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jok3333 [9.3K]
3 years ago
8

Rick serves his community as a firefighter. he works for a full 24 hours and then is off for the next 48 hours before his next 2

4-hour shift. last month, he earned $4,665.60, before taxes, and worked twelve 24-hour shifts. how much money does rick make per hour?
A. $8.10
B. $194.40
C.16.20
D.$24.30
Mathematics
2 answers:
Mrac [35]3 years ago
7 0

Answer: D. $24.30

Step-by-step explanation:

I'm not completely sure.

Ray Of Light [21]3 years ago
7 0
The correct answer is C) $16.20


Here’s the math:
16.20 x 24= $388.80 (24 is the hours he worked in one day so he made $388.80 in one day)
388.80 x 12= $4665.60 (12 is the number of days he worked)
So Rick made $16.20 an hour
Please mark me the brainiest
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A golfball is hit from the ground with an initial velocity of 200 ft/sec. The horizontal distance that the golfball will travel,
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The golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of \pi/4.

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d(\theta)=\frac{v_i^2\sin(2\theta)}{g}\\d'(\theta)=\frac{v_i^2\cos(2\theta)}{g}*(2)\\d'(\theta)=\frac{2v_i^2\cos(2\theta)}{g}

d'(\theta)=0\\\frac{2v_i^2\cos(2\theta)}{g}=0\\\cos(2\theta)=0\\2\theta=\pi/2,3\pi/2,5\pi/2,...\\\theta=\pi/4,3\pi/4,5\pi/4,...

The critical value inside the interval is \pi/4.

d(\theta)=\frac{v_i^2sin(2\theta)}{g}\\d(\pi/4)=\frac{v_i^2sin(2(\pi/4))}{g}\\d(\pi/4)=\frac{v_i^2sin(\pi/2)}{g}\\d(\pi/4)=\frac{v_i^2(1)}{g}\\d(\pi/4)=\frac{(200)^2}{32}\\d(\pi/4)=\frac{40000}{32}\\d(\pi/4)=1250ft

The second step is to find the values of the function at the endpoints of the interval:

d(\theta)=\frac{v_i^2sin(2\theta)}{g}\\\theta=0\\d(0)=\frac{v_i^2sin(2(0))}{g}\\d(0)=\frac{v_i^2(0)}{g}=0ft\\\theta=\pi/2\\d(\pi/2)=\frac{v_i^2sin(2(\pi/2))}{g}\\d(\pi/2)=\frac{v_i^2sin(\pi)}{g}\\d(\pi/2)=\frac{v_i^2(0)}{g}=0ft

The biggest value of f is gived by \pi/4, therefore \pi/4 is the absolute maximum.

In the context of the problem, the golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of \pi/4.

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