Example Question - problem solution
Here are examples of questions we've helped users solve.
Direct Variation Problem Solution
The image features a mathematical problem that reads: "Given \( x \) and \( y \) = 65 when \( x = 5 \), find \( y \) when \( x = 2 \)." This problem describes a direct variation between \( x \) and \( y \), where their product is constant. Initially, when \( x = 5 \), \( y \) is such that \( x \cdot y = 65 \). First, we can find the initial value of \( y \) when \( x = 5 \): \[ 5 \cdot y = 65 \] \[ y = 65 / 5 \] \[ y = 13 \] Now that we know the relationship between \( x \) and \( y \) is such that their product is always 65, we can use this information to find \( y \) when \( x = 2 \): \[ x \cdot y = 65 \] \[ 2 \cdot y = 65 \] \[ y = 65 / 2 \] \[ y = 32.5 \] Therefore, when \( x = 2 \), \( y \) is 32.5.
Direct Variation Problem Solution
The image shows a direct variation problem where \( y \) is directly proportional to \( x \). It states that \( y = 10 \) when \( x = 14 \) and asks for the value of \( y \) when \( x = 21 \). In such problems, the ratio \( y/x \) remains constant, so we have: \[ \frac{y_1}{x_1} = \frac{y_2}{x_2} \] Given \( y_1 = 10 \) and \( x_1 = 14 \), we can express \( y_2 \) in terms of \( x_2 = 21 \) using the direct variation formula: \[ \frac{10}{14} = \frac{y_2}{21} \] To find \( y_2 \), we solve: \[ y_2 = \frac{10}{14} \times 21 \] \[ y_2 = \frac{10 \times 21}{14} \] Now simplify the fraction: \[ y_2 = \frac{10 \times 3}{2} \] \[ y_2 = \frac{30}{2} \] \[ y_2 = 15 \] Hence when \( x = 21 \), \( y = 15 \).
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