# Help Fast!!!

• Mar 28th 2006, 01:59 PM
mathelp
Help Fast!!!
• Mar 29th 2006, 10:27 AM
MathGuru
heres the first one
• Mar 29th 2006, 10:29 AM
MathGuru
For the second one solve the first part for y to get y = 2x-10 and then sub that into the second equation to get:

x + 3(2x-10) = -9, now solve for x and then sub back in to either of the given equations and solve for y also.
• Mar 29th 2006, 10:32 AM
MathGuru
Q#3
9b^2c + 6c
-----------
......3c

Is the same as:

9b^2c + 6c
------ -----
...
3c.......3c

or

3b^2 + 2

does that make sense?
• Mar 29th 2006, 10:41 AM
MathGuru
#5
notice that 27 is 3*(3^2)
notice that 12 is 3*(2^2)

$\displaystyle -4\sqrt27 = -4\sqrt(3*3^2)=3*-4\sqrt3=-12\sqrt3$
• Mar 29th 2006, 11:19 AM
earboth
Quote:

Originally Posted by mathelp

Hello,

to Nr. 5:
$\displaystyle -4 \sqrt{27}+2\sqrt{12}=$
$\displaystyle -4 \sqrt{9 \cdot 3}+2\sqrt{4 \cdot 3}=$
$\displaystyle -4 \cdot 3 \sqrt{ 3}+2 \cdot 2 \sqrt{ 3}=$
$\displaystyle -12 \sqrt{3}+4 \sqrt{3}= -8 \cdot \sqrt{3}$

to 6.:
$\displaystyle \left( \frac{5c^{-3} d^4}{p^{-2}} \right)^{-1}\ \cdot \ \left( \frac{p^7 d^{-1}}{c^2} \right)^2 =$
$\displaystyle 5^{-1} \cdot c^3 \cdot d^{-4} \cdot p^{-2} \cdot p^{14} \cdot d^{-2} \cdot c^{-2} =$
$\displaystyle \frac{cp^{12}}{5d^6}$

Greetings

EB
• Mar 29th 2006, 11:49 AM
mathelp
Thnaks bro so much
• Mar 29th 2006, 12:16 PM
mathelp
can you show me 2,4 how to do?
• Mar 29th 2006, 12:58 PM
MathGuru
#4

Graph both equations on a graphing calculator and the point where they intersect is your answer of the form (x,y)
• Mar 29th 2006, 01:11 PM
ThePerfectHacker
Here is the graphs,