VCE Methods Question Thread!

[snowisawesome]

MMCAS exam 2 2016
MCQ 15
A box contains six red marbles and four blue marbles. Two marbles are drawn from the box, without
replacement.
The probability that they are the same colour is:
a. 1/2
b. 28/45
c. 7/15
d. 3/5
e. 1/3


My working out was like this
p(RR)* p(BB) = ((6/10)*(5/9)) * ((4/10)*(3/9))
= (1/3)*(2/15) = 2/45

but the answer said option c
Could someone plase explain how this question would be worked out?
Thanks 

[Aaron]

MMCAS exam 2 2016
MCQ 15
A box contains six red marbles and four blue marbles. Two marbles are drawn from the box, without
replacement.
The probability that they are the same colour is:
a. 1/2
b. 28/45
c. 7/15
d. 3/5
e. 1/3

Could someone plase explain how this question would be worked out?
Thanks 

What do you think (and why)? At least have a go at it so those who will assist you with this can help you understand misconceptions for future questions of this nature.

[Unsplash]

MMCAS exam 2 2016
MCQ 15
A box contains six red marbles and four blue marbles. Two marbles are drawn from the box, without
replacement.
The probability that they are the same colour is:
a. 1/2
b. 28/45
c. 7/15
d. 3/5
e. 1/3


My working out was like this
p(RR)* p(BB) = ((6/10)*(5/9)) * ((4/10)*(3/9))
= (1/3)*(2/15) = 2/45

but the answer said option c
Could someone plase explain how this question would be worked out?
Thanks 

Just need to add p(RR) and p(BB) rather than multiply. From the question they can both be red OR blue and therefore addition is required. 42/90 simplifies to 7/15.

[snowisawesome]

Just need to add p(RR) and p(BB) rather than multiply. From the question they can both be red OR blue and therefore addition is required. 42/90 simplifies to 7/15.

Thank you very much

[skrt skrt]

Hey guys just started doing transformations and I just need help with this question.

The equation of the image of the graph y=x^2 after a dilation of factor 1/3 from the y-axis followed by a translation of 5 units in the negative direction of the x-axis and 2 units in the positive of the y-axis is:

My steps were:
1.(3x)^2

2.9(x+5)^2

3.9(x+5)^2+2

Not sure if I did the dilation the correct way, please check.

Thanks

[Sine]

Hey guys just started doing transformations and I just need help with this question.

The equation of the image of the graph y=x^2 after a dilation of factor 1/3 from the y-axis followed by a translation of 5 units in the negative direction of the x-axis and 2 units in the positive of the y-axis is:

My steps were:
1.(3x)^2

2.9(x+5)^2

3.9(x+5)^2+2

Not sure if I did the dilation the correct way, please check.

Thanks

yeah it is correct but you should know how to check your answers either by having a solid technique for answering these types of questions or also having an alternative methods to double check.

For the dilation of factor 1/3 from the y-axis

y= x^2
x is replaced with x/(1/3)
y = (x/(1/3))^2
y= (3x)^2 = 9x^2

[snowisawesome]

(x-2)^(1/2)-5

My answer to the range was [-5, infinity) but the answer said it was (-5, infinity]
Can someone please explain why this is the case?

[Sine]

(x-2)^(1/2)-5

My answer to the range was [-5, infinity) but the answer said it was (-5, infinity]
Can someone please explain why this is the case?

answer is (-5, infinity] obviously incorrect since infinity is never inclusive

You should try to get to the point where you are able to check whether your own answer is correct and know when the textbook/other sources are incorrect (even though they are the "answers")

[snowisawesome]

answer is (-5, infinity] obviously incorrect since infinity is never inclusive

You should try to get to the point where you are able to check whether your own answer is correct and know when the textbook/other sources are incorrect (even though they are the "answers")

Wait you said infinity can never be inclusive but then you said the answer was (-5, infinity]

[Shadowxo]

Wait you said infinity can never be inclusive but then you said the answer was (-5, infinity]

I think he meant to say the answer (-5, infinity] was wrong. For that equation the answer should be [-5, infinity)
(I'll edit this later with more information about the solution)

[Sine]

just meant to say that the "answer" (-5, infinity] is wrong

[snowisawesome]

Thanks sine and shadowxo for the help

Had another question
f(x) = (x-2)/(x+1)
find inverse
change y = f(x) into different form
y = (1(x+1)-3)/(x+1)
so it becomes y = 1-3/(x+1) or y = -3/(x+1) + 1
so to find inverse x = -3/(y+1)+1
so (x-1)=-3/(y+1)
so (y+1)=-3/(x-1)
so y = -3/(x-1)-1

Am I correct?
The book's answers said the inverse was y = (x+2)/(1-x)
Is this the same as my answer, and if so, how would I change my answer to the same format of the book's answers?

Thanks

[TheAspiringDoc]

Thanks sine and shadowxo for the help

Had another question
f(x) = (x-2)/(x+1)
find inverse
change y = f(x) into different form
y = (1(x+1)-3)/(x+1)
so it becomes y = 1-3/(x+1) or y = -3/(x+1) + 1
so to find inverse x = -3/(y+1)+1
so (x-1)=-3/(y+1)
so (y+1)=-3/(x-1)
so y = -3/(x-1)-1

Am I correct?
The book's answers said the inverse was y = (x+2)/(1-x)
Is this the same as my answer, and if so, how would I change my answer to the same format of the book's answers?

Thanks

On your CAS, you can use a judge function to tell if one expression is equal to another - I use this heaps!!

[snowisawesome]

On your CAS, you can use a judge function to tell if one expression is equal to another - I use this heaps!!

Thanks

Also, do you know how to solve 1024 = 2^x without a cas?

[Yertle the Turtle]

Thanks sine and shadowxo for the help

Had another question
f(x) = (x-2)/(x+1)
find inverse
change y = f(x) into different form
y = (1(x+1)-3)/(x+1)
so it becomes y = 1-3/(x+1) or y = -3/(x+1) + 1
so to find inverse x = -3/(y+1)+1
so (x-1)=-3/(y+1)
so (y+1)=-3/(x-1)
so y = -3/(x-1)-1

Am I correct?
The book's answers said the inverse was y = (x+2)/(1-x)
Is this the same as my answer, and if so, how would I change my answer to the same format of the book's answers?

Thanks

What you can do here is make -3/(x-1)-1 to -3/(x-1)-(x-1)/(x-1)
=>(-3-x+1)/(x-1)
=>(-2-x)/(x-1)
=>(x+2)/(1-x)
This comes out to the same answer to the book, so you are right. What you need to do is make the whole expression have the same denominator and then you can work it as simple algebra