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ioda
3 years ago
13

If the arc length of a sector in the unit circle is 3π/2 , what is the measure of the angle of the sector? A) 0 radians B) π/2 r

adians C) π radians D) 3π/2 radians
Mathematics
2 answers:
Naily [24]3 years ago
8 0
D the answer is already in radians
ziro4ka [17]3 years ago
6 0

Answer: D) \dfrac{3\pi}{2} radians

Step-by-step explanation:

In geometry, the radius of a unit circle = 1 unit.

Given: The arc length of a sector in the unit circle =\dfrac{3\pi}{2}

We know that the formula to calculate length of arc having central angle x and radius r is given by  :-

l=rx

Substitute r= 1  and  l=\dfrac{3\pi}{2} in the above formula , we get

\dfrac{3\pi}{2}=(1)x\\\\\Rightarrow\ x=\dfrac{3\pi}{2}

Hence, the  measure of the angle of the sector =\dfrac{3\pi}{2} radians

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3 years ago
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ExtremeBDS [4]

Answer:263.9 in squared

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Find the area of the whole circle

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6 0
3 years ago
Choose the abbreviation of the postulate or theorem that supports the conclusion that the triangles are congruent.
Ber [7]

Answer:

Option D.

LA Theorem or Postulates

Explanation:

LA theorem or postulates is leg-acute states that if the leg and an acute angle of one right triangle are congruent to the corresponding leg and acute angle of another right triangle, then the triangles are congruent.

In \triangle ACB and  \triangle DFC

\angle C =\angle F = 90^{\circ}             [Given]

\angle B = \angle E   [Acute angle]      [Given]

BC=EF        [Given]

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\triangle ACB \cong \triangle DFE

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8 0
3 years ago
Read 2 more answers
A force of 3 pounds is required to hold a spring stretched 0.6 feet beyond its natural length. how much work (in foot-pounds) is
ioda

The work done (in foot-pounds) in stretching the spring from its natural length to 0.7 feet beyond its natural length is 1.23 foot-pound

<h3>Data obtained from the question</h3>

From the question given above, the following data were obtained:

  • Force (F) = 3 pounds
  • Extension (e) = 0.6 feet
  • Work done (Wd) =?

<h3>How to determine the spring constant</h3>
  • Force (F) = 3 pounds
  • Extension (e) = 0.6 feet
  • Spring constant (K) =?

F = Ke

Divide both sides by e

K = F/ e

K = 3 / 0.6

K = 5 pound/foot

Thus, the spring constant of the spring is 5 pound/foot

<h3>How to determine the work done</h3>
  • Spring constant (K) = 5 pound/foot
  • Extention (e) = 0.7 feet
  • Work done (Wd) =?

Wd = ½Ke²

Wd = ½ × 5 × 0.7²

Wd = 2.5 × 0.49

Wd = 1.23 foot-pound

Therefore, the work done in stretching the spring 0.7 feet is 1.23 foot-pound

Learn more about spring constant:

brainly.com/question/9199238

#SPJ1

5 0
1 year ago
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