Sets

[whole tangle]

what?

[finite dune]

so do you have a question

[merry notch]

The answer is D but, I need a clear explanation

[finite dune]

hm.. so do you understand intersection and complements

[merry notch]

yes

[finite dune]

can you mark P'

[merry notch]

that's supposed to be element in the universal set, but not in P

[merry notch]

no

[finite dune]

right

[finite dune]

no as in you cannot draw or you cannot mark?

[merry notch]

I cannot mark it. Where am I supposed to mark?

[finite dune]

this is P so everything except yellow will be P'

[merry notch]

ohh okay

[merry notch]

I have primarily done this with numbers

[finite dune]

if you want elements which are not in P and not in R
or $(P\cup R)'$ or $P' \cap R'$ you need to find elements that are neither in P or R

[neat trench]

mercy

[finite dune]

which would be outside both sets P and R

it would be these

now find the elements which are common to Q and $P' \cap R'$

[neat trench]

mercy

[finite dune]

the green ones , they are in Q but do not belong to P or R

you can treat $P' \cap R' \cap Q$ as all elements of Q - those which are common with P or R

[neat trench]

mercy

[merry notch]

Okay

hmmm

thank you so much

[finite dune]

!done

[half igloo]

If you are done with this channel, please mark your problem as solved by typing .close

[merry notch]

I got a question please

[merry notch]

why intersection and not P' U Q U R'

[finite dune]

the question was intersectiom?

oh

what is union ?

AunionB, is the set of elements that are in
A or in
B. or in both.

so P' U R' will be combination of all elements in R', P'

,, P' \cup R' = P' +R' + P' \cap R' = P \cup R - (P \cap R)

green is intersection of both

[neat trench]

mercy

[merry notch]

thank you

@finite dune can you help me solve a simultaneous log equation?

[finite dune]

i can try post it

[finite dune]

do you know log rules

[finite dune]

what have you tried

[merry notch]

let me resolve, I can't find it

Okay

Im supposed to use a substitution method

but I can't figure it out

[finite dune]

apply rule 1

[merry notch]

done

[finite dune]

what do you get

[merry notch]

what is the the next step?

[finite dune]

you had 2logy in second equation so second is wrong,
lets see first, it is correct now you can write $\log_{b} a = x$ as $a=b^x$

[neat trench]

mercy

[finite dune]

oh you corrected it, from here you can split $\log{xy^2} = \log{xy} + \log{y}$

[neat trench]

mercy

[finite dune]

then directly substitute value of logxy

you will find value of logy from there

[merry notch]

7?

[finite dune]

alternately you can divide $\frac{xy^2}{xy} = \frac{10^3}{10^4}$

[neat trench]

mercy

[finite dune]

?

[merry notch]

sorry

[finite dune]

nw

[merry notch]

I would appreciate if you can solve the substitution or the alternative method

[finite dune]

i dont know wgat substitution you are referring to can you provide context

[merry notch]

let me see the solvings

[finite dune]

also which aletrnate method do you not understand?

[merry notch]

Djjsjqkshsh

I totally understand

thanks

thank you @finite dune

[finite dune]

np

you can send image of substitution if you want to solve that way

anyways if your query has solved you can close this by typing .solved or .close

[merry notch]

.solved