how in the world do u solve this?

((2z)2 (x4)3) ÷ ((x2)-2(4z)2)

the final answer must contain positive exponents

Printable View

- Nov 19th 2007, 04:48 AMtdothow in the world
how in the world do u solve this?

((2z)2 (x4)3) ÷ ((x2)-2(4z)2)

the final answer must contain positive exponents - Nov 19th 2007, 05:26 AMSoroban
Hello, tdot!

I must assume you know the basic rules for exponents . . .

Quote:

$\displaystyle \frac{(2z)^2(x^4)^3}{(x^2)^{-2}(4z)^2} $

We have: .$\displaystyle \frac{2^2\cdot z^2\cdot x^{12}}{x^{-4}\cdot4^2\cdot z^2} \;=\;\frac{4\cdot x^{12}\cdot z^2}{16\cdot x^{-4}\cdot z^2} \;=\;\frac{x^{16}}{4}$

- Nov 19th 2007, 11:17 AMtdot
ok. its making more sence. i have other questions, that i tried out but wanna be sure that i have it right.

(2/x2)3 (x5)-2 (x/4)-1

(a2b3c)4 ÷ ( 3abc)2

(1+i)5 (1+i)-8

(y2/3 y1/3 y1/2) ÷ (y-6) - Nov 19th 2007, 10:38 PMearboth
Hello,

1. If you have a new problem start a new thread.

2. Use this sign ^ to write a power (or learn how to write in Latex)

3. Use brackets to separate bases, exponents, factors or summands.

I assume that you meant:

$\displaystyle \left(\frac2{x^2}\right)^3 \cdot (x^5)^{-2} \cdot \left(\frac x4\right)^{-1}$ Transcribe this term into a product:

$\displaystyle (2x^{-2})^3 \cdot (x^5)^{-2} \cdot (4^{-1} x)^{-1}$ Expand the brackets:

$\displaystyle 8 x^{-6} \cdot x^{-10} \cdot 4 x^{-1}=32x^{-17}=\frac{32}{x^{17}}$

= = = = = == = = = = = = = = = = = = = = = = = = = = = = = = = = =

$\displaystyle \frac{(a^2 b^3 c)^4}{(3 a b c)^2}=(a^2 b^3 c)^4 \cdot (3 a b c)^{-2}$ Expand and collect like terms. I've got:

$\displaystyle \frac19 \cdot a^6 \cdot b^{10} \cdot c^2$

= = = = = == = = = = = = = = = = = = = = = = = = = = = = = = = = =

(y^(2/3) y^(1/3) y^(1/2)) ÷ (y^(-6)) ==> $\displaystyle \frac{y^{\frac23} \cdot y^{\frac13} \cdot y^{\frac12} }{y^{-6}}=y^{\frac32} \cdot y^6=y^{\frac{15}2}=\sqrt{y^{15}}$