Multiple choice

If $x = y^{3} + 4y$ and $y = \dfrac {7}{k}$, find the approximate value of $x$ when $k = 21$.

  1. $1.37$
  2. $1.44$
  3. $1.67$
  4. $4.04$
  5. $4.11$
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A Correct answer
Explanation

Given y = 7/k and k = 21, y = 7/21 = 1/3. Substituting y into x = y^3 + 4y gives x = (1/3)^3 + 4(1/3) = 1/27 + 4/3 = 1/27 + 36/27 = 37/27, which is approximately 1.37.

AI explanation

Substitute k = 21 into the equation for y to get y = 7 divided by 21, which equals 1 divided by 3. Substitute y = 1 divided by 3 into the first equation to get x = (1 divided by 3) cubed plus 4 times (1 divided by 3), which simplifies to 1 divided by 27 plus 4 divided by 3. Find a common denominator to add the fractions, yielding 1 divided by 27 plus 36 divided by 27, which equals 37 divided by 27. The result is approximately 1.37.

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