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garri49 [273]
3 years ago
5

Adenan makes a scale drawing of a rectangular flower box. The length is 5 in., and the width is 3 in. Adenan changes the scale o

f the drawing from 1 in. : 4 ft to 1 in. : 6 ft. Which statement about the dimensions of the flower box is true?
The length of the flower box under the new scale is 15 ft.
The length of the flower box under the old scale is 30 ft.
The width of the flower box under the old scale is 12 ft.
The width of the flower box under the new scale is 24 ft.
Mathematics
2 answers:
skelet666 [1.2K]3 years ago
8 0

Answer: C

Step-by-step explanation: because on my test i got it right and i want you to get it right to

Fynjy0 [20]3 years ago
3 0
Length is 5 in, width is 3 in

scale drawing was 1 in to 4 ft

so the old length was : 5 * 4 = 20 ft
and the old width was : 3 * 4 = 12 ft <===

she changed the scale drawing from 1 in to 6 ft

so the new length is : 5 * 6 = 30 ft
and the new width is : 3 * 6 = 18 ft

so ur answer is C
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3 years ago
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°
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  • <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)}

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Answer:

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Percentage of boys =  × 100 = 46.7%

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