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maksim [4K]
3 years ago
9

A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.1 ft/s,

how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 8 ft from the wall? (That is, find the angle's rate of change when the bottom of the ladder is 8 ft from the wall.)
Mathematics
1 answer:
pishuonlain [190]3 years ago
4 0
 <span>let: 
X = the distance of the bottom of the ladder from the wall at any time 
dX/dt = rate of travel of the bottom of the ladder = 1.1 ft/sec 
A = the angle of the ladder with the ground at anytime 
dA/dt = rate of change of the angle in radians per second 

X = 10 cos A 

dX/dt= -10 sin A dA/dt = 1.1 

dA/dt = -1.1/(10 sinA) 

When X = 6; cosA = 6/10; sinA = 8/10 

Therefore: 

dA/dt = -1.1/(10 x 0.8) = -0.1375 radiant per second. </span>
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Answer:

Speed of car B is 33 km/h and speed of car A is 45\,\,km/h

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Let speed of car B be x km/h

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Speed of car A = (12+x) km/h

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Distance travelled by car A in 4 hours = 4(12+x) km = (48+4x) km

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Also, cars A and B are 312km apart after 4 hours.

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5 0
3 years ago
You have a normal distribution of hours per week that music students practice. The mean of the values is 8 and the standard devi
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38%

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6<x<10 implies converting to z

We know z = (x-mean)/sigma = (x-8)/4

Hence 6<x<10 is equivalent to (6-8/4)<Z<(10-8/4)

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