<span> First, reduce the fraction to lowest terms, e.g. 8/6 = 4/3.
Look at the denominator. Split it into its prime factors. If its prime
factors only consist of 2's and 5's, then it will be terminating.
Examples:
16 = 2 x 2 x 2 x 2, so terminating
25 = 5 x 5, so terminating
2000 = 2 x 1000 = 2 x (2 x 5) x (2 x 5) x (2 x 5), so terminating
12 = 2 x 3, so repeating (has prime factor 3, which is not 2 or 5)
13 = 13, so repeating (has prime factor 13, which is not 2 or 5) Hope this helps!!
</span>
Answer:
Final answer is
.
Step-by-step explanation:
Given problem is
.
Now we need to simplify this problem.
![\sqrt[3]{x}\cdot\sqrt[3]{x^2}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%7D%5Ccdot%5Csqrt%5B3%5D%7Bx%5E2%7D)
![\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E1%7D%5Ccdot%5Csqrt%5B3%5D%7Bx%5E2%7D)
Apply formula
![\sqrt[n]{x^p}\cdot\sqrt[n]{x^q}=\sqrt[n]{x^{p+q}}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bx%5Ep%7D%5Ccdot%5Csqrt%5Bn%5D%7Bx%5Eq%7D%3D%5Csqrt%5Bn%5D%7Bx%5E%7Bp%2Bq%7D%7D)
so we get:
![\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=\sqrt[3]{x^{1+2}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E1%7D%5Ccdot%5Csqrt%5B3%5D%7Bx%5E2%7D%3D%5Csqrt%5B3%5D%7Bx%5E%7B1%2B2%7D%7D)
![\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=\sqrt[3]{x^{3}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E1%7D%5Ccdot%5Csqrt%5B3%5D%7Bx%5E2%7D%3D%5Csqrt%5B3%5D%7Bx%5E%7B3%7D%7D)
![\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=x](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E1%7D%5Ccdot%5Csqrt%5B3%5D%7Bx%5E2%7D%3Dx)
Hence final answer is
.
111/210 = about 53%. 52.85714% of 903 is 477.3 students. of course you can't have a fraction of a student so it is 477 students
Answer:
600,000(1+0.03)^20 that’s your equation. use a calculator. remove decimals from your answer because its not possible