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Charra [1.4K]
2 years ago
7

A student determined that the area of the segment of shown above is Asegment = 137.71 ft2. The student's work is shown below. Ev

aluate the student’s answer. Asector = 75.36 ft2 Atriangle = 62.35 ft2 Asegment = Asector + Atriangle Asegment = 137.71 ft2 A. The student’s answer is correct. B. The student did not calculate the area of the sector correctly. C. The student did not calculate the area of the triangle correctly. D. The student did not use the correct formula to calculate the area of the segment.
Mathematics
1 answer:
tankabanditka [31]2 years ago
6 0

Answer:

D. The student did not use the correct formula to calculate the area of the segment.

Step-by-step explanation:

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Q 1 PLEASE HELP ME FIGURE THIS OUT 2
dimulka [17.4K]

Answer: \frac{\pi }{3}

<u>Step-by-step explanation:</u>

cot is \frac{cos}{sin} on the Unit Circle. When is cos = \frac{1}{2} and sin = \frac{\sqrt{3}} {2}?  In other words, where is the coordinate (\frac{1}{2},\frac{\sqrt{3}} {2}) on the Unit Circle?

It occurs at 60° = \frac{\pi }{3} radians

**************************************************************************************

Answer: cot θ

<u>Step-by-step explanation:</u>

  sec (90 - θ) * cos θ

= \frac{1}{cos (90 -\theta)} * cos θ

= \frac{cos\theta}{cos(90 - \theta)}

= \frac{cos\theta}{cos90*cos\theta+sin90*sin\theta}

= \frac{cos\theta}{0*cos\theta+1*sin\theta}

= \frac{cos\theta}{sin\theta}

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**************************************************************************************

Answer: BC = √5, AC = 2√5

<u>Step-by-step explanation:</u>

\frac{\pi} {3} = 60°, which means ΔABC is a 30°-60°-90° triangle so we can use the side length formulas: x - x√3 - 2x.

∠C is the 60°. It matches to side AB so: AB = x√3 = √15   ⇒   x = √5

∠A is the 30°. It matches to side BC so: BC = x = √5

∠B is the 90°. It matches to side AC so: AC = 2x = 2(√5)  = 2√5

**************************************************************************************

Answer: k = 4, RS = 12, QS = 28

<u>Step-by-step explanation:</u>

\frac{QR}{MN} = \frac{RS}{NO}= \frac{QS}{MO}

\frac{24}{6} = \frac{RS}{3}= \frac{QS}{7}

proportionality constant k can be found with \frac{QR}{MN} = \frac{24}{6} = 4

\frac{QR}{MN} = \frac{RS}{NO}

→ 4 = \frac{RS}{3}

→ 4(3) = RS

→ 12 = RS

\frac{QR}{MN} = \frac{QS}{MO}

→ 4 = \frac{QS}{7}

→ 4(7) = QS

→ 28 = QS


3 0
3 years ago
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