Help Fast!!!

**mathelp** · March 28th 2006, 01:59 PM

I need help with few probs very fast plz help!!

1)

2)

3)

4)

5)

6)

**MathGuru** · March 29th 2006, 10:27 AM

heres the first one

**MathGuru** · March 29th 2006, 10:29 AM

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.

**MathGuru** · March 29th 2006, 10:32 AM

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

Is the same as:

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

or

3b^2 + 2

does that make sense?

**MathGuru** · March 29th 2006, 10:41 AM

notice that 27 is 3*(3^2)
notice that 12 is 3*(2^2)

$-4\sqrt27 = -4\sqrt(3*3^2)=3*-4\sqrt3=-12\sqrt3$

**earboth** · March 29th 2006, 11:19 AM

Originally Posted by mathelp

I need help with few probs very fast plz help!!

...
5)

6)

Hello,

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

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

Greetings

EB

**mathelp** · March 29th 2006, 11:49 AM

Thnaks bro so much

**mathelp** · March 29th 2006, 12:16 PM

can you show me 2,4 how to do?

**MathGuru** · March 29th 2006, 12:58 PM

#4

Graph both equations on a graphing calculator and the point where they intersect is your answer of the form (x,y)

**ThePerfectHacker** · March 29th 2006, 01:11 PM

Here is the graphs,

Thread: Help Fast!!!

LinkBack

Thread Tools

Display

Help Fast!!!

Q#3

#5

Similar Math Help Forum Discussions

need help fast

Need help fast!

Need help fast

Need help fast

Can someone help me fast?

Search Tags