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6. water in tank

Question

  1. water in tank

Explanation:

Step1: Find the slope

The slope \( m \) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by \( m=\frac{y_2 - y_1}{x_2 - x_1} \). Here, \((x_1, y_1)=(2, 76)\) and \((x_2, y_2)=(5, 40)\). So, \( m=\frac{40 - 76}{5 - 2}=\frac{- 36}{3}=- 12 \).

Step2: Use point - slope form

The point - slope form of a line is \( y - y_1=m(x - x_1) \). Using the point \((2, 76)\) and \( m=-12 \), we get \( y - 76=-12(x - 2) \).
Expand the right - hand side: \( y - 76=-12x + 24 \).
Add 76 to both sides: \( y=-12x+24 + 76=-12x + 100 \).
(If we want to find when the tank is empty, set \( y = 0 \): \( 0=-12x + 100 \), then \( 12x = 100 \), \( x=\frac{100}{12}=\frac{25}{3}\approx8.33 \) seconds. But since the problem is not fully stated, assuming we need the equation of the line or the rate of change)

Answer:

The slope (rate of water decrease) is \(-12\) ft/s, and the equation of the line is \(y = - 12x+100\) (where \(y\) is water in ft and \(x\) is time in s). If finding when empty, \(x=\frac{25}{3}\approx8.33\) s.