625^1/4-49^1/2+3^-3=

To solve the expression ${625}^{1/4}-{49}^{1/2}+3^{-3}$, we will evaluate each term individually before performing the final addition and subtraction. 1. Evaluate the individual terms

• First Term: ${625}^{1/4}$
  The notation $x^{1/4}$ represents the fourth root of $x$. We look for a number that, when multiplied by itself four times, equals 625.
  $5\times 5\times 5\times 5=625⟹{625}^{1/4}=5$ Second Term: ${49}^{1/2}$
  The notation $x^{1/2}$ represents the square root of $x$.
  $\sqrt{49}=7⟹{49}^{1/2}=7$ Third Term: $3^{-3}$
  A negative exponent indicates the reciprocal of the base raised to the positive power: $a^{-n}=\frac{1}{a^n}$ .
  $3^{-3}=\frac{1}{3^3}=\frac{1}{27}$

2. Substitute the values back into the expression Now, combine the simplified terms: $5-7+\frac{1}{27}$ Perform the subtraction: $-2+\frac{1}{27}$ 3. Final Calculation To combine the integer and the fraction, convert the integer to a fraction with a common denominator of 27: $-2=-\frac{54}{27}$ Now, add the fractions: $-\frac{54}{27}+\frac{1}{27}=-\frac{53}{27}$ As a mixed number, this is: $-1\frac{26}{27}$ In decimal form (rounded to four decimal places): $\approx -1.9630$Final Answer: $-\frac{53}{27}\text{or}-1\frac{26}{27}$ Would you like me to solve another algebraic expression or convert this result into a different format?