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balandron [24]
4 years ago
13

4) The area of a piece of pie in the shape of a sector is 7.1 in. The angle of the sector is 40°. Find the diameter of

Mathematics
1 answer:
elixir [45]4 years ago
4 0

Answer:

Step-by-step explanation:

Area of sector = angle/360 * pi * r^2

7.1 = 40/360 * 22/7 * r^2

7.1 = 1/9 * 22/7 * r^2

63.9 = 22r^2/7

r^2 = (7 * 63.9)/22

r^2 = 20.331818181818

r = square root of 20.331818181818

r = 4.51

d = r * 2 = 4.51 * 2 = 9.02

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I think I know how to solve this, but could someone please check my work for me?
Fiesta28 [93]
Answer is :

 cos⁻¹[(-1-4√7)/(√17, √8) = 173.33° ≈ 173°

5 0
4 years ago
Consider the sequence: 7, 64, 121, 178, 235, ... where u, = 7
Licemer1 [7]

Answer:

It is an arithmetic sequence.

Step-by-step explanation:

64 - 7 = 57

121 - 64 = 57

178 - 121 = 57

We have a common difference of 57, so it is arithmetic.

The explicit formula for the nth term

= 7  +57(n - 1).

7 0
3 years ago
Which of the following examples has a unit rate of 50 miles per hour? Select all that apply.
Usimov [2.4K]

Answer:

I believe your answer would be the following:

A) 250 miles in 5 hours.

C) 200 miles in 4 hours

E) 700 miles in 14 hour

Step-by-step explanation:

Unit rate = 50

A) 250 miles in 5 hours // 50*5 = 250 (CORRECT)

B) 300 miles in 5 hours // 50*5 ≠ 300 (WRONG)

C) 200 miles in 4 hours // 50*4 = 200 (CORRECT)

D) 240 miles in 4 hours // 50*4 ≠ 240 (WRONG)

E) 700 miles in 14 hours // 50*14 = 700 (CORRECT)

Hope this helps! :)

8 0
3 years ago
Read 2 more answers
During a science experiment,the temperature of a solution in beaker 1 was 5 degrees below zero. The temperature of a solution in
GaryK [48]
Answer:
5 degrees

Step-by-step explanation:

We are given that during a science experiment
The temperature of a solution in a beaker 15 degrees below zero
The temperature of solution in a beaker 2 was opposite of the temperature in beaker 1
We have to determine the temperature of beaker 2.
It mean temperature of beaker 1 solution=-5 degrees
According to question
Temperature of beaker 2 solution =5 degrees
Hence, the temperature of solution in a beaker 2=5 degrees
7 0
1 year ago
Root 3 cosec140° - sec140°=4<br>prove that<br><br>​
Lorico [155]

Answer:

Step-by-step explanation:

We are to show that \sqrt{3} cosec140^{0} - sec140^{0} = 4\\

<u>Proof:</u>

From trigonometry identity;

cosec \theta = \frac{1}{sin\theta} \\sec\theta = \frac{1}{cos\theta}

\sqrt{3} cosec140^{0} - sec140^{0} \\= \frac{\sqrt{3} }{sin140} - \frac{1}{cos140} \\= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\

From trigonometry, 2sinAcosA = Sin2A

= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\\\=  \frac{\sqrt{3}cos140-sin140 }{sin280/2}\\=  \frac{4(\sqrt{3}/2cos140-1/2sin140) }{2sin280}\\\\= \frac{4(\sqrt{3}/2cos140-1/2sin140) }{sin280}\\since sin420 = \sqrt{3}/2 \ and \ cos420 = 1/2  \\ then\\\frac{4(sin420cos140-cos420sin140) }{sin280}

Also note that sin(B-C) = sinBcosC - cosBsinC

sin420cos140 - cos420sin140 = sin(420-140)

The resulting equation becomes;

\frac{4(sin(420-140)) }{sin280}

= \frac{4sin280}{sin280}\\ = 4 \ Proved!

3 0
3 years ago
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