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MrRa [10]
2 years ago
9

A baseball diamond is a square with side 90 feet. A batter hits the ball and runs toward first base with a speed of 24 ft/sec. A

t what rate is his distance from second base decreasing when he is halfway to first base?
Mathematics
1 answer:
sweet [91]2 years ago
6 0

9514 1404 393

Answer:

  10.73 ft/s

Step-by-step explanation:

When the runner is halfway to first base, the triangle formed by his position and the first and second bases is a right triangle with legs of 45 ft and 90 ft. The angle between the runner's direction and the line from his position to 2nd base is ...

  angle = arctan(90/45) ≈ 63.43°

The "speed made good" in the direction of 2nd base is the product of the cosine of this angle and the runner's speed toward first.

  speed toward 2nd = cos(63.43°)×(24 ft/s) ≈ 10.73 ft/s

The rate the runner's distance to 2nd base is decreasing is 10.73 ft/sec.

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kati45 [8]

Answer:

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3 0
2 years ago
What is the value of x in parallelogram QRST?<br><br> 16<br> 12<br> 8 <br> 4
Wewaii [24]
This is parallelogram so
8=2x
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3 0
3 years ago
8. Which of the following does NOT deseribe a
topjm [15]

Answer:

B. a decimal that does not terminate and does

not repeat.

7 0
2 years ago
300 mi /20 sec = ? Mi /1 min
snow_lady [41]

Answer:

900

Step-by-step explanation:

To get your answer, you can just multiply 20 seconds by 3 to get 60 seconds, which is a minute.

Since whatever you do on one side you have to do to the other, multiply 300 mi by 3 as well. That would get you 900.

4 0
3 years ago
The weight of an object on Venus is approximately 9/10 of its weight on Earth. The weight of an object
Sergeeva-Olga [200]

An object weighs 140 pounds more on Jupiter than it weighs on Venus. By simple algebraic operations, the required value is calculated.

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Addition, subtraction, multiplication, and division are the basic algebraic operations that are used for many computations.

<h3>Calculation:</h3>

It is given that,

An object's weight on Venus is approximately 9/10 of its weight on earth. I.e., WV = 9/10 WE

An object's weight on Jupiter is approximately 23/10 of its weight on earth. I.e., WJ = 23/10 WE

If an object weighs 100 pounds on earth, then

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The weight of the object on Jupiter is

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8 0
1 year ago
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