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nydimaria [60]
3 years ago
12

Given a sphere with radius r, the formula 4pier^2 gives

Mathematics
2 answers:
zhannawk [14.2K]3 years ago
5 0

Answer:

B.  The surface area.

Step-by-step explanation:

4πr^2 gives the surface area of a sphere.

enot [183]3 years ago
5 0

Answer: the surface area

Step-by-step explanation:

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1719: 573 =   41.46: 23.93

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What value of x makes this inequality true? <br> x + 16 &lt; 43
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Answer:

x < 27

Step-by-step explanation:

x + 16 < 43

move the constant to the right-hand side and change its sign

x < 43 - 16

subtract the numbers

x < 27

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X < 27

alternative form

X ∈ ( - ∞ , 27 ) , { x | x < 27 }

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Mina, bobby, and Julia each have the same number of pencils. 2/6 of Mina's pencils are red, 2/3 of Bobby's pencils are red, and
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A kid at the bus stop is tossing a coin. He tosses the coin 19 times and it lands on heads eight times. If the kid tosses the co
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Step-by-step explanation:

4 0
2 years ago
drag the gray arrow so that it is pointing midway between horizontal and vertical, heading downward and toward the right. extend
alex41 [277]

The speed of the star as compared to the speed of the star relative to earth is the frequency of light emitted by the source.

A star moving away from the observer (say on Earth) will have its emitted light red-shifted (move to lower frequency), i.e., the spectrum moving to left if we consider the intensity of light is plotted as a function of frequency with the frequency increasing from left to right.

A star moving towards the observer (say on Earth) will have its emitted light blue-shifted (move to higher frequency), i.e., the spectrum moving to the right if we consider the intensity of light is plotted as a function of frequency with the frequency increasing from left to right.

A star moving perpendicular to the observer (say on Earth) will have its emitted light will no change with respect to the spectrum of the stationary star.

These are due to the Doppler effect, the equation of frequency detected by the observer as

f_o = f_s\sqrt{\frac{1-\beta }{1+\beta } } where \beta = \frac{v}{c}, v the relative velocity of the moving source (the star here), f^s is the frequency of light emitted by the source.

Hence the answer is the speed of the star as compared to the speed of the star relative to earth is the frequency of light emitted by the source.

To learn more about frequency click here brainly.com/question/26177128

#SPJ4

7 0
9 months ago
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