Math

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abajan
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#21 Postby abajan » Tue May 01, 2007 6:43 pm

Whether you like it or not, mathematics is important.

One just needs to observe how helpless cashiers become when their registers go down.
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lurkey
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#22 Postby lurkey » Tue May 01, 2007 7:08 pm

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alicia-w
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#23 Postby alicia-w » Wed May 02, 2007 8:53 am

fact789 wrote:That means I dont have to get the work, I just have to do it.


huh? please tell me you arent exempt from your English exam too!
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JonathanBelles
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#24 Postby JonathanBelles » Wed May 02, 2007 12:56 pm

Actually I am exempt from my english exam with the second highest grade in the class.

I dont have to understand the work, just do it.
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wxmann_91
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Re: Math

#25 Postby wxmann_91 » Wed May 02, 2007 11:54 pm

fact789 wrote:I hate math:

1) logbase5 (x-1) +log base5 (x-2) - Logbase5 (x+6)= 0
A) solve for X
B) show your work
log properties to start. subtracting logs equals dividing the stuff that's "logged" (don't know exact term). So in this case it will be logbase5 [(x-1)(x-2)]/(x+6). After that just definition of logs, 5^0 = [(x-1)(x-2)]/(x+6)

2) If 1+2i is one root of X^3 +x +10 = 0, find the other two roots.
A) show your work
Goodness I wish I still remember imaginary roots from last year. And I hate third-degree or higher polynomials. I think you'd use conjugate base/zero (I forgot name, thinking chemistry here oops) theorum. So you'd know the other root is 1-2i. Although I don't know the rest.

3) If F(x) is a linear function and f(1)=2 and f(3)=7, find an equation for f(x) in terms of x.
A) show your work
In calc we've learned a nice formula, y-f(x)=m(x-x0). Find the slope, f(b)-f(a) divided by b-a, and then plug in any one of the two points.
Or, using mx+b formula, find slope and plug a point in to find y-intercept, then plug in values.


4) (2x-5)/4 < (4-3x)/ 2.
A) solve for X
B) show your work
In this case I would set them equal, then do cross products and solve for x, then do sign chart, plug in x values and see if they're positive or negative

5) find the sum and product of the roots of the equation x^3 - 19x - 30 =0
A) show your work
I hate solving polynomials with powers greater than 3 (did I already say that?). I'd guess you could use the rational zero theorum (or whatever it's called) to get the possible factors, use synthetic division, leaving a quadratic.

This is supposed to be review from ALG1, but that was 3 years ago and I dont remember it.

Thank You in advance! You're welcome, hope this helps
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#26 Postby pojo » Thu May 03, 2007 1:57 am

ouch.... my brain is hurting!

and I'm good at math!
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#27 Postby tropicana » Thu May 03, 2007 8:19 am

and im bad at math, and my brain is still hurting just looking at those problems, days after i first saw them.
-justin-
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