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Need to practice these for next year, so will get on to them soon  thanks
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Your answers for #1 and #2 are a bit off! I haven't looked at #3!
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On December 09 2011 21:12 Day[9] wrote:
Your answers for #1 and #2 are a bit off! I haven't looked at #3!
A bit off how ? Incorrect numbers ? o.O
+ Show Spoiler +OMFG Day9 replied to my post! Bronze to diamond thanks to you, and my mother thinks you're funny<3
(Please do a daily on how to spin pens pleaaaaase :D)
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is the first question y=(3x+2)/(x-4)?
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On December 09 2011 21:23 plutonius wrote:
is the first question y=(3x+2)/(x-4)?
Yep, it is like this...
3x+2
(divided by)
x-4
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3# is not a problem, all you need to do is to apply the law of cosines. Result is a bit off by the way
+ Show Spoiler +cos-1(4.2/6.4) = 48.9855°
edit : if you like maths problems, you should read evanthebouncy!'s blog entries !
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On December 09 2011 21:22 achristes wrote:Show nested quote +On December 09 2011 21:12 Day[9] wrote:
Your answers for #1 and #2 are a bit off! I haven't looked at #3! A bit off how ? Incorrect numbers ? o.O + Show Spoiler +OMFG Day9 replied to my post! Bronze to diamond thanks to you, and my mother thinks you're funny<3
(Please do a daily on how to spin pens pleaaaaase :D)
In #1 take a look at the denominator.. Suffice it to say, strange things happen when the denominator approaches zero..
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On December 09 2011 21:25 endy wrote:3# is not a problem, all you need to do is to apply the law of cosines. Result is a bit off by the way + Show Spoiler +cos-1(4.2/6.4) = 48.9855°
Maybe not a problem, but we just finished a chapter about trigonometry and I thought it was cool
I did (after calculating BC)
+ Show Spoiler +4.8^2=6.4^2+4.2^2-2*6.4*4.2*cosA
A=cos-1 (6.4^2+4.2^2-4.8^2/2*6.4*4.2)
A=48.5888
If it's off I blame my calculator! (casio fx-9860GII)
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For the second question, I'm assuming you are finding the roots or values of x that yield y = 0.
+ Show Spoiler +Applying the quadratic formula:
a = 2
b = 7
c = -4
x = [- b + root(b^2-4ac)]/2a
x = [- b - root(b^2-4ac)]/2a
x = 1/2,-4
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Oh, shit, big blunder by me. Editing OP.
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Unfortunately, all of your answers are incorrect.
1.
y = 3x + 2 / x - 4. I'll assume that you didn't post the function with missing brackets. What's written is y = (3 x) + (2 / x) - 4
+ Show Spoiler +
Asymptotes are line that the function creeps closer and closer to but never reaches. To determine a horizontal asymptote, simply look at what the function does as x tends to positive and negative infinity. An easy test is to compute f(100), f(1000), f(10000), etc... and similarly for the negative numbers, to find the value that the function tends to. In this case, the function is ever increasing (for large positive x) or decreasing (large negative x), so there is no horizontal asymptote.
A vertical asymptote can be found by finding a value of x where the function is not defined. In this case, at x = 0, the function is not defined (division by zero in the second term). Fill in small value of x and see if the function goes to infinity to test if it is indeed an asymptote. As you fill in f(0.01), f(0.001), etc.. you'll notice that the function values explode. So the vertical asymptote is x = 0.
Note that there is also an oblique asymptote: y = 3 x.
edit: You edited the first question. Here the updated solution:
1. edited
(3x + 2) / (x - 4)
+ Show Spoiler +
For large positive or negative values of x, the constant terms in the numerator and denominator no longer matter much, so the function will tend to 3x / x = 3. Consequently, the horizontal asymptote is y = 3.
The vertical asymptote will most likely lie where the function is undefined: At x = 4. Testing by filling in values very close to x = 4, f(4.001), f(4.0001), etc... we see the function-value blow up as we approach x = 4. So the vertical asymptote is x = 4.
2.
2x^2 + 7x - 4 = 0
+ Show Spoiler +
The solutions are given by the "abc" formula:
x = -b / (2 a) + sqrt(b^2 - 4 a c) / (2 a)
If you fill in the relevant values (a = 2, b = 7, c = -4) and work it out, you get that x = 0.5 or x = -4.
3.
+ Show Spoiler +
You don't need the cosine law or anything like that. What you need to do is to recognize that you have a triangle with a right angle whose side have a ratio of 1 : sqrt(3) : 2. AC is the hypothenuse (not sure about the spelling of that word in english) and is twice as long as AB. A property of 1 : sqrt(3) : 2 triangles is that the angles are 30, 60 and 90 degrees. In this case, angle A is 60 degrees.
Draw a picture to make it more clear.
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Here's how to find horizontal asymptotes for when you have a polynomial divided by another polynomial:
Look at the biggest exponent in both the numerator and denominator. Then:
1. If the numerator's biggest exponent is larger than the denominator's biggest exponent, then there's no asymptote.
2. If the numerator's biggest exponent is smaller than the denominator's biggest exponent, then there's a horizontal asymptote at zero.
3. If the biggest exponents match, then take the leading coefficients (the coefficients of the x-values that have the highest exponents) and find the numerator's coefficient divided by the denominator's coefficient. There's an asymptote at that number.
(You can literally ignore all other smaller terms when evaluating the horizontal asymptote of a polynomial divided by another polynomial.)
Vertical asymptote of a polynomial divided by another polynomial is wherever you're dividing by zero.
Good luck!
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On December 09 2011 20:41 achristes wrote:
Mathematical english is really different (IMO) than mathematical norwegian, sorry in advance for any confusion.
When i did my thesis, I had to learn mathematical french because the area i was working in only had one good elementary textbook and of course it was only in french at the time. One good thing about maths is that there is definitely truth in the cliché that mathematics is a universal language
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On December 09 2011 22:44 sigma_x wrote:Show nested quote +On December 09 2011 20:41 achristes wrote:
Mathematical english is really different (IMO) than mathematical norwegian, sorry in advance for any confusion.
When i did my thesis, I had to learn mathematical french because the area i was working in only had one good elementary textbook and of course it was only in french at the time. One good thing about maths is that there is definitely truth in the cliché that mathematics is a universal language
Wow, I would be really mad if I had to learn french for any subject ^^
It's universal enough, but the expressions are a little bit different, for example
ligning = eqaution
andregradsligning = quadratic eqation
potens = power
standardform = scientific notation
pytagorassetningen = pythagorean theoren
etc
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What is this like 9th grade?
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On December 10 2011 00:02 jamesr12 wrote:
What is this like 9th grade?
Almost, + 2 years and then you're right. Anything wrong with that ?
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How would you even do 4 with digital help unless you have something abusive like a TI-89/NSpire or some overkill stuff like WAlpha/Mathematica/MATLAB....?
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On December 10 2011 01:06 ymir233 wrote:
How would you even do 4 with digital help unless you have something abusive like a TI-89/NSpire or some overkill stuff like WAlpha/Mathematica/MATLAB....?
By separating the problems
+ Show Spoiler +a^1/2*a^2*(a^1/3)^-2
----------------------------
a^2/3
(a^1/3)^-2 = a^-2/3
a^1/2*a^2
------------------
a^2/3*a^2/3
a^2/2
--------
a^4/3
It is what I think is the answer. Although i'm not completely sure.
And FFS, did excactly what I did just a few hours ago, posting answers without double checking >.<
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EPT
