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Hello! I know it's been a while since I've been on but I need some help. 

Plz simplify (x^4)^5/(x^3)^3 ;  (a^2b^3/ab^2) and (2/3)^-3×(a^3/b)^-2

Sorry I don't know how to make it easier to read!

 Mar 25, 2015

Best Answer 

 #2
avatar+3693 
+10

Hey! Long time no see! How ya been?

I figured out all three expressions:

On the first problem, you would multiply the exponents...

$${\frac{{\left({{\mathtt{x}}}^{{\mathtt{4}}}\right)}^{{\mathtt{5}}}}{{\left({{\mathtt{X}}}^{{\mathtt{3}}}\right)}^{{\mathtt{3}}}}} = {\frac{{{\mathtt{x}}}^{{\mathtt{20}}}}{{{\mathtt{x}}}^{{\mathtt{9}}}}}$$

Now, since you are dividing, you would now subtract the exponents (11). Therefore, the answer to the simplified expression is x^11.

 

On the second problem, the only thing you have ot do to simplify it is cancel out the terms in the numerator and the denominator.

$${\frac{{{\mathtt{a}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{{\mathtt{b}}}^{{\mathtt{3}}}}{{{\mathtt{ab}}}^{{\mathtt{2}}}}} = {\frac{{{\mathtt{ab}}}^{{\mathtt{3}}}}{{{\mathtt{b}}}^{{\mathtt{2}}}}}$$

After that, you can continue to simplify the expression by cancelling out the"b" variables:

$${\frac{{{\mathtt{ab}}}^{{\mathtt{3}}}}{{{\mathtt{b}}}^{{\mathtt{2}}}}} = {\mathtt{ab}}$$

As a result, the answer to the simplified expression is ab.

 

Finally, on the thrid problem, you must get rid of the negative exponents. To do this, you would use the reciprocal of all the fractions.

$${\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{b}}}{{{\mathtt{a}}}^{{\mathtt{3}}}}}\right)}^{{\mathtt{2}}}$$

Then simplify the fractions by multiplying the expoinents into the fractions. So, it would be:

$$\left({\frac{{\mathtt{27}}}{{\mathtt{8}}}}{\mathtt{\,\times\,}}{\frac{{{\mathtt{b}}}^{{\mathtt{2}}}}{{{\mathtt{a}}}^{{\mathtt{6}}}}}\right)$$

After that, you would multiply it out and get your answer:

$${\frac{\left({\mathtt{27}}{\mathtt{\,\times\,}}{{\mathtt{b}}}^{{\mathtt{2}}}\right)}{\left({\mathtt{8}}{\mathtt{\,\times\,}}{{\mathtt{a}}}^{{\mathtt{6}}}\right)}}$$ -> -> -> 27b^2/8a^6

 

I hope this helped!

 Mar 25, 2015
 #1
avatar+128079 
+10

 (x^4)^5/(x^3)^3=  x^20 / x^9  = x^11

-------------------------------------------------------------------------------------------------------------------------

 (a^2b^3/ab^2)....I'm going to assume this is  (a^2b^3)/(ab^2).... if so, we have

(a^2)/(a) * (b^3)/(b^2) =

a * b  =

ab

-------------------------------------------------------------------------------------------------------------------------

(2/3)^-3×(a^3/b)^-2.......notice that we can use this"trick' to get rid of the negative exponents.....

(3/2)^3 * (b/a^3)^2    ....so we have...

(27/8) * (b^2 / a^6) =

(27b^2)/(8a^6)

 

  

 Mar 25, 2015
 #2
avatar+3693 
+10
Best Answer

Hey! Long time no see! How ya been?

I figured out all three expressions:

On the first problem, you would multiply the exponents...

$${\frac{{\left({{\mathtt{x}}}^{{\mathtt{4}}}\right)}^{{\mathtt{5}}}}{{\left({{\mathtt{X}}}^{{\mathtt{3}}}\right)}^{{\mathtt{3}}}}} = {\frac{{{\mathtt{x}}}^{{\mathtt{20}}}}{{{\mathtt{x}}}^{{\mathtt{9}}}}}$$

Now, since you are dividing, you would now subtract the exponents (11). Therefore, the answer to the simplified expression is x^11.

 

On the second problem, the only thing you have ot do to simplify it is cancel out the terms in the numerator and the denominator.

$${\frac{{{\mathtt{a}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{{\mathtt{b}}}^{{\mathtt{3}}}}{{{\mathtt{ab}}}^{{\mathtt{2}}}}} = {\frac{{{\mathtt{ab}}}^{{\mathtt{3}}}}{{{\mathtt{b}}}^{{\mathtt{2}}}}}$$

After that, you can continue to simplify the expression by cancelling out the"b" variables:

$${\frac{{{\mathtt{ab}}}^{{\mathtt{3}}}}{{{\mathtt{b}}}^{{\mathtt{2}}}}} = {\mathtt{ab}}$$

As a result, the answer to the simplified expression is ab.

 

Finally, on the thrid problem, you must get rid of the negative exponents. To do this, you would use the reciprocal of all the fractions.

$${\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{b}}}{{{\mathtt{a}}}^{{\mathtt{3}}}}}\right)}^{{\mathtt{2}}}$$

Then simplify the fractions by multiplying the expoinents into the fractions. So, it would be:

$$\left({\frac{{\mathtt{27}}}{{\mathtt{8}}}}{\mathtt{\,\times\,}}{\frac{{{\mathtt{b}}}^{{\mathtt{2}}}}{{{\mathtt{a}}}^{{\mathtt{6}}}}}\right)$$

After that, you would multiply it out and get your answer:

$${\frac{\left({\mathtt{27}}{\mathtt{\,\times\,}}{{\mathtt{b}}}^{{\mathtt{2}}}\right)}{\left({\mathtt{8}}{\mathtt{\,\times\,}}{{\mathtt{a}}}^{{\mathtt{6}}}\right)}}$$ -> -> -> 27b^2/8a^6

 

I hope this helped!

BrittanyJ Mar 25, 2015

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