Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Evaluate.

(root(3)(-5))/(40^((1)/(3)))=

Evaluate.\newline534013= \frac{\sqrt[3]{-5}}{40^{\frac{1}{3}}}=

Full solution

Q. Evaluate.\newline534013= \frac{\sqrt[3]{-5}}{40^{\frac{1}{3}}}=
  1. Understand the expression: Understand the expression.\newlineWe need to evaluate the expression (53)/(40(1)/(3))(\sqrt[3]{-5})/(40^{(1)/(3)}).\newlineThis means we need to find the cube root of 5-5 and divide it by the cube root of 4040.
  2. Evaluate the cube root of 5-5: Evaluate the cube root of 5-5.\newlineThe cube root of 5-5 is the number that, when multiplied by itself three times, gives 5-5.\newlineThe cube root of 5-5 is 53-\sqrt[3]{5} (since 53×53×53=5-\sqrt[3]{5} \times -\sqrt[3]{5} \times -\sqrt[3]{5} = -5).
  3. Evaluate the cube root of 4040: Evaluate the cube root of 4040.\newlineThe cube root of 4040 is the number that, when multiplied by itself three times, gives 4040.\newlineThe cube root of 4040 is 403.\sqrt[3]{40}.
  4. Divide the cube root of 5-5 by the cube root of 4040: Divide the cube root of 5-5 by the cube root of 4040. Now we divide 53-\sqrt[3]{5} by 403\sqrt[3]{40}. The expression becomes (53)/(403)(-\sqrt[3]{5})/(\sqrt[3]{40}).
  5. Simplify the expression: Simplify the expression.\newlineSince both the numerator and the denominator are under a cube root, we can combine them under a single cube root.\newlineThe expression becomes 5403\sqrt[3]{\frac{-5}{40}}.
  6. Simplify the fraction inside the cube root: Simplify the fraction inside the cube root.\newlineThe fraction (5)/40(-5)/40 can be simplified to 1/8-1/8.\newlineSo the expression now is 1/83\sqrt[3]{-1/8}.
  7. Evaluate the cube root of 18-\frac{1}{8}: Evaluate the cube root of 18-\frac{1}{8}. The cube root of 18-\frac{1}{8} is the number that, when multiplied by itself three times, gives 18-\frac{1}{8}. The cube root of 18-\frac{1}{8} is 12-\frac{1}{2} (since (12)×(12)×(12)=18(-\frac{1}{2}) \times (-\frac{1}{2}) \times (-\frac{1}{2}) = -\frac{1}{8}).

More problems from Powers with negative bases

QuestionGet tutor helpright-arrow

Posted 7 months ago

QuestionGet tutor helpright-arrow

Posted 9 months ago