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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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If 2(6 + 3x) + 1 = - 5 + 3(x - 1), what is the value of x?
Stels [109]

Answer: x = -7

Step-by-step explanation:

2(6 + 3x) + 1 = - 5 + 3(x - 1)

(2) (6) + (2) (3x) + 1 = -5 + (3) (x) + (3) (-1)

12 + 6x + 1 = - 5 + 3x + - 1

( 6x ) + ( 12 - 1 ) = ( 3x ) + ( -5 + -3 )

6x + 13 = 3x - 8

6x + 13 - 3x = 3x - 8 - 3x

3x + 13 = -8

3x + 13 - 13 = -8 - 13

3x = -21

3x/3 -21/3

X = -7

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OBAMA

Step-by-step explanation:

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I’m confused and don’t have an angle tool to even try to do the question
yanalaym [24]
<h3>Answer:   40</h3>

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

Explanation:

JQ is longer than QN. We can see this visually, but the rule for something like this is the segment from the vertex to the centroid is longer compared to the segment that spans from the centroid to the midpoint.

See the diagram below.

The ratio of these two lengths is 2:1, meaning that JQ is twice as long compared to QN. This is one property of the segments that form when we construct the centroid (recall that the centroid is the intersection of the medians)

We know that JN = 60

Let x = JQ and y = QN

The ratio of x to y is x/y and this is 2/1

x/y = 2/1

1*x = y*2

x = 2y

Now use the segment addition postulate

JQ + QN = JN

x + y = 60

2y + y = 60

3y = 60

y = 60/3

y = 20

QN = 20

JQ = 2*y = 2*QN = 2*20 = 40

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

We have

JQ = 40 and QN = 20

We see that JQ is twice as larger as QN and that JQ + QN is equal to 60.

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