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Nataly [62]
2 years ago
7

light travels 3000000 scs if earth is 150 million km away from sun how many minutes it will take the light to come from sun to e

arth
Mathematics
2 answers:
sweet-ann [11.9K]2 years ago
5 0
  • Speed of light=3×10^8m/s=3×10^5km/s
  • Distance=150million km=150×10^6km

Now

\\ \rm\longmapsto Speed=\dfrac{Distance}{Time}

\\ \rm\longmapsto Time=\dfrac{Distance}{Speed}

\\ \rm\longmapsto Time=\dfrac{150\times 10^6}{3\times 10^5}

\\ \rm\longmapsto Time=50\times 10^1

\\ \rm\longmapsto Time=500s

\\ \rm\longmapsto Time=8.33min

  • Can be written as 8min 20s.
Airida [17]2 years ago
4 0

Answer:

speed \: of \: light  = 3 \times 10 {}^{8} m \: per \: second. \\ distance = 150million \: km \\  = 150 \times 10 {}^{6} km. \\  \\ now \\ speed =  \frac{distance}{time}  \\  \frac{150 \times 10 {}^{6} }{3 \times 10 {}^{5} }  \\ 50 \times 10 {}^{1}  \\ 500sec \\ 8.33min. \\ it \: can \: be \: written \: as \\ 8min \: 20sec.

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Write the formula for an exponential function with initial value 300 and decay factor 0.73
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Answer: 300*0.73^x

Step-by-step explanation:

300 is the initial value and every time x goes up by 1, the value get multiplied by 0.73

when x=3

then the value would be 300*0.73*0.73*0.73 or 300*0.73^3

then for x amount of times the equation would be 300*(0.73 x times ) or 300*0.73^x

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3 years ago
Dillon divided a 3 ⅓ pound bag of pears among his 5 friends. How many pounds of pears did each friend receive?
Salsk061 [2.6K]

Answer:

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3 years ago
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ΔABC is dilated by a scale factor of 0.25 with the origin as the center of dilation, resulting in the image ΔA′B′C′. If A = (-1,
Leni [432]

Answer:

5

Step-by-step explanation:

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B' =  

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I did mine in fraction form, because it will prove to be more useful in future mathematics.


B' = (1/2  ,  1/4)


Repeat the process with C.


C' =

(14 x .25),(17 x .25)


C' =

(7/2  ,  17/4)




You only need to focus on B and C because you are finding the length of B'C'.


The formula for distance is the square root of x to the sub of 2 minus x to the sub of 1 squared minus y to the sub of 2 minus y to the sub of 1 square.


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y2 - y1 = 17/4 - 1/4 = 16/4  =  4 squared = 16


16 + 9  = 25


Square root of 25 is 5.


Therefore, the distance is 5.


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