# hi i really need help on this stuff

• Jan 27th 2009, 08:41 PM
KyrieKaye
hi i really need help on this stuff
i dont really suck at math...

i blame my teacher... he doesnt know how to teach...
he doesnt explain the problems very well..

im switching my teacher next week.. bleh....

please i really need help =3

2) standard form 9i - 3i^2

6)rational zero test
2x^3-2x^2+2x-10

8) (1/2)^x=228

9) if logw = logx+1/2logp then w = ?

10) if f(x) = x^2-x+1 then f(2a) = ?

12) x^2+x-3=0 root = ? < --- not sure
• Jan 28th 2009, 06:02 AM
earboth
Quote:

Originally Posted by KyrieKaye
...

6)rational zero test
2x^3-2x^2+2x-10

8) (1/2)^x=228

9) if logw = logx+1/2logp then w = ?

10) if f(x) = x^2-x+1 then f(2a) = ?

12) x^2+x-3=0 root = ? < --- not sure

to #6:

Plug in the values $\pm1,\ \pm2,\ \pm5,\ \pm10$ and check if the term becomes zero.

to #8:

$\left(\frac12\right)^x=228~\implies~\left(2\right) ^{-x}=228~\implies~ -x=\log_2(228)~\implies~ x=-\log_2(228)$

But if the original problem reads:

$\left(\frac12\right)^x=128~\implies~\left(2\right) ^{-x}=2^7~\implies~ -x=7~\implies~ x=-7$

to #9:
Use the log rules:
$a\log(x)=\log(x^a)$

$\log(a) + \log(b) = \log(a\cdot b)$

The final result should be: $w = \sqrt{px^2}$

to #10:
You are asked to plug in 2a instead of x:

$f(2a) = (2a)^2-(2a)+1$

to #12:
Solve $x^2+x-3=0$ . Use the formula to solve a quadratic equation.

The equation

$ax^2+bx+c=0$ has the solution $x=\dfrac{-b \pm\sqrt{b^2-4ac}}{2a}$
• Jan 28th 2009, 09:19 AM
KyrieKaye
thank you soo much...
its much clearer to me now