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(f+90)/(20) >= 2

f+9020≥2 \frac{f+90}{20} \geq 2

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Full solution

Q. f+9020≥2 \frac{f+90}{20} \geq 2
  1. Multiply by 2020: To solve the inequality (f+90)/20≥2(f+90)/20 \geq 2, we need to isolate ff on one side of the inequality.\newlineWe start by multiplying both sides of the inequality by 2020 to eliminate the denominator.\newline(f+90)/20×20≥2×20(f+90)/20 \times 20 \geq 2 \times 20
  2. Simplify: After multiplying both sides by 2020, we get: f+90≥40f + 90 \geq 40
  3. Subtract 9090: Next, we subtract 9090 from both sides of the inequality to solve for ff.\newlinef+90−90≥40−90f + 90 - 90 \geq 40 - 90
  4. Isolate ff: After subtracting 9090 from both sides, we get:\newlinef≥−50f \geq -50
  5. Final Answer: We have isolated ff and found the range of values that satisfy the inequality.\newlineThe final answer is ff must be greater than or equal to −50-50.

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