Pls help me with this problem mam confused
2 answers:
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Answer:
1) The windsurfer is approximately 580 ft from each lifeguard stand.
2) The distance between the Earth and Mercury is approximately 61 million miles.
Step-by-step explanation:
The image other two situations described is presented in the attached image to this answer.
1) From the attached image, the windsurfer forms a right angled triangle with each of the lifeguard stand and the midpoint between the two lifeguard stands.
Hence, the angle at the top of the triangle is half of 30°, 15° and the distance from the midpoint of the lifeguard stands to the lifeguard stands is 300/2 = 150 ft.
Let the required distance of the windsurfer from each of the lifeguard stands be x.
Using trigonometric relations,
Sin 15° = (150/x)
x = 150 ÷ sin 15° = 150 ÷ 0.2588
x = 579.6 ft = 580 ft to the nearest whole number
2) From the image, the Sun, Earth and Mercury form a triangle.
Let the possible distance between the Earth and Mercury be y.
Using cosine rule,
y² = 36² + 93² - (2×36×93×cos 22°)
y² = 3,736.5769098208
y = √3,736.5769098208 = 61.13 million miles = 61 million miles to the nearest million miles.
Hope this Helps!!!
f(x) + n - shift a graph of f n units up
f(x) - n - shift a graph of f n units down
f(x + n) - shift a graph of f n units left
f(x - n) - shift a graph of f n units right.
f(x) = x³, g(x) = (x - 2)³ - 3 = f(x - 2) - 3
2 units right and 3 units down.
