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expeople1 [14]
3 years ago
15

Can SOMEONE ANSWER these two questions ? A dog runs from point O to B. The dog reaches A at 2.3 s into the sprint. Then the dog

reaches C at 4.1 s. What is the average velocity of the dog from A to C?
A test car travels in a straight line along the -axis. The graph in the figure shows the car’s position as a function of time. At which point is the velocity zero?
will give brainliest

Mathematics
1 answer:
frosja888 [35]3 years ago
6 0

<u>dog</u>

Average velocity is given by the change in position divided by the time required for that change.

v = ∆x/∆t = (4 m)/(4.1 s -2.3 s) = 4/1.8 m/s ≈ 2.2 m/s

<u>car</u>

The instantaneous velocity is zero where the graph is horizontal (no change in position while time marches on—zero slope). Marked points C and G are such points. The appropriate choice is point C.

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The horizontal distance is the difference of x-coordinates: 6 - 4 = 2.

The vertical distance is the difference of y-coordinates: -1 - (-2) = 1.

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2 years ago
What is the multiplicative inverse of 2 1/2
anzhelika [568]

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2 years ago
5.815 + 6.021 as a mixed number
tia_tia [17]
<h2><u>Answer</u></h2>

11 209/250

<h2><u>Explanation</u></h2>

5.815 + 6.021

First add the numbers the way they are.

          5.815

       <u> +6.021</u>

      <u>   11.836</u>

The whole number is 11.

11.836 = 11 + 0.836

0.836 = 836/1000

Dividing byy 4 we get,

0.836 = 836/1000 = 209/250

∴ 11.836  = 11 209/250

3 0
2 years ago
Read 2 more answers
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