+0

+1
198
4

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
+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
+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
+2

Thank you!

sinclairdragon428  Jun 12, 2019
#4
+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