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lana66690 [7]
3 years ago
5

PLSSSSSSSSHELPMEPLSSSSSS

Mathematics
2 answers:
andreev551 [17]3 years ago
8 0
23 miles



15-2.35 = 12.65
12.65/ .55= 23
Fittoniya [83]3 years ago
4 0
It has to be 23 miles fam
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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
Gwar [14]

\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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2 years ago
A half-acre building lot is five times as long as it is wide. What are its dimensions?
Gnoma [55]

we know that

A half-acre is equal to----------> 2,023.43\ m^{2}

Let

x---------> the length side of the building lot

y--------> the width side of the building lot

A------> Area of the building lot

x=5y-------> equation 1

A=2,023.43\ m^{2}

A=x*y

2,023.43=x*y --------> equation 2

substitute equation 1 in equation 2

2,023.43=[5y]*y

5y^{2}= 2,023.43\\y^{2}= 404.686\\y= 20.11\ m

<u>Find the value of x</u>

x=5y---> x=5*20.11--> x=100.55 m

therefore

<u>the answer is</u>

the dimensions of the building lot are 100.55 m* 20.11 m

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Answer:

i think 2-3 min.

Step-by-step explanation: i wish it's help u

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The swimmer will take 67.17 second.
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Ilya [14]

Hey!

Hope this helps...

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The correct answers are A, B, D, and E

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