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s344n2d4d5 [400]
3 years ago
15

the earth rotates about its axis once every 23 hours, 56 minutes and 4 seconds. Approximate the number of radians the earth rota

tes in one second.
Mathematics
1 answer:
marishachu [46]3 years ago
4 0

Answer:

\frac{\pi}{43082}\text{ radians per second}

Step-by-step explanation:

Given,

Time taken in one rotation of earth = 23 hours, 56 minutes and 4 seconds.

Since, 1 minute = 60 seconds and 1 hour = 3600 seconds,

⇒ Time taken in one rotation of earth = (23 × 3600 + 56 × 60 + 4) seconds

= 86164  seconds,

Now,  the number of radians in one rotation = 2π,

That is, 86164 seconds = 2π radians

\implies 1\text{ second }=\frac{2\pi}{86164}=\frac{\pi}{43082}\text{ radians}

Hence, the number of radians in one second is \frac{\pi}{43082}

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Which of these is not a characteristic of a perpendicular bisector?
Arte-miy333 [17]

Answer:

The perpendicular bisector meets any line at an angle which is not 90°

Step-by-step explanation:

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3 years ago
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((((((ASAP))))))On Tuesday afternoon at camp, Finn did archery and sailing before dinner. Archery started at 12:10 P.M. Finn spe
Andrei [34K]

Answer:

2:15

Step-by-step explanation:

legit just mush it up and add or sum .-.

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3 years ago
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In a basketball tournament, team A scored 6 more points than 3 times as many points as team B scored. Team C scored 45 more poin
velikii [3]

Answer:

"Team C scored 130 points."

Step-by-step explanation:

Let x represent how many points team B scored.

Team A: 3x + 6

Team B: x

Team C: x + 45

Total: 476 points

476 = (x) + (3x + 6) + (x + 45)

476 = x + 3x + 6 + x + 45

476 = 5x + 51

425 = 5x

x = 85

Team A:

3(85) + 6

= 261 pts

Team B:

= 85 pts

Team C:

(85) + 45

= 130 points

4 0
3 years ago
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the figure below shows a circle centre of radius 10 cm the chord PQ=16cm calculate the area of the shaded region​
m_a_m_a [10]

Step-by-step explanation:

OPQ originally forms a sector, the formula for sector is

\frac{x}{360} \pi {r}^{2}

where x is the degrees of rotation between the two radii.

We know three sides length and is trying to find an angle between the radii so we can use law of cosines which states that

16 =  \sqrt{10 {}^{2} + 10 {}^{2}   - 2(100) \times  \cos(o) }

This isn't the standard formula, it's for this problem

16 =  \sqrt{200 - 200 \times  \cos(o) }

16 =  \sqrt{200  - 200 \cos(o) }

256 = 200 - 200 \cos(o)

56 =  - 200 \cos(o)

-  \frac{7}{25}  =  \cos(o)

\cos {}^{ - 1} (  - \frac{7}{25} )  =  \cos {}^{ - 1} ( \cos(o) )

106.26 = o

So we found our angle of rotation, which is 106.26

Now, we do the sector formula.

\frac{106.26}{360} (100)\pi

\frac{10626}{360} \pi

is the area of sector.

Now let find the area of triangle, we can use Heron formula,

The area of a triangle is

\sqrt{s(s - a)(s - b)(s - c)}

where s is the semi-perimeter.

To find s, add all the side lengths up, 10,10,16 and divide it by 2.

Which is

\frac{36}{2}  = 18

\sqrt{18(18 - 10)(18 - 10)(18 - 16)}

\sqrt{18(8)(8)(2)}

\sqrt{(36)(64)}

6  \times 8 = 48

So our area of the triangle is 48. Now, to find the shaded area subtract the main area,( the sector of the circle) by the area of the triangle so we get

\frac{10626}{360}  - 48

Which is an approximate or

44.73

6 0
2 years ago
Match the solution set given in inequality notation with the solution set given in interval motion x > 7.8
Maslowich

Answer:

(7.8, ∞)

Step-by-step explanation:

Given the inequality :

x > 7.8 ; the inequality can be explicitly interpreted as x greater than 8 ; this means that the inequality holds for all values of x above 7.8

The interval for which the inequality holds true is :

x > 7. 8 - - - > (7.8, ∞)

7 0
3 years ago
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