a tank holds 1,000 gallons of water, which drains from the bottom of the tank in half an hour. the values in…

a tank holds 1,000 gallons of water, which drains from the bottom of the tank in half an hour. the values in the table show the volume v of water remaining in the tank (in gallons) after t minutes.\n| t (min) | 5 | 10 | 15 | 20 | 25 | 30 |\n| v (gal) | 686 | 443 | 255 | 115 | 28 | 0 |\n(a) if p is the point (15, 255) on the graph of v, find the slopes of the secant lines pq when q is the point on the graph with t = 5, 10, 20, 25, and 30. (round your answers to one decimal place.)\n| q | slope |\n| (5, 686) | |\n| (10, 443) | |\n| (20, 115) | |\n| (25, 28) | |\n| (30, 0) | |\n(b) estimate the slope of the tangent line at p by averaging the slopes of the two adjacent secant lines corresponding to the two points closest to p. (round your answer to one decimal place.)

a tank holds 1,000 gallons of water, which drains from the bottom of the tank in half an hour. the values in the table show the volume v of water remaining in the tank (in gallons) after t minutes.\n| t (min) | 5 | 10 | 15 | 20 | 25 | 30 |\n| v (gal) | 686 | 443 | 255 | 115 | 28 | 0 |\n(a) if p is the point (15, 255) on the graph of v, find the slopes of the secant lines pq when q is the point on the graph with t = 5, 10, 20, 25, and 30. (round your answers to one decimal place.)\n| q | slope |\n| (5, 686) | |\n| (10, 443) | |\n| (20, 115) | |\n| (25, 28) | |\n| (30, 0) | |\n(b) estimate the slope of the tangent line at p by averaging the slopes of the two adjacent secant lines corresponding to the two points closest to p. (round your answer to one decimal place.)

Answer

Explanation:

Step1: Recall slope formula

The slope of a secant line between two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by $m=\frac{y_2 - y_1}{x_2 - x_1}$. Point $P=(15,255)$.

Step2: Calculate slope for $Q=(5,686)$

$m_1=\frac{686 - 255}{5 - 15}=\frac{431}{- 10}=-43.1$

Step3: Calculate slope for $Q=(10,443)$

$m_2=\frac{443 - 255}{10 - 15}=\frac{188}{-5}=-37.6$

Step4: Calculate slope for $Q=(20,115)$

$m_3=\frac{115 - 255}{20 - 15}=\frac{-140}{5}=-28.0$

Step5: Calculate slope for $Q=(25,28)$

$m_4=\frac{28 - 255}{25 - 15}=\frac{-227}{10}=-22.7$

Step6: Calculate slope for $Q=(30,0)$

$m_5=\frac{0 - 255}{30 - 15}=\frac{-255}{15}=-17.0$

Step7: Estimate slope of tangent line

The two points closest to $P=(15,255)$ are $(10,443)$ and $(20,115)$. The average of their slopes is $\frac{-37.6+( - 28.0)}{2}=\frac{-65.6}{2}=-32.8$

Answer:

For part (a):

  • $(5,686)$: $-43.1$
  • $(10,443)$: $-37.6$
  • $(20,115)$: $-28.0$
  • $(25,28)$: $-22.7$
  • $(30,0)$: $-17.0$ For part (b): $-32.8$