Let's solve each question step-by-step.

Question 1

$Z$ varies jointly as $x$ and $y$. If $z = 3$ when $x = 3$ and $y = 15$, find $z$ when $x = 6$ and $y = 9$.

1. Since $Z$ varies jointly as $x$ and $y$, we can express this as: $z = k \cdot x \cdot y$ where $k$ is the constant of variation.

2. Substitute $z = 3$, $x = 3$, and $y = 15$ to find $k$: $3 = k \cdot 3 \cdot 15$ $3 = 45k$ $k = \frac{3}{45} = \frac{1}{15}$

3. Now that we have $k = \frac{1}{15}$, substitute $x = 6$ and $y = 9$ to find the new $z$: $z = \frac{1}{15} \cdot 6 \cdot 9$ $z = \frac{1}{15} \cdot 54$ $z = \frac{54}{15} = 3.6$

Answer for Question 1: $z = 3.6$

Question 2

The area $A$ of a triangle varies jointly as the base $b$ and the altitude $h$ of the triangle. If $A = 65 \, \text{cm}^2$ when $b = 10 \, \text{cm}$ and $h = 13 \, \text{cm}$, find the area of a triangle whose base is $8 \, \text{cm}$ and whose altitude is $11 \, \text{cm}$.

1. Since $A$ varies jointly as $b$ and $h$, we have: $A = k \cdot b \cdot h$ where $k$ is the constant of variation.

2. Substitute $A = 65$, $b = 10$, and $h = 13$ to find $k$: $65 = k \cdot 10 \cdot 13$ $65 = 130k$ $k = \frac{65}{130} = \frac{1}{2}$

3. Now that we have $k = \frac{1}{2}$, substitute $b = 8$ and $h = 11$ to find the new $A$: $A = \frac{1}{2} \cdot 8 \cdot 11$ $A = \frac{1}{2} \cdot 88$ $A = 44 \, \text{cm}^2$

Answer for Question 2: $A = 44 \, \text{cm}^2$

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Related Questions

1. How do you find the constant of variation in joint variation problems?
2. How does joint variation differ from direct and inverse variation?
3. What would happen if the values of $x$ or $y$ were doubled in the first question?
4. How can joint variation be applied to real-world situations?
5. How is the area formula for a triangle related to joint variation?

Tip: For joint variation problems, always start by isolating the constant $k$ using the given values, then use $k$ to find the unknown values in different scenarios.