3U Maths Question Thread

[jamonwindeyer]

like how would you draw the inverse relation of something like that then restrict the domain

Hey! I reckon it's the actual restriction of the domain that might be confusing you? As Rui said, you do it before you sketch, doing it after is just not quite a sensible thing to do. To restrict the domain, you just take enough of the function out so that it will pass the horizontal line test. For a parabola, this means only considering one arm of the parabola (if you draw only one arm of a parabola, then it becomes invertible). Often this restriction will be given to you! If it isn't, you just have to try and make a sensible choice given the scenario

[beau77bro]

help please hahaha. i dont get how to show it, help would be very much appreciated.

[RuiAce]

help please hahaha. i dont get how to show it, help would be very much appreciated

[beau77bro]

OMG I JUST REALISED THAT IS X OMG I DONT NEED TO FIND IT AHHAHAHAHAHA I WAS LIKE TRYING TO INTEGRATE AND I WAS LIKE THIS CANT BE RIGHT. THANKYOU RUI, its already in x so the definition already applies.

[itssona]

heey for this:
six people enter a boat where there are 8 seats - 4 on the left and 4 on the right. In how many ways can they be seated if:
a) no restrictions
b) two particular people A and B want to be on the left and one person on the right

so for a is it 8C6?

and for b, 4C2 multiplied by 4C1?

[Mahan]

heey for this:
six people enter a boat where there are 8 seats - 4 on the left and 4 on the right. In how many ways can they be seated if:
a) no restrictions
b) two particular people A and B want to be on the left and one person on the right

so for a is it 8C6?

and for b, 4C2 multiplied by 4C1?

for a) the answer is 8C6*6!, because when you choose 6 seats there are 6!ways to arrange six people on six seats.

For b) The rule of thumb in counting question is to always take care of things that have more restrictions. Let's say A wants to sit on the left but B wants to sit on the right. There are four empty seats for A to choose from and when A choses his/her seat then B has also four choices for the seats on the right. Now, you technically have 4 people and three seats on the left and three seats on the right. This is the same as part a but with 6 seats, because there is no restriction on the rest of them.
So the answer is 4*4*6C4*4!

[itssona]

for a) the answer is 8C6*6!, because when you choose 6 seats there are 6!ways to arrange six people on six seats.

For b) The rule of thumb in counting question is to always take care of things that have more restrictions. Let's say A wants to sit on the left but B wants to sit on the right. There are four empty seats for A to choose from and when A choses his/her seat then B has also four choices for the seats on the right. Now, you technically have 4 people and three seats on the left and three seats on the right. This is the same as part a but with 6 seats, because there is no restriction on the rest of them.
So the answer is 4*4*6C4*4!

omg THANK YOU!!

[legorgo18]

Dont know if its the wording thats getting me confused (page 1 question...)

ABC is an acute-angled triangle. Squares BAHK and CAXY are constructed on AB and AC respectively and outside the triangle. The diagonals HB and XC meet when produced at P. Prove that angle BPC + angle BAC= 90

Can someone perhaps draw a diagram then i should know whats going on. Tyvm

[Shadowxo]

Dont know if its the wording thats getting me confused (page 1 question...)

ABC is an acute-angled triangle. Squares BAHK and CAXY are constructed on AB and AC respectively and outside the triangle. The diagonals HB and XC meet when produced at P. Prove that angle BPC + angle BAC= 90

Can someone perhaps draw a diagram then i should know whats going on. Tyvm

Here's a diagram, hope it helps

[bsdfjnlkasn]

If someone could please help me identify where I went wrong I'd really appreciate it! Thank you

[jamonwindeyer]

(Image removed from quote.)

If someone could please help me identify where I went wrong I'd really appreciate it! Thank you

Could just be not changing your limits of integration?

Edit: Woops, you changed it back to \(x\), nevermind! Using those values in your integrated expression wrt \(\theta\) gives the correct answer, give me a sec...

Next Edit: Think I got it

[bsdfjnlkasn]

Could just be not changing your limits of integration?

Edit: Woops, you changed it back to \(x\), nevermind! Using those values in your integrated expression wrt \(\theta\) gives the correct answer, give me a sec...

Next Edit: Think I got it

Thank you Jamon!! Silly me

[legorgo18]

Help pls with this

An external common tangent to two non-intersecting circles touches the first circle at C and the second at D. An internal common tangent touches the first circle at F and the second circle at H. If CD and FH meet the line of centres at E and G respectively, prove that CE/FG = DE/GH

I think im getting lost in the words... again ty

[jamonwindeyer]

Help pls with this

An external common tangent to two non-intersecting circles touches the first circle at C and the second at D. An internal common tangent touches the first circle at F and the second circle at H. If CD and FH meet the line of centres at E and G respectively, prove that CE/FG = DE/GH

I think im getting lost in the words... again ty

Hey! This confuses me too, to my understanding a common external tangent doesn't touch the line joining the centre of two circles, by definition... They didn't attach a diagram by any chance?

[RuiAce]

Help pls with this

An external common tangent to two non-intersecting circles touches the first circle at C and the second at D. An internal common tangent touches the first circle at F and the second circle at H. If CD and FH meet the line of centres at E and G respectively, prove that CE/FG = DE/GH

I think im getting lost in the words... again ty

Hey! This confuses me too, to my understanding a common external tangent doesn't touch the line joining the centre of two circles, by definition... They didn't attach a diagram by any chance?

Yeah, a quick Google search shows that this question is just a contradiction. CD should never meet the line of centres unless it is also an internal tangent