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erma4kov [3.2K]
3 years ago
13

The figure is made up of two shapes, a semicircle and a rectangle.What is the exact perimeter of the figure?

Mathematics
2 answers:
grandymaker [24]3 years ago
7 0
Perimeter of circular area = πr = 3.14 * 10 = 31.4 mm

Perimeter of rectangular area = 30 + 20 + 30 = 80 mm

Total Perimeter = 80 + 31.4 = 111.4

In short, Your Answer would be 111.4 mm

Hope this helps!
Svetllana [295]3 years ago
7 0
Perimeter of a rectangle =2(w+l)=2(50)=100mm

perimeter of a semicircle =circumference of a semicircle=2π2/2=2π*10/2

here d=20 so r=d/2=10mm

perimeter of a semicircle=10πmm=31.4 mm

perimeter of a whole fig is 31.4+100=131.4mm



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a car gets 27 miles per gallon in the city and 35 miles per gallon on the highway. what range of miles can the car travel on 3 g
Verdich [7]
The answer is 81 < < 105

Explanation: Since the car gets 27 mpg in the city and gets 35 mpg on the highway, you can set it up in the inequality 27 < < 35 and this represents the range a car can drive for 1 gallon, so to find the range for 3 gallons, you multiply this inequality by 3.

27(3)<<35(3)
27(3)=81
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Then you can insert the numbers into the inequality to get 81<<105.
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Please dont ignore, Need help!!! Use the law of sines/cosines to find..
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Answer:

16. Angle C is approximately 13.0 degrees.

17. The length of segment BC is approximately 45.0.

18. Angle B is approximately 26.0 degrees.

15. The length of segment DF "e" is approximately 12.9.

Step-by-step explanation:

<h3>16</h3>

By the law of sine, the sine of interior angles of a triangle are proportional to the length of the side opposite to that angle.

For triangle ABC:

  • \sin{A} = \sin{103\textdegree{}},
  • The opposite side of angle A a = BC = 26,
  • The angle C is to be found, and
  • The length of the side opposite to angle C c = AB = 6.

\displaystyle \frac{\sin{C}}{\sin{A}} = \frac{c}{a}.

\displaystyle \sin{C} = \frac{c}{a}\cdot \sin{A} = \frac{6}{26}\times \sin{103\textdegree}.

\displaystyle C = \sin^{-1}{(\sin{C}}) = \sin^{-1}{\left(\frac{c}{a}\cdot \sin{A}\right)} = \sin^{-1}{\left(\frac{6}{26}\times \sin{103\textdegree}}\right)} = 13.0\textdegree{}.

Note that the inverse sine function here \sin^{-1}() is also known as arcsin.

<h3>17</h3>

By the law of cosine,

c^{2} = a^{2} + b^{2} - 2\;a\cdot b\cdot \cos{C},

where

  • a, b, and c are the lengths of sides of triangle ABC, and
  • \cos{C} is the cosine of angle C.

For triangle ABC:

  • b = 21,
  • c = 30,
  • The length of a (segment BC) is to be found, and
  • The cosine of angle A is \cos{123\textdegree}.

Therefore, replace C in the equation with A, and the law of cosine will become:

a^{2} = b^{2} + c^{2} - 2\;b\cdot c\cdot \cos{A}.

\displaystyle \begin{aligned}a &= \sqrt{b^{2} + c^{2} - 2\;b\cdot c\cdot \cos{A}}\\&=\sqrt{21^{2} + 30^{2} - 2\times 21\times 30 \times \cos{123\textdegree}}\\&=45.0 \end{aligned}.

<h3>18</h3>

For triangle ABC:

  • a = 14,
  • b = 9,
  • c = 6, and
  • Angle B is to be found.

Start by finding the cosine of angle B. Apply the law of cosine.

b^{2} = a^{2} + c^{2} - 2\;a\cdot c\cdot \cos{B}.

\displaystyle \cos{B} = \frac{a^{2} + c^{2} - b^{2}}{2\;a\cdot c}.

\displaystyle B = \cos^{-1}{\left(\frac{a^{2} + c^{2} - b^{2}}{2\;a\cdot c}\right)} = \cos^{-1}{\left(\frac{14^{2} + 6^{2} - 9^{2}}{2\times 14\times 6}\right)} = 26.0\textdegree.

<h3>15</h3>

For triangle DEF:

  • The length of segment DF is to be found,
  • The length of segment EF is 9,
  • The sine of angle E is \sin{64\textdegree}}, and
  • The sine of angle D is \sin{39\textdegree}.

Apply the law of sine:

\displaystyle \frac{DF}{EF} = \frac{\sin{E}}{\sin{D}}

\displaystyle DF = \frac{\sin{E}}{\sin{D}}\cdot EF = \frac{\sin{64\textdegree}}{39\textdegree} \times 9 = 12.9.

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Step-by-step explanation:

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Answer:

Every person in the US.

Step-by-step explanation:

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Answer:

1/5

Step-by-step explanation:

Both Amir and Tamara ate 2/5 each.

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If 4/5 of the candy bar was eaten, 1/5 of the candy bar remains because

5/5 - 4/5 = 1/5

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