Answer:
It is terminating decimal expansion.
Step-by-step explanation:
17/3125 is a terminating decimal expansion.
REASON:
If the factors of the denominator are in the form of 2^n 5^m then the rational number is a terminating decimal expansion otherwise it is recurring. Here n and m are non negative integers.
Proof:
Lets have a look on the solution of the given term.
17/1325
We will break the denominator in the factors:
If we multiply 5 five times than it will give us 1325.
17/1325 = 17/5^5 * 2^0 = 17/2^0 * 5^5
We know that any number with exponent zero = 1
∴ 2^0 = 1
So it satisfies our explanation that the factors of the denominator are in the form of 2^n * 5^m and n and m are non negative integers.
Thus this term has a terminating decimal expansion....
The expression is equivalent to ![\sqrt[6]{2}](https://tex.z-dn.net/?f=%5Csqrt%5B6%5D%7B2%7D)
.
<h2>
Given that</h2>
Expression; ![\dfrac{\sqrt{2}}{\sqrt[3]{2} }](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B%5Csqrt%5B3%5D%7B2%7D%20%7D)
<h3>We have to determine</h3>
The equivalent expression to the given expression.
<h3>According to the question</h3>
To determine the equivalent relation following all the steps given below.
Expression;
The equivalent expression is;
![= \dfrac{\sqrt{2}}{\sqrt[3]{2} }\\\\= \dfrac{2^{\frac{1}{2}}}{2^{\frac{1}{3}}}\\\\ = 2^{\frac{1}{2}-\frac{1}{3}}\\\\ = 2 ^{\frac{3-2}{6}}\\\\= 2^{\frac{1}{6}}\\\\= \sqrt[6]{2}](https://tex.z-dn.net/?f=%3D%20%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B%5Csqrt%5B3%5D%7B2%7D%20%7D%5C%5C%5C%5C%3D%20%5Cdfrac%7B2%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%7D%7B2%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%7D%5C%5C%5C%5C%20%3D%202%5E%7B%5Cfrac%7B1%7D%7B2%7D-%5Cfrac%7B1%7D%7B3%7D%7D%5C%5C%5C%5C%20%3D%202%20%5E%7B%5Cfrac%7B3-2%7D%7B6%7D%7D%5C%5C%5C%5C%3D%202%5E%7B%5Cfrac%7B1%7D%7B6%7D%7D%5C%5C%5C%5C%3D%20%5Csqrt%5B6%5D%7B2%7D)
Hence, the expression is equivalent to .
To know more about Expression click the link given below.
brainly.com/question/16450385
