A climber is standing at the top of mount kazbek, approximately 3.1 miles above sea level. the radius of the earth is 3959 miles
. what is the climber's distance to the horizon? enter your answer as a decimal in the box. round only your final answer to the nearest tenth. mi
2 answers:
In the figure below, the radius, AB=AD=3959 mi, the climber is at position C. BC=3.1 mi. CD=x mi, is the distance from the climber to the horizon. Thus to solve for x we proceed as follows:
AC=3959+3.1=3962.1 mi
To evaluate for x we use the Pythagorean theorem, this is given by:
c^2=a^2+b^2
c=AC is the hypotenuse
a and b are the legs
plugging in the values we shall have:
3962.1^2=x^2+3959^2
x^2=3961.1^2-3959^2
x^2=24555.41
hence
x=156.7 mi
Answer: 156.7 mi
AC=3959+3.1=3962.1 mi
To evaluate for x we use the Pythagorean theorem, this is given by:
c^2=a^2+b^2
c=AC is the hypotenuse
a and b are the legs
plugging in the values we shall have:
3962.1^2=x^2+3959^2
x^2=3961.1^2-3959^2
x^2=24555.41
hence
x=156.7 mi
Answer: 156.7 mi
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Solve for n:
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5n + 3n + 6 = -3n + 3n - 10
8n + 6 = -10
Subtract 6 from both sides:
8n + 6 - 6 = -10 - 6
8n = -16
Divide both sides by 8:
8n/8 = -16/8
n = -2.