The problem in the image states:

"P varies jointly as $T$ and the square of $Q$, and $P = 39$ when $T = 18$ and $Q = 6$. Find $P$ when $T = 4$ and $Q = 5$."

Step 1: Write the equation for joint variation

Since $P$ varies jointly as $T$ and $Q^2$, the relationship can be expressed as: $P = k \cdot T \cdot Q^2$ where $k$ is the constant of proportionality.

Step 2: Find the constant $k$

We are given that $P = 39$, $T = 18$, and $Q = 6$. Plug these values into the equation to solve for $k$: $39 = k \cdot 18 \cdot 6^2$ $39 = k \cdot 18 \cdot 36$ $39 = 648k$ $k = \frac{39}{648} = \frac{1}{16.615} = \frac{1}{16.615} \approx 0.0602$

Step 3: Find $P$ when $T = 4$ and $Q = 5$

Now that we have the value of $k$, substitute $k$, $T = 4$, and $Q = 5$ into the equation to find $P$: $P = 0.0602 \times 4 \times 5^2$ $P = 0.0602 \times 4 \times 25$ $P = 0.0602 \times 100$ $P = 6.02$

Final Answer

So, $P$ is approximately 6.02 when $T = 4$ and $Q = 5$.

Would you like further details or have any questions?

5 Related Questions:

1. How do you determine the constant of proportionality in joint variation problems?
2. What is the difference between direct, inverse, and joint variation?
3. How would the problem change if $P$ varied inversely with $Q^2$?
4. Can joint variation include more than two variables? If so, how?
5. How would you solve for $P$ if only $T$ or $Q$ were given, but not both?

Tip:

Always double-check your calculations when dealing with constants of proportionality, as a small mistake can lead to an incorrect final answer.