Advanced Placement Calculus 2016 Graphical Numerical Algebraic Fifth Edition Student Edition
5th Edition
ISBN: 9780133311617
Author: Prentice Hall
Publisher: Prentice Hall
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I need help solving normal line.
A cardiac monitor is used to measure the heart rate of a patient after surgery. It complies the number of heartbeats after t minutes. When data in the table are graphed, the slope of the tangent line represents the heart rate in beats per minute.
t (min)
Heartbeats
(b) t-38 and t - 42
(c) t-40 and t - 42
36
(d) t-42 and t-44
38
2,503 2,639
40
2,773
The monitor estimates this value by calculating the slope of a secant line. Use the data to estimate the patient's heart rate after 42 minutes using the secant line between the points with the given values of t. (Round your answers to one decimal place.)
(a) t-36 and t-42
42
44
2,903 3,033
Part 2. Math column A to column B by finding the slope of tangent line at any given point and write the letter of
your answer on the space provided.
Column A
Column B
A. y = 2x - 7
1. y 4x2- 4x +1 at point (1,1)
B. y = x-9
2x-5
_2. y =
at point (1,5)
7-2x
C. y = *
3. у %3D х +3
at point (4,1)
D. y = 2x - 10
4. y = x² - 4
at point (-2,-4)
E. y = 2
5. у %3D Зх — 5
at point (-2,3)
F. y = 3x - 5
6. y = 4x2 + 2
at point (0,0)
G. y = -4x-8
7. y = 2x- 10
at point (4,2)
H. y = x+ 3
8. y =
x+4
at point (-2,-)
X-4
I. y = 5x -9
%3D
9.y x-9
at point (3,1)
J. y = 4x- 3
10. y = x2 - 6
at point (1,2)
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- Answer the ff.arrow_forwardOnly need parts f. and g. Needed info in first picture. Questions in second picture.arrow_forwardFind the equation of the tangent and normal line to the following curves at a given specified point. Write the equation in slope-intercept and general equation form. Graph the curve, the tangent line, the normal line and the point of intersection of the lines and the curve.arrow_forward
- Can you help me number 6. Finding normal.arrow_forwardOhm's law states that the voltage drop Vacross an ideal resistor is linearly proportional to the current i flowing through the resistor as V= iR. Where R is the resistance. However, real resistors may not always obey Ohm's law. Suppose that you perform some very precise experiments to measure the voltage drop and the corresponding current for a resistor. The following results suggest a curvilinear relationship rather than the straight line represented by Ohm's law. i -1 - 0.5 - 0.25 0.25 0.5 1 V -637 -96.5 -20.25 20.5 96.5 637 Instead of the typical linear regression method for analyzing such experimental data, fit a curve to the data to quantify the relationship. Compute V for i = 0.1 using Polynomial Interpolation.arrow_forwardOhm's law states that the voltage drop Vacross an ideal resistor is linearly proportional to the current i flowing through the resistor as V= iR. Where R is the resistance. However, real resistors may not always obey Ohm's law. Suppose that you perform some very precise experiments to measure the voltage drop and the corresponding current for a resistor. The following results suggest a curvilinear relationship rather than the straight line represented by Ohm's law. i -1 - 0.5 - 0.25 0.25 0.5 1 V -637 -96.5 -20.25 20.5 96.5 637 Instead of the typical linear regression method for analyzing such experimental data, fit a curve to the data to quantify the relationship. Compute V for i = 0.1 using Newton's Divided Difference Method.arrow_forward
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