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Aleks [24]
3 years ago
7

Find the​ y-coordinates of the points that are 10 units away from the point (-7, 6) that have an​ x-coordinate of 1.

Mathematics
1 answer:
prohojiy [21]3 years ago
7 0

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-7}~,~\stackrel{y_1}{6})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{y})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ 10=\sqrt{[1-(-7)]^2+[y-6]^2}\implies 10^2=[1-(-7)]^2+[y-6]^2 \\\\\\ 100=(1+7)^2+(y-6)(y-6)\implies 100=8^2+\stackrel{F~O~I~L}{(y^2-12y+36)} \\\\\\ 100=64+36+y^2-12y\implies 100=100+y^2-12y \\\\\\ 0=y^2-12y\implies 0=y(y-12)\implies y= \begin{cases} 0\\ 12 \end{cases}

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8 0
3 years ago
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mina [271]

Answer:

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2 years ago
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8 0
4 years ago
A plane takes off from an airport on a bearing of 270° and travels at a speed of 320 mph. after sometime and flight, space the p
ratelena [41]

The speed of the plane after it encounters the wind is C.285mph

<h3>How to calculate the speed of the plane when it encounters the wind?</h3>

Since the plane takes off from an airport on a bearing of 270° and travels at a speed of 320 mph it's velocity is v = (320cos270°)i + (320sin270°)j

= (320 × 0)i + (320 × -1)j

= 0i - 320j

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Also, the plane encounters a 35 mph wind blowing directly north. The velocity of the wind is v' = 35j mph

So, the velocity of the plane after it encounters the wind is the resultant velocity, V = v + v'

= -320j mph + 35j mph

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So, the speed of the plane after it encounters the wind is the magnitude of V = |-285j| mph

= 285 mph

So, the speed of the plane after it encounters the wind is C.285mph

Learn more about speed of plane here:

brainly.com/question/3387746

#SPJ1

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Answer:

3 hr(s)

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3 years ago
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