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mariarad [96]
3 years ago
10

How long is the arc intersected by a central angle of StartFraction 5 pi Over 3 EndFraction radians in a circle with a radius of

2 ft? Round your answer to the nearest tenth. Use 3.14 for Pi.
Mathematics
1 answer:
aniked [119]3 years ago
3 0

Answer:

10.5 ft

Step-by-step explanation:

When central angle is given in radians, the formula for arc length = r × θ

θ = central angle = 5π/3 radians

π = 3.14

r = 2 ft

Hence,

Arc length = 2 ×( 5 × 3.14)/3

= 10.466666667 ft

Approximately to the nearest tenth = 10.5 ft

Therefore, the length of the arc is 10.5 ft

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which of the following is equivalent to 5/12 ÷ 8/10? A) 5/12 x 8/10 B) 12/5 x 8/10 C) 5/12 x 10/8 D) 12/5 x 8/10
Romashka-Z-Leto [24]

5/12 ÷ 8/10

=5/12 x 10/8

Answer

C) 5/12 x 10/8

6 0
4 years ago
Please help me anyone i need this in 10 minutes
Radda [10]

Given:

The rate of interest on three accounts are 7%, 8%, 9%.

She has twice as much money invested at 8% as she does in 7%.

She has three times as much at 9% as she has at 7%.

Total interest for the year is $150.

To find:

Amount invested on each rate.

Solution:

Let x be the amount invested at 7%. Then,

The amount invested at 8% = 2x

The amount invested at 9% = 3x

Total interest for the year is $150.

x\times \dfrac{7}{100}+2x\times \dfrac{8}{100}+3x\times \dfrac{9}{100}=150

Multiply both sides by 100.

7x+16x+27x=15000

50x=15000

Divide both sides by 50.

x=\dfrac{15000}{50}

x=300

The amount invested at 7% is 300.

The amount invested at 8% is

2(300)=600

The amount invested at 9% is

3(300)=900

Therefore, the stockbroker invested $300 at 7%, $600 at 8%, and $900 at 9%.

6 0
3 years ago
Fission tracks are trails found in uranium-bearing minerals, left by fragments released during fission events. An article report
Harlamova29_29 [7]

Answer:

Mean track length for this rock specimen is between 10.463 and 13.537

Step-by-step explanation:

99% confidence interval for the mean track length for rock specimen can be calculated using the formula:

M±\frac{t*s}{\sqrt{N}} where

  • M is the average track length (12 μm) in the report
  • t is the two tailed t-score in 99% confidence interval (2.977)
  • s is the standard deviation of track lengths in the report (2 μm)
  • N is the total number of tracks (15)

putting these numbers in the formula, we get confidence interval in 99% confidence as:

12±\frac{2.977*2}{\sqrt{15}} =12±1.537

Therefore, mean track length for this rock specimen is between 10.463 and 13.537

4 0
3 years ago
Without solving, what can you tell about the solution to
Sidana [21]

Answer:

Look at this picture so you will get your answer

Step-by-step explanation

4 0
2 years ago
In this graph, the y-intercept of the line is<br> The equation of the line is y=<br> X
Advocard [28]

Answer:

-2

Step-by-step explanation:

that is were the x is zero

3 0
4 years ago
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