Which value is equivalent to 7 multiplied by 3 multiplied by 2 whole over 7 multiplied by 5, the whole raised to the power of 2 multiplied by 7 to the power of 0 over 5 to the power of negative 3, whole to the power of 3 multiplied by 5 to the power of negative 9?

Respuesta :

Given problem is 7 multiplied by 3 multiplied by 2 whole over 7 multiplied by 5, the whole raised to the power of 2 multiplied by 7 to the power of 0 over 5 to the power of negative 3, whole to the power of 3 multiplied by 5 to the power of negative 9. which can be written as:

[tex]\left(\frac{7\cdot3\cdot2}{7\cdot5}\right)^2\cdot\left(\frac{7^0}{5^{-3}}\right)^3\cdot5^{-9}[/tex]

Now we have to find the equivalent value for that so we will simplify it

[tex]\left(\frac{7\cdot3\cdot2}{7\cdot5}\right)^2\cdot\left(\frac{7^0}{5^{-3}}\right)^3\cdot5^{-9}[/tex]


[tex]=\left(\frac{3\cdot2}{5}\right)^2\cdot\left(\frac{7^0}{5^{-3}}\right)^3\cdot5^{-9}[/tex]


[tex]=\left(\frac{3\cdot2}{5}\right)^2\cdot\left(\frac{7^0}{5^{-3}}\right)^3\cdot5^{-9}[/tex]


[tex]=\left(\frac{3\cdot2}{5}\right)^2\cdot\left(\frac{1}{5^{-3}}\right)^3\cdot5^{-9}[/tex]


[tex]=\left(\frac{3\cdot2}{5}\right)^2\cdot\left(5^3\right)^3\cdot\frac{1}{5^9}[/tex]


[tex]=\left(\frac{3\cdot2}{5}\right)^2\cdot5^9\cdot\frac{1}{5^9}[/tex]


[tex]=\left(\frac{3\cdot2}{5}\right)^2\cdot1[/tex]


[tex]=\left(\frac{3\cdot2}{5}\right)^2[/tex]


[tex]=\left(\frac{6}{5}\right)^2[/tex]

[tex]=\frac{36}{25}[/tex]

Hence final answer is [tex]\frac{36}{25}[/tex].


Answer:

[tex]=1\frac{11}{25}[/tex]

Step-by-step explanation:

To understand this word problem and form an equation, we will go step by step.

A value: [tex]x[/tex]

is equivalent to 7 multiplied by 3 multiplied by 2: [tex]x = 7*3*5[/tex]

whole over 7 multiplied by 5: [tex]x=\frac{7*3*2}{7*5}[/tex]

the whole raised to the power of 2: [tex]x=(\frac{7*3*2}{7*5})^2[/tex]

multiplied by 7 to the power of 0:  [tex]x=(\frac{7*3*2}{7*5})^2 * 7^0[/tex]

over 5 to the power of negative 3: [tex](\frac{7*3*2}{7*5} )^2*\frac{7^0}{5^{-3}}[/tex]

multiplied by 5 to the power of negative 9:  [tex](\frac{7*3*2}{7*5} )^2*\frac{7^0}{5^{-3}}*5^{-9}[/tex]

So we have  [tex](\frac{7*3*2}{7*5} )^2*\frac{7^0}{5^{-3}}*5^{-9}[/tex]

[tex]=(\frac{3*2}{5} )^2*(\frac{1}{5^{-3}})^3 *\frac{1}{5^{9}}[/tex]

[tex]=(\frac{3*2}{5} )^2*5^9*\frac{1}{5^{9}}[/tex]

[tex]=(\frac{3*2}{5} )^2[/tex]

[tex]=(\frac{6}{5} )^2[/tex]

[tex]= \frac{36}{25}[/tex]

[tex]=1\frac{11}{25}[/tex]