For a certain insulation material, heat flow rate , in watts; surface area , in square meters; temperature difference , in degrees Celsius; and thermal resistance are related by where , , , and are positive. When , which equation correctly expresses in terms of and ?

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Super-Hard SAT Math Question 53 | Free SAT Prep | Aniko

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Question 53·200 Super-Hard SAT Math Questions·Advanced Math

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For a certain insulation material, heat flow rate $H$ , in watts; surface area $A$ , in square meters; temperature difference $D$ , in degrees Celsius; and thermal resistance $R$ are related by

\frac{A^2D}{H}=\sqrt[3]{H^2R}

where $A$ , $D$ , $H$ , and $R$ are positive. When $R=48$ , which equation correctly expresses $H$ in terms of $A$ and $D$ ?

Convert radicals to rational exponents so you can combine powers cleanly. After substituting the given constant, clear any fractions involving the target variable, then add exponents when multiplying like bases (for example, $H\cdot H^{2/3}=H^{5/3}$ ). Finally, isolate the power of the variable and raise both sides to the reciprocal power to solve, rewriting the result as a single radical expression.

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Substitute first

Replace $R$ with $48$ in the given equation before trying to solve for $H$ .

Turn the cube root into exponents

Use $\sqrt[3]{x}=x^{1/3}$ so you can combine powers of $H$ more easily.

Combine the $H$ factors carefully

After you multiply both sides by $H$ , you should get a product like $H\cdot H^{2/3}$ . Add the exponents.

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Pick positive test values for $A$ and $D$

Choose simple values like $A=2$ and $D=3$ (any positive values work).

Enter the substituted equation as two expressions

After using $R=48$ , enter these two expressions (with your chosen values for $A$ and $D$ ):

$\text{LHS}=\frac{A^2D}{H}$
$\text{RHS}=(48H^2)^{1/3}$

Test each answer choice by defining $H$

For one option at a time, define $H$ using that expression (with your chosen $A$ and $D$ ). Then compare the numeric values of $\text{LHS}$ and $\text{RHS}$ (or graph $y=\text{LHS}$ and $y=\text{RHS}$ as horizontal lines).

Identify the matching expression

The correct option is the one that makes $\text{LHS}$ and $\text{RHS}$ equal (or makes the two lines coincide) for your test values.

Step-by-step Explanation

Substitute the given value of $R$

Plug in $R=48$ :

\frac{A^2D}{H}=\sqrt[3]{48H^2}

Rewrite the cube root using exponents

Rewrite the cube root as a power of $\tfrac{1}{3}$ :

\sqrt[3]{48H^2}=(48H^2)^{1/3}=48^{1/3}H^{2/3}

So the equation is

\frac{A^2D}{H}=48^{1/3}H^{2/3}

Clear the denominator and combine powers of $H$

Multiply both sides by $H$ :

A^2D=48^{1/3}H\cdot H^{2/3}=48^{1/3}H^{5/3}

Now isolate $H^{5/3}$ :

H^{5/3}=\frac{A^2D}{48^{1/3}}

Solve for $H$ by raising both sides to the reciprocal power

Raise both sides to the power $\tfrac{3}{5}$ :

H=\left(\frac{A^2D}{48^{1/3}}\right)^{3/5}

Simplify the exponents:

$(A^2D)^{3/5}=A^{6/5}D^{3/5}$
$(48^{1/3})^{3/5}=48^{1/5}$

H=\frac{A^{6/5}D^{3/5}}{48^{1/5}}=\left(\frac{A^6D^3}{48}\right)^{1/5}=\sqrt[5]{\frac{A^6D^3}{48}}

Therefore, the correct choice is $H=\sqrt[5]{\frac{A^6D^3}{48}}$ .

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Step-by-step Explanation

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