Erica Sutherland earns a yearly salary go 45,650, how much would Erica earn in a monthly paycheck
2 answers:
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Using the given information, the width of the river is 505.96 m
<h3>Bearing and Distances</h3>From the question, we are to determine the width of the river
Consider the given diagram,
Let F be the first position of the biker
S be the second position
and C be the position of the church
First, we will determine the distance covered from F to S
Speed = 160m/min
Time = 4 minutes
Using the formula,
Distance = Speed × Time
Distance covered = 160 × 4
Distance covered = 640 m
That is, /FS/ = 640 m
Now, consider ΔFSC
∠F = 90° - 70° = 20°
∠S = 90° + 56° = 146°
and ∠C = 180° - 20° - 146° = 14°
By the Law of Sines
∴
/FC/ = 1479.33 m
Now,
By SOH CAH TOA
NOTE: d is the width of the river
∴ d = /FC/ × cos70°
d = 1479.33 × cos70°
d = 505.96 m
Hence, the width of the river is 505.96 m
Learn more on Bearing and distances here: brainly.com/question/19351991
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Answer:
a)
x = -1/√3 = -0.58 is a local maximum
x = 1/√3 = 0.58 is a local minimum
b)
(-1/√3 , 1/√3 ) DECREASING
(-∞, -1/√3) U (1/√3, ∞) INCREASING
Step-by-step explanation:
To answer all of the questions we must obtain the derivative of the fucntion:
If
U(x) = 4(x^3 - x)
then
U'(x) = 4(3x^2 - 1)
U''(x) = 4(3*2 x) = 24 x
U'''(x) = 24
The local maxima and minima of the function U(x) can be found when U'(x) = 0
this occurs when :
3x^2 = 1
that is:
x = ±1/√3
We will know if they are a minimum or a maximum evaluating this points on the second derivative (you can look for it as <u><em>Second derivative test</em></u>), if the result is positive the point corresponds to a minimum and if it is negative it will be a maximum.
Here it is easy to determine wheather is a maximum or a minimum because the second derivative is 24x, therefore if x is negative or positive so the second derivative will be.
a)
x = -1/√3 = -0.58 is a local maximum
x = 1/√3 = 0.58 is a local minimum
b)
the intervals at which the function is increasing { decreasing } is given when the first derivative is positive { negative }
The first derivative will be positive when:
3x^2 > 1
|x| > 1/√3 --> x > 1/√3 and x < -1/√3
The first derivatice will be negative when:
3x^2 < 1
|x| < 1/√3 --> x < 1/√3 and x > -1/√3
Therefore the intervals are:
(-1/√3 , 1/√3 ) DECREASING
(-∞, -1/√3) U (1/√3, ∞) INCREASING
<u><em>** The attached image is a plot of the fucntion where you can see part of the intervals and the local maximum and minimum</em></u>
