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Nina [5.8K]
2 years ago
11

HELP:. In baseball, the distance from the pitcher's mound to the batter is 60.5 feet. A pitcher can throw the baseball at 121 fe

et per second (82.5 miles per hour). An eye blink is 0.1 second. In how many blinks of an eye does it take for the baseball to travel from the pitcher to the batter?
A. 0.5 blinks
B.1 blink
C.2 blinks
D.5 blinks
E.10 blinks
Mathematics
1 answer:
zheka24 [161]2 years ago
3 0
<h3>Answer:  D. 5 blinks</h3>

=========================================================

Explanation:

First we'll need to find out how long it takes for the ball to go from the pitcher to the batter.

It needs to travel 60.5 feet and does so at a speed of 121 ft/sec

distance = speed*time

time = distance/speed

time = (60.5 ft)/(121 ft per sec)

time = 0.5 seconds

In half a second, the ball is from the pitcher to the batter.

------------

An eye blink is 0.1 second which means we can form this equation

0.1 second = 1 blink

We multiply both sides by 5 to get the "0.1 second" to turn into "0.5 seconds".

So,

0.1 second = 1 blink

5*(0.1 second) = 5*(1 blink)

0.5 seconds = <u>5 blinks</u> which points us to choice D.

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Both the formula can be applied to the expression((x^{3} x^{-6} )^{2}) during the first step while solving it.

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Comparing the terms of (x^{3} x^{-6} ) with p^a \times p^b

p=x, a =3, b=-6

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So, (x^{3} x^{-6} )^{2} is reduced to (x^{-3} )^{2}

<u>Applying formula (2):</u>

Comparing the terms of (x^{3} x^{-6} )^{2} with (p^a \times q^b)^c

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(x^{-3} )^{2}

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