About how much farther is it to drive than to walk directly from building A to building B? Round to the nearest whole number.
1 answer:
Answer:
A. 183 meters
Step-by-step explanation:
Building A and building B are 500 meters apart. There is no road between them, so to drive from building A to building B, it is necessary to first drive to building C and then to building B. About how much farther is it to drive than to walk directly from building A to building B? Round to the nearest whole number. A) 183 meters B) 250 meters C) 366 meters D) 683 meters
Find distance BC
Cos (60°)=BC / AB (Adjacent divided by the hypotenuse)
Cos (60°)=1/2
BC=a
AB=500
Cos (60°)=BC / AB
1/2=a/500
1/2 * 500=a
250=a
a=250m
Find distance AC
Sin(60°)=AC/AB (opposite side divided by hypotenuse)
Sin(60°)=√3/2
AC=b
AB=500
Sin(60°)=AC/AB
√3/2=b/500
√3/2 * 500=b
250√3=b
b=433m
Distance AC and BC=AC+BC
433m+250m=683m
Subtract the distance AB from AC+BC
= 683m - 500m
=183m
Answer is A. 183 meters
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Answer:
The maximum velocity of the object in the time interval [0, 4] is 10 ft/s.
Step-by-step explanation:
Given : The position of an object along a vertical line is given by , where s is measured in feet and t is measured in seconds.
To find : What is the maximum velocity of the object in the time interval [0, 4]?
Solution :
The velocity is rate of change of distance w.r.t time.
Distance in terms of t is given by,
Derivate w.r.t. time,
It is a quadratic function so its maximum is at vertex of the function.
The x point of the function is given by,
Where, a=-3, b=6 and c=7
As 1 lie between interval [0,4]
Substitute t=1 in the function,
Th maximum velocity is 10 ft/s.
Therefore, the maximum velocity of the object in the time interval [0, 4] is 10 ft/s.