I have few probs i need to help with here are cupple i did them on photoshop heres the link to it CLICK HERE TO SEE THE PROBS

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- Mar 5th 2006, 08:27 AM #1

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## Need Help!

I have few probs i need to help with here are cupple i did them on photoshop heres the link to it CLICK HERE TO SEE THE PROBS

- Mar 5th 2006, 10:13 AM #2

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Problem 3]

$\displaystyle \frac{x-3}{4}-\frac{x}{7}=7$

Multiply by Common Denominator to get,

$\displaystyle 28\cdot\frac{x-3}{4}-28\cdot\frac{x}{7}=28\cdot 7$, thus,

$\displaystyle 7(x-3)-4x=196$

Thus,

$\displaystyle 7x-21-4x=196$

Thus,

$\displaystyle 3x-21=196$

Thus,

$\displaystyle 3x=217$

Thus,

$\displaystyle x=\frac{217}{3}=72.\bar 3$

- Mar 5th 2006, 10:38 AM #3

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Problem 4]

$\displaystyle \frac{m^4(n^{-3})^2}{(m^{-2})^{-3}n^0} \left(\frac{m^{-4}n^8}{n^8m^{-2}} \right)^2$

Open parantheses,

$\displaystyle \frac{m^4n^{-6}}{m^6n^0}\cdot \frac{m^{-8}n^{16}}{n^{16}m^{-4}} \right)$

Use rule of exponents,

$\displaystyle m^{-2}n^{-6}m^{-4}n^0$

Use rule of exponents again ($\displaystyle n^0=1$),

$\displaystyle m^{-6}n^{-6}$

Q.E.D.

- Mar 5th 2006, 10:40 AM #4

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- Mar 5th 2006, 10:45 AM #5Originally Posted by
**autopimp**

Solve by substitution:

$\displaystyle 2x+3y=7$

$\displaystyle x-2y=7$

In the substitution method we solve one of the equations for one unknown and plug it into the other...this creates an equation with only one unknown in it. I tend to pick the simplest variable to solve for. In this case that looks to be the x in the second equation.

So solving the second equation for x I get:

$\displaystyle x=2y+7$

Plugging this value of x into the first equation I get:

$\displaystyle 2(2y+7)+3y=7$.

I'll let you take it from here. You may plug this value of y into any of the original equations, or more simply, into your x equation. I get (x,y) = (5,-1).

-Dan

- Mar 5th 2006, 10:57 AM #6Originally Posted by
**autopimp**

to problem nr. 6: Transform the quotient into a product:

$\displaystyle \frac{2x-10}{x^2-25}\ \cdot \ \frac{x^2-4x-5}{x^2-2x-15}$

factorize all numerators and denominators:

$\displaystyle \frac{2(x-5)}{(x+5)(x-5)}\ \cdot \ \frac{(x+1)(x-5)}{(x+3)(x-5)}$

simplify and you'll get:

$\displaystyle \frac{2(x+1)}{(x+5)(x+3)}$

Greetings

EB

- Mar 5th 2006, 11:03 AM #7Originally Posted by
**autopimp**

to problem nr. 8:

Calculate the roots partially:

$\displaystyle 3\cdot \sqrt{32}-5\cdot \sqrt{50}-8 \cdot \sqrt{45} + 7\cdot \sqrt{20}=$

$\displaystyle 3\cdot 4\sqrt{2}-5\cdot 5\sqrt{2}-8 \cdot 3\sqrt{5} + 7\cdot 2 \sqrt{5}=$

$\displaystyle -13 \sqrt{2}-10 \sqrt{5}$

Greetings

EB