Question

Question content area top
Part 1
Suppose that S varies directly as the 4​/5 power of​ T, and that Sequals48 when Tequals32. Find S when Tequals243.
Question content area bottom
Part 1
Sequals
  
enter your response here

Answer 1

To solve the problem, we start with the direct variation relationship, which can be expressed mathematically as:

\[
S = k T^{4/5}
\]

where \( k \) is the constant of proportionality.

From the problem, we know that \( S = 48 \) when \( T = 32 \). We can substitute these values into the equation to find \( k \):

\[
48 = k (32^{4/5})
\]

Now, we calculate \( 32^{4/5} \):

1. Calculate \( 32^{1/5} \):
\[
32^{1/5} = 2 \quad (\text{since } 2^5 = 32)
\]

2. Now calculate \( 32^{4/5} \):
\[
32^{4/5} = (32^{1/5})^4 = 2^4 = 16
\]

Now substitute this back into the equation to solve for \( k \):

\[
48 = k (16)
\]

To find \( k \):
\[
k = \frac{48}{16} = 3
\]

Now we have the direct variation equation:

\[
S = 3 T^{4/5}
\]

Next, we want to find \( S \) when \( T = 243 \):

\[
S = 3 (243^{4/5})
\]

To calculate \( 243^{4/5} \):

1. Calculate \( 243^{1/5} \):
\[
243^{1/5} = 3 \quad (\text{since } 3^5 = 243)
\]

2. Now calculate \( 243^{4/5} \):
\[
243^{4/5} = (243^{1/5})^4 = 3^4 = 81
\]

Now substitute this back into the equation for \( S \):

\[
S = 3 \times 81 = 243
\]

Thus, when \( T = 243 \):

\[
S = 243
\]

So the final answer is:

\[
\boxed{243}
\]