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tekilochka [14]
2 years ago
14

Isaiah brought two dozen donuts to a class party. Half of the donuts had vanilla frosting with sprinkles, 2 had strawberry frost

ing with cream filling, and the rest had chocolate frosting with sprinkles. What fraction of the donuts had sprinkles? There are 12 donuts in a dozen.
Mathematics
1 answer:
enot [183]2 years ago
7 0

Answer:

22/24

ALL TOGETHER there is 24 donuts.

take half of 24, or 24 divided by 2, and you get 12.

so we know 12 of the donuts are vanilla with sprinkles.

then 2 had strawberry frosting and the rest had chocolate and sprinkles.

we took 12 away so we only have twelve left and if 2 of those 12 are strawberry with no sprinkles and the rest are chocolate WITH sprinkles, this means 10 are chocolate with sprinkles.

so 10 chocolate and sprinkles +12 vanilla and sprinkles=22 donuts with sprinkles out of 24 donuts total

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A golfer hits an errant tee shot that lands in the rough. A marker in the center of the fairway is 150 yards from the center of
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\bold{\huge{\underline{ Solution}}}

<h3><u>Given </u><u>:</u><u>-</u></h3>

  • A marker in the center of the fairway is 150 yards away from the centre of the green
  • While standing on the marker and facing the green, the golfer turns 100° towards his ball
  • Then he peces off 30 yards to his ball

<h3><u>To </u><u>Find </u><u>:</u><u>-</u></h3>

  • <u>We </u><u>have </u><u>to </u><u>find </u><u>the </u><u>distance </u><u>between </u><u>the </u><u>golf </u><u>ball </u><u>and </u><u>the </u><u>center </u><u>of </u><u>the </u><u>green </u><u>.</u>

<h3><u>Let's </u><u> </u><u>Begin </u><u>:</u><u>-</u></h3>

Let assume that the distance between the golf ball and central of green is x

<u>Here</u><u>, </u>

  • Distance between marker and centre of green is 150 yards
  • <u>That </u><u>is</u><u>, </u>Height = 150 yards
  • For facing the green , The golfer turns 100° towards his ball
  • <u>That </u><u>is</u><u>, </u>Angle = 100°
  • The golfer peces off 30 yards to his ball
  • <u>That </u><u>is</u><u>, </u>Base = 30 yards

<u>According </u><u>to </u><u>the </u><u>law </u><u>of </u><u>cosine </u><u>:</u><u>-</u>

\bold{\red{ a^{2} = b^{2} + c^{2} - 2ABcos}}{\bold{\red{\theta}}}

  • Here, a = perpendicular height
  • b = base
  • c = hypotenuse
  • cos theta = Angle of cosine

<u>So</u><u>, </u><u> </u><u>For </u><u>Hypotenuse </u><u>law </u><u>of </u><u>cosine </u><u>will </u><u>be </u><u>:</u><u>-</u>

\sf{ c^{2} = a^{2} + b^{2} - 2ABcos}{\sf{\theta}}

<u>Subsitute </u><u>the </u><u>required </u><u>values</u><u>, </u>

\sf{ x^{2} = (150)^{2} + (30)^{2} - 2(150)(30)cos}{\sf{100°}}

\sf{ x^{2} = 22500 + 900 - 900cos}{\sf{\times{\dfrac{5π}{9}}}}

\sf{ x^{2} = 22500 + 900 - 900( - 0.174)}

\sf{ x^{2} = 22500 + 900 + 156.6}

\sf{ x^{2} = 23556.6}

\bold{ x = 153.48\: yards }

Hence, The distance between the ball and the center of green is 153.48 or 153.5 yards

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