VCE Methods Question Thread!

[amanaazim]

can someone solve this

sketch the region described by x+3y<6 (by the way the inequality sign is actually greater than or equal to 6 i just don't know how to put that sign someone teach me)

[Sine]

can someone solve this

sketch the region described by x+3y<6 (by the way the inequality sign is actually greater than or equal to 6 i just don't know how to put that sign someone teach me)

Rearrange to have a inequality with y > or y <
Sketch this line and then use a test point (the origin is easy - if the line does not go through the origin) to see which side of the graph you need to shade.

[TheEagle]

can someone solve this

sketch the region described by x+3y<6 (by the way the inequality sign is actually greater than or equal to 6 i just don't know how to put that sign someone teach me)

Rearrange to make y the subject
3y<6-x
y<2- 1/3x

Since its "y is less than 2 - 1/3x", we must shade below our "y" or in other words, shade below the line.
If it's y>2 - 1/3x, then shade above the line since it's "greater than y"

Also, if you want to express the less than or equal to sign on computer, just type both:  <=
i.e:   y>=x     is read as "y is greater than or equal to x"

Otherwise, you can use Sine's method of substituting points to find the boundary.

[amanaazim]

can someone solve this question for me:
MCQ
how long would a plumber have visited for if her total charge was $104, given a service call costs $27 and her usual charge is $35 per hour?

A- 2 hours
B- 2 hours and 12 minutes
C- 2 hours and 15 minutes
D- 2 hours and 2 minutes
E- 2 hours and 20 minutes

[fun_jirachi]

can someone solve this question for me:
MCQ
how long would a plumber have visited for if her total charge was $104, given a service call costs $27 and her usual charge is $35 per hour?

A- 2 hours
B- 2 hours and 12 minutes
C- 2 hours and 15 minutes
D- 2 hours and 2 minutes
E- 2 hours and 20 minutes

Hey there!

What would your automatic approach to this question be? We'd like to see some working because asking us to solve questions for you doesn't help you learn as much as guiding you - you're not going to have this forum in the exam, after all. It's just a small thing to help us help you!

Something to note is that the answer is an amount of time, while there is a rate of change with respect to time in the question. That's a subtle hint that needs to be noticed every time in these sorts of questions.

Hope this helps

[colline]

can someone solve this question for me:
MCQ
how long would a plumber have visited for if her total charge was $104, given a service call costs $27 and her usual charge is $35 per hour?

A- 2 hours
B- 2 hours and 12 minutes
C- 2 hours and 15 minutes
D- 2 hours and 2 minutes
E- 2 hours and 20 minutes

Try putting the info given in the question into an algebra equation, with whatever you're solving for being 'x'. In this case x would be the time (in hours) she visited for. If she charges $35 an hour and visited for 'x' hours, then her charge for this would be 35*x. Then you must add the service call fee, which is $27, so her total charge is 27 + 35x which the question states is equal to 104. Then all you have to do is solve for x, then convert your answer into hours and minutes.

[Itbelikethatsometimes]

Pretty confused with this:
Find the y-coordinates of the points of intersection of the parabola with the equation y=x²-a and the circle with equation x²+y²=4.

I'm not sure if i'm suppose to simultaneous then use discrim?

[TheEagle]

Pretty confused with this:
Find the y-coordinates of the points of intersection of the parabola with the equation y=x²-a and the circle with equation x²+y²=4.

I'm not sure if i'm suppose to simultaneous then use discrim?

I would solve it like a normal intersection of two graphs; equate them
Since we know  y=x^2-a
then we can substitute it in the circle equation
therefore, x^2 + (x^2-a)^2 = 4
x^2 + x^4 - 2ax^2 + a^2 = 4
x^4 + x^2 - 2ax^2 = 4 - a^2
solve for x, you should get for solutions (since it's a circle being intersected with a parabola)
you can the substitute in the first equation, y=x^2 - a , in order to get the y coordinates.

Having done that, I'd highly suggest beginning with y instead of x. As shown above, I begun with x and it's going to take a bit of time. Always read the question if it's asking for x or y. That will help determine what to begin with.

[TheEagle]

Just in case you don't get what I mean by "starting with y", I'll do it below;

We know   1.) y= x^2 - a
we also know 2.) x^2 +y^2=4

2.) can be written as:  x^2 = 4 - y^2
now all we need to do is substitute it in 1.)
therefore,
y= (4-y^2) - a
now we just solve for Y.

This way is obviously way quicker although the first method will still get you to the right answer

[Geoo]

Hi, can someone give me instructions on how to find the derivative on a casio classpad calculator.
I can't find instructions anywhere, thank you

[ArtyDreams]

Hi, can someone give me instructions on how to find the derivative on a casio classpad calculator.
I can't find instructions anywhere, thank you

Keyboard - Math 2 - 3rd row, 2nd button should have an icon with
d/d[] []
Just type 'x' into the box in the denominator (or whatever variable you are derivating) and then write the expression in the next box.

Hope this helps! Sorry, I really dont know how to format things here properly. I can add a pic if you need.

[Geoo]

Keyboard - Math 2 - 3rd row, 2nd button should have an icon with
d/d[] []
Just type 'x' into the box in the denominator (or whatever variable you are derivating) and then write the expression in the next box.

Hope this helps! Sorry, I really dont know how to format things here properly. I can add a pic if you need.

This helped so thank you

[Rose34]

Can someone please help me with this question:
its from the topic y= √x

For each of the following rules, sketch the corresponding graph, giving the axis intercepts when they exist, the set of x-values for which the rule is defined and the set of y-values which the rule takes:
y= −2√3−x  Note: (the square root is for both 3 and -x)


Thanks in advance

[TheEagle]

Can someone please help me with this question:
its from the topic y= √x

For each of the following rules, sketch the corresponding graph, giving the axis intercepts when they exist, the set of x-values for which the rule is defined and the set of y-values which the rule takes:
y= −2√3−x  Note: (the square root is for both 3 and -x)


Thanks in advance

the domain (x values): since the square root needs to be greater than or equal to zero, the domain is therefore [3, infinity)


x intercept: when y=0. solve for x
y intercept when x=0. solve y

[LachlanBarr8]

Hey guys,

Could i please have some clarification with the following question(s), working out by hand would be much appreciated, im pretty sure i have the right answers just want to double check

Q1: Consider the function defined by f(x) = 2(x-3)^2.
a)   Find the rule for the inverse.
b)   Restrict the domain of f to the form of [a, infinity] so that the inverse is also a function.
c)   State the rules for the restricted f and f^-1 using function notation.
d)   Show that f(f^-1(x)) = x

Q2: For: mx +2y = n and 3x + 6y = -1, find m and n for which the equations have:
a) a unique solution
b) an infinite number of solutions
b) no solution