Time and Date Stamps (logged): 12:39:21 05-21-2012 °±Ÿ²¸Ÿ±°¯´Ÿ±°Ÿ±¯°± Precalculus I

Precalculus I Major Quiz


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Test Problems:

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Problem Number 1

Problem:  Obtain a quadratic depth vs. clock time model if depths of 42.40351 cm, 32.5878 cm and 27.55286 cm are observed clock times t = 13.06683, 26.13366 and 39.20049 seconds.

 

Problem: The quadratic depth vs. clock time model corresponding to depths of 42.40351 cm, 32.5878 cm and 27.55286 cm at clock times t = 13.06683, 26.13366 and 39.20049 seconds is depth(t) = .014 t2 + -1.3 t + 57. Use the model to determine whether the depth will ever reach zero.

 

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Problem Number 2

Problem:  Evaluate f( 41.97753) and f( 2a + b ) for the function y = f(t) = .014 t^2 + -2.2 t + 96. What is the value of the function for clock time t = 62.96629?  What equation would you solve to determine the value of t for which f(t) = 34.4806? (You need not actually evaluate the equation).

Problem:  Sketch a graph representing the linear function family y = m x + b for m = .46, with b varying over all positive real numbers.

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Problem Number 3

Find the vertex and the zeros of y = .024 t2 + -1.7 t + 89.

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Problem Number 4

Make tables and sketch graphs of y = x, y = x 2 , y = x -1 and y = 2 x , for x = -3 to x = 3.

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Problem Number 5

At clock time t = 13 sec the depth of water in a container is 64 cm, while at clock time t = 35 sec the depth is 26 cm. Plot the corresponding points on a graph of depth vs. clock time and determine the slope of the straight line segment connecting these points. Explain why this slope represents the average rate at which the water depth changes over this time interval.

For the quadratic function y = f(t) = .02 t2 + -2.69 t + 96, determine the average rate of change of y with respect to t, between clock times t = 35 and t = 40.