I wrong again

I think I'm wrong

[tardy heart]

b = 2, but a isn't 1.

[coral rover]

b = 2 because the parabola is 2 high

I remembered that

[tardy heart]

If a was 1, the vertex would be at (-1, 2), not (1, 2).

[coral rover]

how do you work out a?

oh

a is -1

[tardy heart]

Yes.

[coral rover]

you saying that reminded me you just take the sign and change it

[tardy heart]

Yes.
Remember that y = f(x), when translated by {a, b}, becomes y = f(x - a) + b.

[coral rover]

for this question is all you have to do is write x = 3 and x = -7

[tardy heart]

Yes.

[coral rover]

this is gibberish to me

[coral rover]

lmao I expanded the brackets in the homework for some reason

this is the next question

how do you find which angle is which, for ADC would you go to A and then to D and then to C and would the angle be at d?

is the angle I need to find next to the letter D

I think I can explain simpler if you want me to

[tardy heart]

Note that tangents are perpendicular to radii. So, BAD and BCD are both right angles.

[coral rover]

yeah

what I just said

I have the answer

because of how you said that

but how do you work out the angles of a kite

is what throws me off here

[tardy heart]

We need two facts here:

1. BAD and BCD are right angles.
2. Sum of angles of any quadrilateral is 360°.

[coral rover]

so it's just 142 + 90 x 2 = ADC

so it's like 28?

ADC is 28 degrees

[tardy heart]

No. Rather, ∠ADC + 2*90° + 142° = 360°.

[tardy heart]

I think you made an error in 180° - 142°.

[coral rover]

I was doing it a different way but yes I have

its 38

[tardy heart]

Yes.

[coral rover]

the only other question I wrong I think I know how to do it but my working was super weird

[tardy heart]

Let's see.

[coral rover]

you would do 360/10 and then do 180 - 36*2 right

and then that would be the answer

[tardy heart]

Rather, just 180° - 36°.
It's good to remember that a regular n-gon has angle ((n - 2)/n)180°. So, in this case, as n = 10, the angle will be (8/10)*180° = 144°.

[coral rover]

I didn't understand any of that, the second part you said. Why would it just be 180 - 36 when theres 2 corners of triangle total shaded

if you get what I mean

[tardy heart]

The marked angles are the same. Thus, when you subtract 36° from 180°, you already get the needed angle.

[coral rover]

oh yea

I understnad

thats all of it

thanks os muchh for the help

[tardy heart]

You're welcome!

[coral rover]

now I understand the major contributer role

you have been the most helpful interaction on thhis server

I've ever hahd

had lol

[tardy heart]

Well, thanks 😄