Questions thread

[kenhung123]

You would be a good tutor. Excellent explanations! A+

[Souljette_93]

1st one looks like a boring and very long expansion and simplification -_-

I'm not sure about this for 1 but the 1st part,(p+3)^5 will have a part that is to the power of
5,4,3,2,1 and a constant. If you were to times that to the other part ( 2p-5)
the ^5 would go to ^6 and ^5. So the ^4 part would have ^5 and ^4 ( 1 part here) and ^3 would have ^4 and ^3.

so maybe it could just be:
= 1 part that will have p^4
and = 2nd part
so I'd say -75p^4 + 180p^4 = 105 as the coefficient.


2nd one: All you want is the 2nd expansion, so using the binomial theorem, you would have but you're only after the co-efficient so you don't need the a.
This leads to     leaving n on top.
I end up with that thing above or simplifying it to ( by breaking into The answer is n = 6 but I don't see how to solve it from there ( maybe you need a calc). Maybe someone else can continue from here.

Thank you!
PS: i used the 'solve' function in the calculator. So yeah i don't think i need to know it manually.
TC,
Souljette<3

[Hielly]

Hey
Three points have coordinates A(1, 7), B(7, 5) and C(0,−2). Find:
a the equation of the perpendicular bisector of AB
b the point of intersection of this perpendicular bisector and BC.

for a, i found that m=3
do you sub either the A or B to find the equation? i did both and it doesnt work.

thanks

[qshyrn]

Hey
Three points have coordinates A(1, 7), B(7, 5) and C(0,−2). Find:
a the equation of the perpendicular bisector of AB
b the point of intersection of this perpendicular bisector and BC.

for a, i found that m=3
do you sub either the A or B to find the equation? i did both and it doesnt work.

thanks

for a. first you find the mid point of AB adding the x values of A and B then dividing by 2, then doing the same for the y values.
you then find the gradient of AB, then you can work out the gradient of the perpendicular bisector (when multiplied together they equal -1).  now you've got the gradient and 1 point to work with and therefore you can find the equation.

for b. just find the equation of BC, then you can find where it intersects by equating to the equation of AB

[Ilovemathsmeth]

How do you do in equations such as
{x:y>x^2+4}
Do you think of it like: when is the graph above y=0 or something?

Greater/less than or equal to type equations need a complete line, where < or > type equations need a dotted line. I think it's good to just draw the function x^2 + 4 and find where y > 0. In that case, this is for all x is R. Don't forget to shade above the function as y is greater than the x^2 + 4 for other y values other than those that lie on the line.

[Ilovemathsmeth]

And also (since the square root function always give a positive answer)
(This caught out a lot of people in the Methods CAS 2009 Exam apparently)

Ahh that was mentioned in a tiny section in the Essentials text book. Those who did not take notice of that section/or didn't cover it in practice exam style questions would have been caught out. It's quite a useful little rule in simplifying some root (function squared) questions

[Ilovemathsmeth]

Hielly - you wouldn't use either A or B, because a perpendicular bisector cuts the line AB directly in half making an angle of 90 degrees with the line as it is perpendicular. This means you need to find the midpoint of line AB and then use those coordinates in finding the equation of the perpendicular bisector.

[kenhung123]

How do you do in equations such as
{x:y>x^2+4}
Do you think of it like: when is the graph above y=0 or something?

Greater/less than or equal to type equations need a complete line, where < or > type equations need a dotted line. I think it's good to just draw the function x^2 + 4 and find where y > 0. In that case, this is for all x is R. Don't forget to shade above the function as y is greater than the x^2 + 4 for other y values other than those that lie on the line.

Why when solving y<x^2 a domain while y<x+5 shading in the graph?

[TrueTears]

How do you do in equations such as
{x:y>x^2+4}
Do you think of it like: when is the graph above y=0 or something?

Greater/less than or equal to type equations need a complete line, where < or > type equations need a dotted line. I think it's good to just draw the function x^2 + 4 and find where y > 0. In that case, this is for all x is R. Don't forget to shade above the function as y is greater than the x^2 + 4 for other y values other than those that lie on the line.

Why when solving y<x^2 a domain while y<x+5 shading in the graph?

When sketching them you must shade in the area required or area not required depending on the question.

When asked to solve for x, you must give a domain.

[Hielly]

Hey, going back to simultaneous equations. I dont understand solutions such as no solutions,infinite solutions, and unique solution. Can someone please explain the concept of this? The essential textbook only shows you how to input the numbers in the calc and get the answer.
For example
Consider the simultaneous equations
mx + 2y = 8
4x − (2 − m)y = 2m
a Find the values of m for which there are:
i )no solutions
ii) infinitely many solutions
iii)unique solution
b) Solve the equations in terms of m, for suitable values of m.

Thanks

[TrueTears]

Solving these 2 simultaneous equations yields:





i) For no solutions, as that will leave x and y both undefined. Sketch the graph, these 2 lines are now parallel, they have the same gradient.

ii) For infinitely many solutions, the 2 lines must be the same line.

Thus

iii) For unique solutions and

[Hielly]

Thanks,
Also for this
a)Solve the simultaneous equations 2x − 3y = 4 and x + ky = 2, where k is a constant.
so elimination
equation becomes y(-3-2k)=0
so y=0 k=-3/2

why does the answer say k does not equal to k=-3/2 ?

[TrueTears]

and

But you must remember why? Because otherwise it would have infinite solutions and I assume the question wanted an unique solution.

[Hielly]

thanks

x=y^2 + 2y

make y the subject. How woule i do this?


thanks

[TrueTears]