Absolute value inequalities are often used to represent measurement error. For example, suppose that a machine is calibrated to dispense 8 fl oz of orange juice with a measurement error of on more than 0.05 fl oz. If x represents the actual amount of orange juice poured into the bottle, then x is a solution to the inequality x − 8 ≤ 0.05. For Exercises 79-82. a. Write an absolute value inequality to represent each statement. b. Solve the inequality. Write the solution set in interval notation. The variation between the measured value v and 16 oz is less than 0.01 oz.
Absolute value inequalities are often used to represent measurement error. For example, suppose that a machine is calibrated to dispense 8 fl oz of orange juice with a measurement error of on more than 0.05 fl oz. If x represents the actual amount of orange juice poured into the bottle, then x is a solution to the inequality x − 8 ≤ 0.05. For Exercises 79-82. a. Write an absolute value inequality to represent each statement. b. Solve the inequality. Write the solution set in interval notation. The variation between the measured value v and 16 oz is less than 0.01 oz.
Solution Summary: The author determines the absolute value inequality for represent the statement the variation between measured value vand 16 oz
Absolute value inequalities are often used to represent measurement error. For example, suppose that a machine is calibrated to dispense 8 fl oz of orange juice with a measurement error of on more than 0.05 fl oz. If
x
represents the actual amount of orange juice poured into the bottle, then
x
is a solution to the inequality
x
−
8
≤
0.05.
For Exercises 79-82.
a. Write an absolute value inequality to represent each statement.
b. Solve the inequality. Write the solution set in interval notation.
The variation between the measured value
v
and
16
oz is less than 0.01 oz.
17. Suppose we know that the graph below is the graph of a solution to dy/dt = f(t).
(a) How much of the slope field can
you sketch from this information?
[Hint: Note that the differential
equation depends only on t.]
(b) What can you say about the solu-
tion with y(0) = 2? (For example,
can you sketch the graph of this so-
lution?)
y(0) = 1
y
AN
(b) Find the (instantaneous) rate of change of y at x = 5.
In the previous part, we found the average rate of change for several intervals of decreasing size starting at x = 5. The instantaneous rate of
change of fat x = 5 is the limit of the average rate of change over the interval [x, x + h] as h approaches 0. This is given by the derivative in the
following limit.
lim
h→0
-
f(x + h) − f(x)
h
The first step to find this limit is to compute f(x + h). Recall that this means replacing the input variable x with the expression x + h in the rule
defining f.
f(x + h) = (x + h)² - 5(x+ h)
=
2xh+h2_
x² + 2xh + h² 5✔
-
5
)x - 5h
Step 4
-
The second step for finding the derivative of fat x is to find the difference f(x + h) − f(x).
-
f(x + h) f(x) =
= (x²
x² + 2xh + h² -
])-
=
2x
+ h² - 5h
])x-5h) - (x² - 5x)
=
]) (2x + h - 5)
Macbook Pro
Evaluate the integral using integration by parts.
Sx² cos
(9x) dx
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