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xxMikexx [17]
3 years ago
10

Can anyone help me with this problem??

Mathematics
1 answer:
Arlecino [84]3 years ago
4 0

9514 1404 393

Answer:

  (a)  4/3

  (b)  y -3 = 4/3(x -1)

  (c)  y -3 = -3/4(x -1)

  (d)  r = 5

Step-by-step explanation:

a) The slope is given by the slope formula:

  m = (y2 -y1)/(x2 -x1)

  m = (7 -3)/(4 -1) = 4/3

__

b) The radius is normal to the circle. The point-slope form of the equation for a line can be useful here:

  y -k = m(x -h) . . . . . line with slope m through point (h, k)

For slope 4/3, the line through point (1, 3) will have the equation ...

  y -3 = 4/3(x -1) . . . . point-slope equation of the normal

__

c) The tangent is perpendicular to the radius. It will have a slope that is the opposite reciprocal of the slope of the radius: -1/(4/3) = -3/4.

  y -3 = -3/4(x -1) . . . . point-slope equation of the tangent

__

d) The radius can be found from the distance formula.

  d = √((x2 -x1)² +(y2 -y1)²)

  d = √((4 -1)² +(7 -3)²) = √(3² +4²) = √25 = 5

The radius of the circle is 5.

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Sitting on a park bench, you see a swing that is 100 feet away and a slide that is 80 feet away. The angle between them 30°. Wha
Vedmedyk [2.9K]

Answer:

B:  50 feet

Step-by-step explanation:

Let A represents the location of the swing, B represents the location of bench and C represents the position of slide,

According to the question,

AB = 100 ft,

BC = 80 ft,

m∠ABC = 30°,

By the cosine law,

AC^2=AB^2+BC^2-2\times AB\times BC\times cos 30^{\circ}

=100^2+80^2-2\times 100\times 80\times \frac{\sqrt{3}}{2}

=10000+6400-8000\sqrt{3}

=2543.59353945

\implies AC=50.434051388\approx 50

Hence, the approximately distance between the swing and the slide is 50 ft.

Option 'B' is correct.

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