ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: MrsStark on January 29, 2015, 05:26:09 pm
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hey ! so this is probs easy but I it don't get it atm
x/(a-b) + 2x/(a+b) = 1/ (a^2 - b^2)
If you can help me, thanks so much!
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solve for x
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 + 2x(a-b)}{(a-b)(a+b)} = \frac{x(3a - b)}{a^2 - b^2})
Compare the two fractions, and you will see that
.
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 + 2x(a-b)}{(a-b)(a+b)} = \frac{x(3a - b)}{a^2 - b^2})
Compare the two fractions, and you will see that
.
Thank you :)
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solve for x:
p^2(1-x) - 2pqx= q^2(1+x)
solve for x:
(p-qx)/t + p = (qx-t)/p
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solve for x:
p^2(1-x) - 2pqx= q^2(1+x)
solve for x:
(p-qx)/t + p = (qx-t)/p
I tend to find these questions annoying because they're so easy to screw up! It's just a bunch of messy algebra. I suggest you try the first one yourself, but I will do the second one for you.
Solve for x: }{t} + p = \frac{(qx-t)}{p})
}{t} - \frac{(qx-t)}{p} = -p)
p-t(qx-t)}{tp} = -p)
Now expand:

Multiply both sides by
:

Collate like terms:

Simplify and solve:
))
} )
Hopefully I didn't make any silly algebra mistakes.
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I tend to find these questions annoying because they're so easy to screw up! It's just a bunch of messy algebra. I suggest you try the first one yourself, but I will do the second one for you.
Solve for x: }{t} + p = \frac{(qx-t)}{p})
}{t} - \frac{(qx-t)}{p} = -p)
p-t(qx-t)}{tp} = -p)
Now expand:

Multiply both sides by
:

Collate like terms:

Simplify and solve:
))
} )
Hopefully I didn't make any silly algebra mistakes.
Yeah I hate these qs too lol its simple but easy to mess up I know !!
Yup, I tried the first one again and I got it yay!
Thanks for helping! and no I don't think you made any mistakes :)
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Glad I could help!
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Revising for my functions sac and stumbled on this question!
The cost of a taxi trip in a particular city is $4 up to and including 2km. After 2km the passenger pays an additional $2 per km. find the function f which describes this method of payment and sketch the graph of y= f(x), where x is the number of km travelled ( use a continuous model)
okay so you don't have to draw the graph for me but I have trouble with worded questions and don't get how the function is derived
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Revising for my functions sac and stumbled on this question!
The cost of a taxi trip in a particular city is $4 up to and including 2km. After 2km the passenger pays an additional $2 per km. find the function f which describes this method of payment and sketch the graph of y= f(x), where x is the number of km travelled ( use a continuous model)
okay so you don't have to draw the graph for me but I have trouble with worded questions and don't get how the function is derived
Let x=number of kilometres travelled, then f(x) describes how much the person owes per kilometre.
For the first 2km, f(x)=4 (hopefully easy to see). After that, f(x) will raise $2 per kilometre, so it will be f(x)=2x+4 - BUT, we must remember to take off the 2 kilometres already travelled, giving us f(x)=2(x-2)+4=2x-4+4=2x
So, we get:
f(x)=4, 0<=x<2
f(x)=2x, x>=2
You should write this as a proper hybrid function.
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How do you find the inverse of a 3 x 3 matrix using the cas? it doesn't really show it in my textbook..
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How do you find the inverse of a 3 x 3 matrix using the cas? it doesn't really show it in my textbook..
Just put it to the power of -1.
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Just put it to the power of -1.
yeah I have been but I get a strange answer :s according to the back of the book answers its wrong.. don't know whats up with that..
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Put the matrix here, I'll tell you what the inverse should be and we'll compare to the book. Book could be wrong.
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Put the matrix here, I'll tell you what the inverse should be and we'll compare to the book. Book could be wrong.
figured it out, my book is secondhand and its been written on and rubbed with rubber and I just misread a number -.- oh dear. I got the answer its all good, thank you for wanting to help :)
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hey can someone please help me solve for x?
2^(x) X 3^(x+1)= 10
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hey can someone please help me solve for x?
2^(x) X 3^(x+1)= 10
2^x X 3^x X 3 = 10 (splitting up 3^x+1)
2^x X 3^x = 10/3
(2 X 3)^x = 10/3 (Power rule: (ab)^n = a^n x b^n)
therefore, log6(10/3) = x
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2^x X 3^x X 3 = 10 (splitting up 3^x+1)
2^x X 3^x = 10/3
(2 X 3)^x = 10/3 (Power rule: (ab)^n = a^n x b^n)
therefore, log6(10/3) = x
(2 X 3)^x = 10/3
6^x = 10/3
log6^x = log(10/3)
x X log6 = log(10/3)
x = (log(10/3))/(log6)
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(2 X 3)^x = 10/3
6^x = 10/3
log6^x = log(10/3)
x X log6 = log(10/3)
x = (log(10/3))/(log6)
They are the exact same answer - watch what happens when I change from base 6 to base 10 (with wobblywobbly's answer being base 6, yours being base 10):