5. A golfer, teeing off on an easy par 3 hole from point A, hits the ball to point B. The second stroke movies the ball from point B to C (on the green), Ten feet away from the hole (Point D). Write the component form for
AB, BC, AD
AB , BC , AD

5b. If the golfer makes a ten foot putt from point C into the hole (at point D), then how far did the ball travel en route from A to B to C to D? ( Hint: use the pythagorean triples 3,4,5; 5,12,13; and 8,15,17 and a scale of 1 unit = 10 feet.
?

5c.How far would the ball travel for a hole in one (A to D)?
_____?

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Respuesta :

DeanR DeanR · Answer 1 · 2019-12-14T14:28:30+01:00

5a

AB = B - A = (12,5) - (0,0) = (12,5)

BC = C - B = (16, 8) - (12, 5) = (4, 3)

AD = D - A = (15,8)

5b

CD = D - C = (15, 8) - (16, 8) = (-1, 0)

We have triples 12²+5²=13² and 4²+3²=5² so lengths

AB=13, BC=5, CD=1

Total distance = 10(13 + 5 + 1) = 190 feet

5c

That's the AD triple 15²+8²=17²

so 17 units or

170 feet

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