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dybincka [34]
4 years ago
15

It takes Kay 20 minutes to drive to work traveiling 45 mph. Two minutes after she left home this morning, her husband, Dan, star

ted out with her briefcase, which she had forgotten. If Dan arrived at Kay's office just as she did, how fast did he drive?
Mathematics
1 answer:
BARSIC [14]4 years ago
8 0

Answer:

Kay's husband drove at a speed of 50 mph

Step-by-step explanation:

This is a problem of simple motion.

First of all we must calculate how far Kay traveled to her job, and then estimate the speed with which her husband traveled later.

d=vt

v=45 mph

t= 20 minutes/60 min/hour = 0.333 h (to be consistent with the units)

d= 45mph*0.333h= 15 miles

If Kay took 20 minutes to get to work and her husband left home two minutes after her and they both arrived at the same time, it means he took 18 minutes to travel the same distance.

To calculate the speed with which Kate's husband made the tour, we will use the same initial formula and isolate the value of "V"

d=vt; so

v=\frac{d}{t}

d= 15 miles

t= 18 minutes/60 min/hour = 0.30 h  (to be consistent with the units)

v=\frac{d}{t}=\frac{15 miles}{0.3 h}=50mph

Kay's husband drove at a speed of 50 mph

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The point-slope form:

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The optimal height h of the letters of a message printed on pavement is given by the formula <img src="https://tex.z-dn.net/?f=h
choli [55]

Answer:

The value of h is 42.956 approximately.

Step-by-step explanation:

Consider the provided formula h=\dfrac{0.00252 d^{2.27}}{e}.

Here d is the distance of the driver from the letters and e is the height of the​ driver's eye above the pavement. All of the distances are in meters.

We need to find the value of h where the value of d = 92.4 m, e = 1.7 m.

Substitute d = 92.4 m, e = 1.7 m in above formula and solve for h.

h=\dfrac{0.00252\left(92.4\right)^{2.27}}{1.7}

h\approx\dfrac{0.00252\left(28978.4648\right)}{1.7}

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h\approx42.956

Hence, the value of h is 42.956 approximately.

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3 years ago
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