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An inverse variation includes the points (25,2)(25,\,2) and (5,n)(5,\,n). Find nn. \newlineWrite and solve an inverse variation equation to find the answer.\newlinen=____n = \,\_\_\_\_

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Q. An inverse variation includes the points (25,2)(25,\,2) and (5,n)(5,\,n). Find nn. \newlineWrite and solve an inverse variation equation to find the answer.\newlinen=____n = \,\_\_\_\_
  1. Find kk: Inverse variation means y=kxy = \frac{k}{x}. We got two points (25,2)(25, 2) and (5,n)(5, n), so let's find kk first using the first point.
  2. Calculate kk: Using the point (25,2)(25, 2), we get 2=k252 = \frac{k}{25}. Multiply both sides by 2525 to find kk.
  3. Find nn: So, 2×25=k2 \times 25 = k. That means k=50k = 50.
  4. Substitute kk: Now we got kk, let's use the second point (5,n)(5, n) to find nn. The equation is n=k5n = \frac{k}{5}.
  5. Calculate nn: Substitute k=50k = 50 into the equation, so n=505n = \frac{50}{5}.
  6. Calculate nn: Substitute k=50k = 50 into the equation, so n=505n = \frac{50}{5}.Calculate nn. n=505n = \frac{50}{5} gives us n=10n = 10.

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