+0  
 
+1
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avatar+397 

deleted.

 Jun 12, 2019
edited by sinclairdragon428  Jun 12, 2019
edited by sinclairdragon428  Jun 12, 2019
edited by sinclairdragon428  Jun 12, 2019
edited by sinclairdragon428  Nov 20, 2019
 #1
avatar+219 
+7

\(\frac{24t^3}{15t^4}*\frac{5t^8}{3t^6}\)

 

Simplify \(\frac{24t^3}{15t^4}\)

 

Both 24 and 15 have a common factor of 3

\(\frac{24t^3}{15t^4} = \frac{8t^3}{5t^4}\)

 

\(\frac{8t^3}{5t^4}*\frac{5t^8}{3t^6}\)

Apply exponent rule: \(\frac{x^a}{x^b} = \) 1/x^b-a

 

\(\frac{8}{5t}\)

 

Continue

\(\frac{8}{5t}*\frac{5t^8}{3t^6}\)

 

Apply exponent rule: \(\frac{x^a}{x^b} = \) x^a-b

 

\( \frac{5t^2}{3}\)

 

Continue

\(\frac{8}{5t} * \frac{5t^2}{3}\)

 

\(\frac{8*5t^2}{5t*3}\)

 

\(\frac{8t^2}{t*3}\)

 

\(\frac{8t}{3}\)

 

Your answer would be \(\frac{8t}{3}\)

Hope this helps ;P

 Jun 12, 2019
edited by EmeraldWonder  Jun 12, 2019
edited by EmeraldWonder  Jun 12, 2019
 #2
avatar+8884 
+4

EmeraldWonder, there are a couple of small errors in your work at the end because  \(\frac{8\,*\,5t^2}{5t\,*\,3}\ \neq\ \frac{8t^2}{3}\)

 

Here is another way to work this problem:

 

\(\ \phantom{=\quad}\dfrac{24t^3}{15t^4}\cdot \dfrac{5t^8}{3t^6}\)

 

\(=\quad\dfrac{24\,\cdot\,t^3\,\cdot\,5\,\cdot\,t^8}{15\,\cdot\,t^4\,\cdot\,3\,\cdot\,t^6}\)

 

\(=\quad\dfrac{120\,\cdot\,t^3\,\cdot\,t^8}{45\,\cdot\,t^4\,\cdot\,t^6}\)          because we can multiply numbers in any order and   24 · 5 = 120   and   15 · 3 = 45

 

\(=\quad\dfrac{120\,\cdot\,t^{11}}{45\,\cdot\,t^{10}}\)          because   t3 · t8  =  t t t · t t t t t t t t  =  t(3 + 8)  =  t11   and   t4 · t6  =  t(4 + 6)  =  t10

 

\(=\quad\dfrac{120\,\cdot\,t^{10}\cdot\,t}{45\,\cdot\,t^{10}}\)          because   t11  =  t t t t t t t t t t t  =  t10 · t

 

\(=\quad\dfrac{120\,\cdot\,t}{45}\)          when  t ≠ 0  because we can cancel the common factor of  t10  in the numerator and denominator.

 

\(=\quad\dfrac{8t}{3}\)          because   \(\frac{120}{45}\)   reduces to   \(\frac83\)_

 Jun 12, 2019
 #3
avatar+397 
+2


Thank you!

sinclairdragon428  Jun 12, 2019
 #4
avatar+219 
+5

I forgot to remove the square at the end. I redited my work to put the correct end result, thank you for correcting me, I see my error

 

;P

EmeraldWonder  Jun 12, 2019

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