VCE Methods Question Thread!

[lzxnl]

1 - cos^2 x = sin^2 x
That should help

[MightyBeh]

How does 1 (attached) simplify to 2 (attached)? What am I missing here? Thanks 

Hint first: Pythagorean Identity

Rest is in here

Pythagorean identity is which means you can substitute the LHS into the expression you were given, like so:


The cos^2 in the demonimator cancels out so you have:

Which further simplifies to

Which is the reciprocal of the tan identity, so the answer is 1/tan(theta)

[YellowTongue]

Thanks lzxnl and MightyBeh. I can't believe I missed that...

[Syndicate]

Hey guys,

How can I solve:
Tickets for a concert are available at two prices. The more expensive ticket is $30 more than the cheaper one. Find the cost of each type of ticket if a group can buy 10 more of the cheaper tickets than the expensive ones for $1800.

Thanks

[IntelxD]

Hey guys,

How can I solve:
Tickets for a concert are available at two prices. The more expensive ticket is $30 more than the cheaper one. Find the cost of each type of ticket if a group can buy 10 more of the cheaper tickets than the expensive ones for $1800.

Thanks

Let x be the price of a cheap ticket. Therefore, the price of an expensive ticket is x+30. We are told that they can purchase an extra 10 tickets for 1800$ if they are to purchase cheap tickets. Since the total cost is 1800$, the number of cheap tickets purchased can be expressed as 1800/x. Similarly, the number of expensive tickets would be 1800/(x+30). Now using the fact that there are 10 more cheap tickets then expensive tickets, we can setup the following equation: 1800/x = 1800/(x+30) + 10. We now have one equation with one variable which can be solved using algebra or a CAS calculator.

[Syndicate]

Let x be the price of a cheap ticket. Therefore, the price of an expensive ticket is x+30. We are told that they can purchase an extra 10 tickets for 1800$ if they are to purchase cheap tickets. Since the total cost is 1800$, the number of cheap tickets purchased can be expressed as 1800/x. Similarly, the number of expensive tickets would be 1800/(x+30). Now using the fact that there are 10 more cheap tickets then expensive tickets, we can setup the following equation: 1800/x = 1800/(x+30) + 10. We now have one equation with one variable which can be solved using algebra or a CAS calculator.

EDIT: Thanks I get it now 

Thanks 

[Sarah P]

Find the value of m for which the simultaneous equations;

mx-2y=4
x+(m-3)y=m

has infinitely many solutions


i know that i need to find the gradient and y-intercept for which both equations are equal, but i just can't get it!

i've worked out so far that the gradient is 2/3, but that's as far as i can get! did i do something wrong or am i on the right track? any help will be great! its been annoying me for far too long

(the answer for 'm' is 2 in the book)

[Sarah P]

Find the value of m for which the simultaneous equations;

mx-2y=4
x+(m-3)y=m

has infinitely many solutions


i know that i need to find the gradient and y-intercept for which both equations are equal, but i just can't get it!

i've worked out so far that the gradient is 2/3, but that's as far as i can get! did i do something wrong or am i on the right track? any help will be great! its been annoying me for far too long

(the answer for 'm' is 2 in the book)

sorry i wrote that incorrectly! i meant to say that for m (NOT gradient) i got 2/3 which is incorrect.

[qazser]

sorry i wrote that incorrectly! i meant to say that for m (NOT gradient) i got 2/3 which is incorrect.

Its back a few pages in this topic.

Here's a similar question, might help

Infinitely many solutions: m1 = m2 and c1 = c2

No solutions: m1 = m2 and c1 does not equal c2

First step, rearrange both the equations in terms of y.

(1) y = -3x/m + 5/m
(2) y = -(m+2)x/5 + m/5

Make m1 = m2, and then solve for m.

-3/m = -(m+2)/5

m^2 + 2m - 15 = 0

(m + 5)(m-3) = 0

Therefore, m = -5 or m = 3

Sub in these values to find the y intercept. Where for infinite solutions c1 = c2, and no solutions c1 does not equal c2.

a) infinitely many solutions: m = -5 (because 5/-5 = -5/5)
b) no solutions: m = 3 (because 5/3 does not equal 3/5)

[Sarah P]

Its back a few pages in this topic.

Here's a similar question, might help

Infinitely many solutions: m1 = m2 and c1 = c2

No solutions: m1 = m2 and c1 does not equal c2

First step, rearrange both the equations in terms of y.

(1) y = -3x/m + 5/m
(2) y = -(m+2)x/5 + m/5

Make m1 = m2, and then solve for m.

-3/m = -(m+2)/5

m^2 + 2m - 15 = 0

(m + 5)(m-3) = 0

Therefore, m = -5 or m = 3

Sub in these values to find the y intercept. Where for infinite solutions c1 = c2, and no solutions c1 does not equal c2.

a) infinitely many solutions: m = -5 (because 5/-5 = -5/5)
b) no solutions: m = 3 (because 5/3 does not equal 3/5)

Thank you so much for the help! turns out i made a mistake in my working out and this helped me find it 

[upandgo]

hi all! is rational root theorem in the study design?

i know it's in the textbook but my maths teacher told my class to ignore it 

[keltingmeith]

hi all! is rational root theorem in the study design?

i know it's in the textbook but my maths teacher told my class to ignore it 

It's a new addition to this year - your teacher may have skipped it because in previous years, this sort of stuff wasn't tested much anyway. At the end of the day, VCAA are UNHIGHLY likely to say, "factorise this using the rational root theorem", and more likely to just say, "factorise this", just because it's actually kind of hard to show your working for the rational root theorem. (also the fact that VCAA generally don't force you to use specific methods, contrary to popular belief)

[upandgo]

It's a new addition to this year - your teacher may have skipped it because in previous years, this sort of stuff wasn't tested much anyway. At the end of the day, VCAA are UNHIGHLY likely to say, "factorise this using the rational root theorem", and more likely to just say, "factorise this", just because it's actually kind of hard to show your working for the rational root theorem. (also the fact that VCAA generally don't force you to use specific methods, contrary to popular belief)

thanks for clearing that up!  i'll probably learn it anyway though just in case

[exit]

thanks for clearing that up!  i'll probably learn it anyway though just in case

Good idea, I think it would only be needed in an non-calc exam and I can't imagine a lot of people getting it right in the unlikely instance of it appearing.

[Stewart98]

Hey Fellas, i've got a question to ask!

Find the values of a and b such that the graph of y=alog2(x+b) goes through the points (8,10) and (32,14)
Thanks