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fiasKO [112]
3 years ago
5

PLEASE HELP! Find the distance between each pair of points. Round your answers to the nearest tenth.

Mathematics
1 answer:
strojnjashka [21]3 years ago
6 0
A.
Distance = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\\\\  = \sqrt{\left( -4 - \left( -5 \right) \right)^2 + \left( -2 - \left( -3 \right) \right)^2} \\ \\  = \sqrt{ 1 + 1} \\ \\ = \sqrt{ 2 }

b. 
Distance = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\\\\ = \sqrt{\left( -4 - \frac{ 7 }{ 2 } \right)^2 + \left( \frac{ 5 }{ 2 } - 1 \right)^2} \\  \\= \sqrt{ \frac{ 225 }{ 4 } + \frac{ 9 }{ 4 }} \\ \\=  3 \frac{\sqrt{ 26 }}{ 2 }
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$ 720

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Normal time $ 15

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So I have another trigonometry question PLZ HELP ME I'm clueless at this rate I have like a little idea but I'm lost, SO PLZZZ!
Sergio039 [100]

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!!!

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2 years ago
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