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KiRa [710]
3 years ago
6

Light travels at a speed of about 186,000 miles per second. How far does light travel in 5 second? Use repeated addition to solv

e.
Mathematics
1 answer:
11Alexandr11 [23.1K]3 years ago
8 0

Answer:

Step-by-step explanation:

The speed of light in a vacuum is 186,282 miles per second (299,792 kilometres per second), and in theory nothing can travel faster than light. In miles per hour, light speed is, well, a lot: about 670,616,629 mph. If you could travel at the speed of light, you could go around the Earth 7.5 times in one second.

186 ,282 x 5  = 931 , 410  for 5 seconds

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What is 27 is what % of 60?
natta225 [31]
Let 27 be x% of 60

so,
x/100 *60 = 27
x = 27*100/60 = 270/6 = 45

So,
27 is <em>45%</em> of 60.
6 0
3 years ago
If the scale factor for the dilation shown is 3, which is the length of B'C'?
Arturiano [62]

Answer:

15

Step-by-step explanation:

Scale Factor = 3

Using Formula:

Scale Factor = LengthOfDilatedTraingle/LengthOfActualTriangle

Scale Factor =  B'C'/B'C'

B'C' = 3 * 5

B'C' = 15

5 0
2 years ago
Find the x-values that are solutions of the equation 5cot^2x-15=0
julia-pushkina [17]
First solve for the trig function 'cot'
cot^2 x = \frac{15}{5} = 3

Next take the sqrt of both sides (include plus/minus)
cot x = \pm \sqrt{3}

Now take reciprocal of both sides, this will change trig function to 'tan'
(cot = 1/tan)

tan x = \pm \frac{1}{\sqrt{3}} = \pm \frac{\sqrt{3}}{3}

Finally use the unit circle or inverse tan on your calculator to find x.
There will be 4 solutions, one for each quadrant.

x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}
3 0
3 years ago
What is the simplified form of the expression? k3 [k^7/3] -5
iragen [17]
Simplify the following:
(k^3 k^7)/3 - 5

Combine powers. (k^3 k^7)/3 = k^(7 + 3)/3:
k^(7 + 3)/3 - 5

7 + 3 = 10:
k^10/3 - 5

Put each term in k^10/3 - 5 over the common denominator 3: k^10/3 - 5 = k^10/3 - 15/3:
k^10/3 - 15/3

k^10/3 - 15/3 = (k^10 - 15)/3:
Answer: (k^10 - 15)/3
3 0
3 years ago
Samantha ran on a fitness track for 20 ​min, and then walked for 15 min. Her running rate was 230 ft per minute faster than her
Mnenie [13.5K]

Answer:

  560 ft/min

Step-by-step explanation:

Let r represent Samantha's running rate in feet per minute. Then r-230 is her walking rate. We can use the relation ...

   distance = speed × time

to write an equation for the total distance Samantha went.

  20r + 15(r -230) = 16150

  35r - 3450 = 16150 . . . . eliminate parentheses, collect terms

  35r = 19600 . . . . . . . . . . add 3450

  r = 560 . . . . . . . . . . . . . . divide by the coefficient of r

Samantha ran 560 feet per minute.

5 0
3 years ago
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