Math Problem Statement

m = (z.f - g) / (z.f) * 100

formula to calculate z

Solution

To isolate zz in the equation m=z.fgz.f×100m = \frac{z.f - g}{z.f} \times 100, follow these steps:

  1. Start with the given equation: m=z.fgz.f×100m = \frac{z.f - g}{z.f} \times 100

  2. Divide both sides by 100 to simplify: m100=z.fgz.f\frac{m}{100} = \frac{z.f - g}{z.f}

  3. Rewrite the right side of the equation: m100=1gz.f\frac{m}{100} = 1 - \frac{g}{z.f}

  4. Rearrange to isolate the fraction involving zz: m1001=gz.f\frac{m}{100} - 1 = -\frac{g}{z.f}

  5. Simplify the left side: m100100=gz.f\frac{m - 100}{100} = -\frac{g}{z.f}

  6. Multiply both sides by -1 to clear the negative sign: 100m100=gz.f\frac{100 - m}{100} = \frac{g}{z.f}

  7. Multiply both sides by z.fz.f: z.f×100m100=gz.f \times \frac{100 - m}{100} = g

  8. Divide both sides by 100m100\frac{100 - m}{100} to solve for z.fz.f: z.f=g×100100mz.f = \frac{g \times 100}{100 - m}

  9. Finally, divide both sides by ff to solve for zz: z=g×100f(100m)z = \frac{g \times 100}{f (100 - m)}

Thus, the formula to calculate zz is: z=g×100f(100m)z = \frac{g \times 100}{f (100 - m)}

Would you like further details or have any questions?

Here are some related questions that might interest you:

  1. How can you derive a similar formula if the variable ff was not present?
  2. What are common mistakes to avoid when isolating variables in an equation?
  3. How can you verify the accuracy of your derived formula?
  4. How does the formula change if mm is in percentage format?
  5. What happens to the value of zz if mm approaches 100?
  6. Can you apply this formula to a real-world problem? If so, how?
  7. How does altering gg affect the value of zz?
  8. What mathematical concepts are important for understanding this derivation?

Tip: Always double-check your algebraic manipulations to avoid mistakes in isolating variables.

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Math Problem Analysis

Mathematical Concepts

Algebra
Equations
Percentages

Formulas

Percentage formula
Isolating variables in equations

Theorems

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Suitable Grade Level

Advanced High School