The volume ( \( v \) ) of water in a tank is proportional to se where \( c \) is the rate at which water is kept into the tank and constant is \( k \). 1. Find the value of \( k \) if \( v=60 \) and \( c \neq s ~ 36 \) ii. Use the value of \( k \) in i. to find the value of \( c \) when \( v \) is 90

It appears there might have been some typographical errors in your question. Based on the context, I'll interpret the problem as follows: - **Volume (\( v \))** of water in a tank is **proportional** to the **rate (\( c \))** at which water is added to the tank. - The relationship can be expressed as: \[ v = k \cdot c \] where \( k \) is the **constant of proportionality**. Given this, let's address each part of your question. --- ## 1. Find the value of \( k \) when \( v = 60 \) and \( c = 36 \). Using the proportionality equation: \[ v = k \cdot c \] Plug in the known values: \[ 60 = k \cdot 36 \] Solve for \( k \): \[ k = \frac{60}{36} = \frac{5}{3} \approx 1.\overline{6} \] **So, the constant \( k = \frac{5}{3} \).** --- ## 2. Find the value of \( c \) when \( v = 90 \) using the value of \( k \) found in part 1. Again, using the proportionality equation: \[ v = k \cdot c \] We know \( v = 90 \) and \( k = \frac{5}{3} \). Plug these values in: \[ 90 = \frac{5}{3} \cdot c \] Solve for \( c \): \[ c = 90 \times \frac{3}{5} = 54 \] **So, the rate \( c = 54 \).** --- ## Summary 1. **Constant \( k \):** \( \frac{5}{3} \) 2. **Rate \( c \) when \( v = 90 \):** \( 54 \) If there are additional variables or if the relationship involves more factors (e.g., time), please provide further details so I can assist you more accurately.