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This worksheet challenges students to isolate specific variables within complex algebraic and physics formulas.

Math worksheet for solving variables by rearranging algebraic equations.

Math worksheet for solving variables by rearranging algebraic equations.

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Show Answer Key & Explanations Step-by-step solution for: Manipulating Equations in term of a given Variable worksheets library
To solve the given equations for different unknowns, we will rearrange each equation step by step. Let's go through each problem:

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1. \( v = \frac{D}{T} \)


#### Solve for \( D \):
\[ v = \frac{D}{T} \]
Multiply both sides by \( T \):
\[ D = vT \]

#### Solve for \( T \):
\[ v = \frac{D}{T} \]
Multiply both sides by \( T \):
\[ vT = D \]
Divide both sides by \( v \):
\[ T = \frac{D}{v} \]

Answers:
- \( D = vT \)
- \( T = \frac{D}{v} \)

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2. \( A = BC + D \)


#### Solve for \( B \):
\[ A = BC + D \]
Subtract \( D \) from both sides:
\[ A - D = BC \]
Divide both sides by \( C \):
\[ B = \frac{A - D}{C} \]

#### Solve for \( C \):
\[ A = BC + D \]
Subtract \( D \) from both sides:
\[ A - D = BC \]
Divide both sides by \( B \):
\[ C = \frac{A - D}{B} \]

#### Solve for \( D \):
\[ A = BC + D \]
Subtract \( BC \) from both sides:
\[ D = A - BC \]

Answers:
- \( B = \frac{A - D}{C} \)
- \( C = \frac{A - D}{B} \)
- \( D = A - BC \)

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3. \( \frac{A}{B} = \frac{C}{D} \)


#### Solve for \( A \):
\[ \frac{A}{B} = \frac{C}{D} \]
Multiply both sides by \( B \):
\[ A = \frac{BC}{D} \]

#### Solve for \( B \):
\[ \frac{A}{B} = \frac{C}{D} \]
Multiply both sides by \( B \):
\[ A = \frac{BC}{D} \]
Multiply both sides by \( D \):
\[ AD = BC \]
Divide both sides by \( C \):
\[ B = \frac{AD}{C} \]

#### Solve for \( C \):
\[ \frac{A}{B} = \frac{C}{D} \]
Multiply both sides by \( D \):
\[ \frac{AD}{B} = C \]
\[ C = \frac{AD}{B} \]

#### Solve for \( D \):
\[ \frac{A}{B} = \frac{C}{D} \]
Multiply both sides by \( D \):
\[ \frac{AD}{B} = C \]
Multiply both sides by \( B \):
\[ AD = BC \]
Divide both sides by \( A \):
\[ D = \frac{BC}{A} \]

Answers:
- \( A = \frac{BC}{D} \)
- \( B = \frac{AD}{C} \)
- \( C = \frac{AD}{B} \)
- \( D = \frac{BC}{A} \)

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4. \( A = V^2 \cdot 2D \)


#### Solve for \( V \):
\[ A = V^2 \cdot 2D \]
Divide both sides by \( 2D \):
\[ \frac{A}{2D} = V^2 \]
Take the square root of both sides:
\[ V = \sqrt{\frac{A}{2D}} \]

#### Solve for \( D \):
\[ A = V^2 \cdot 2D \]
Divide both sides by \( 2V^2 \):
\[ D = \frac{A}{2V^2} \]

Answers:
- \( V = \sqrt{\frac{A}{2D}} \)
- \( D = \frac{A}{2V^2} \)

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5. \( E = M \cdot C^2 \)


#### Solve for \( C \):
\[ E = M \cdot C^2 \]
Divide both sides by \( M \):
\[ \frac{E}{M} = C^2 \]
Take the square root of both sides:
\[ C = \sqrt{\frac{E}{M}} \]

#### Solve for \( M \):
\[ E = M \cdot C^2 \]
Divide both sides by \( C^2 \):
\[ M = \frac{E}{C^2} \]

Answers:
- \( C = \sqrt{\frac{E}{M}} \)
- \( M = \frac{E}{C^2} \)

---

6. \( V = \sqrt{2AS} \)


#### Solve for \( A \):
\[ V = \sqrt{2AS} \]
Square both sides:
\[ V^2 = 2AS \]
Divide both sides by \( 2S \):
\[ A = \frac{V^2}{2S} \]

#### Solve for \( S \):
\[ V = \sqrt{2AS} \]
Square both sides:
\[ V^2 = 2AS \]
Divide both sides by \( 2A \):
\[ S = \frac{V^2}{2A} \]

Answers:
- \( A = \frac{V^2}{2S} \)
- \( S = \frac{V^2}{2A} \)

---

7. \( D = \frac{1}{2} A^2 T^2 \)


#### Solve for \( A \):
\[ D = \frac{1}{2} A^2 T^2 \]
Multiply both sides by 2:
\[ 2D = A^2 T^2 \]
Divide both sides by \( T^2 \):
\[ \frac{2D}{T^2} = A^2 \]
Take the square root of both sides:
\[ A = \sqrt{\frac{2D}{T^2}} \]
\[ A = \frac{\sqrt{2D}}{T} \]

#### Solve for \( T \):
\[ D = \frac{1}{2} A^2 T^2 \]
Multiply both sides by 2:
\[ 2D = A^2 T^2 \]
Divide both sides by \( A^2 \):
\[ \frac{2D}{A^2} = T^2 \]
Take the square root of both sides:
\[ T = \sqrt{\frac{2D}{A^2}} \]
\[ T = \frac{\sqrt{2D}}{A} \]

Answers:
- \( A = \frac{\sqrt{2D}}{T} \)
- \( T = \frac{\sqrt{2D}}{A} \)

---

Final Answers:


1. \( D = vT \), \( T = \frac{D}{v} \)
2. \( B = \frac{A - D}{C} \), \( C = \frac{A - D}{B} \), \( D = A - BC \)
3. \( A = \frac{BC}{D} \), \( B = \frac{AD}{C} \), \( C = \frac{AD}{B} \), \( D = \frac{BC}{A} \)
4. \( V = \sqrt{\frac{A}{2D}} \), \( D = \frac{A}{2V^2} \)
5. \( C = \sqrt{\frac{E}{M}} \), \( M = \frac{E}{C^2} \)
6. \( A = \frac{V^2}{2S} \), \( S = \frac{V^2}{2A} \)
7. \( A = \frac{\sqrt{2D}}{T} \), \( T = \frac{\sqrt{2D}}{A} \)

\[
\boxed{
\begin{aligned}
1. & \quad D = vT, \, T = \frac{D}{v} \\
2. & \quad B = \frac{A - D}{C}, \, C = \frac{A - D}{B}, \, D = A - BC \\
3. & \quad A = \frac{BC}{D}, \, B = \frac{AD}{C}, \, C = \frac{AD}{B}, \, D = \frac{BC}{A} \\
4. & \quad V = \sqrt{\frac{A}{2D}}, \, D = \frac{A}{2V^2} \\
5. & \quad C = \sqrt{\frac{E}{M}}, \, M = \frac{E}{C^2} \\
6. & \quad A = \frac{V^2}{2S}, \, S = \frac{V^2}{2A} \\
7. & \quad A = \frac{\sqrt{2D}}{T}, \, T = \frac{\sqrt{2D}}{A}
\end{aligned}
}
\]
Parent Tip: Review the logic above to help your child master the concept of manipulating formulas worksheet.
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