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Tanya [424]
1 year ago
8

As Juan continued his hitting practice showing off his abilities, one of the balls flew over the center field stands and into th

e parking lot. “Did you see that shot”, he yelled at the girls. “The ball hung in the air for at least 10 seconds”, he exclaimed.
The formula for this hit is:
h(x−) = 16+x2 8+3x 4 where h is the height of the ball and x is the number of seconds the ball is in the air.

I need help with part B

Mathematics
1 answer:
WARRIOR [948]1 year ago
4 0

Answer:

  A. Find h(10) = 0 . . . (it is not)

  B. h(5.235) = 0; hang time was about 5.235 seconds

Step-by-step explanation:

The hang time of the ball is the value of x in h(x) such that h(x) = 0.

__

<h3>A.</h3>

Juan can show the hang time is 10 seconds by finding h(10) = 0. The value of h(10) is ...

  h(10) = -16(10²) +83(10) +4 = -1600 +834 = -766 ≠ 0

Juan cannot prove the hang time was 10 seconds.

__

<h3>B.</h3>

The hang time is the solution to ...

  h(x) = 0

  -16x² +83x +4 = 0 . . . . . use the given expression for h(x)

  x² -5.1875x = 0.25 . . . . divide by -16

  x² -5.1875x +6.7275390625 = 6.9775390625 . . . . add (5.1875/2)²

  (x -2.59375)² = 6.9775390625 . . . . . . . . . . rewrite as a square

  x = 2.59375 ±√6.9775390625 ≈ {-0.048, 5.235} . . . . square root, find x

The hang time of the ball was about 5.235 seconds.

__

We like a graphing calculator for finding a quick solution to questions like this.

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<h2>length=26 m</h2>
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Read 2 more answers
Vectors: Find the vector from the point A=(−1,−7,3) to the point B=(3,−2,9).
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Answer:

\overrightarrow {AB} = (4,5,6)

1. 〈x,y,z〉=〈−1,−7,3〉+t〈−4,−5,−6〉 F

2. 〈x,y,z〉=〈−1,−7,3〉+t〈3,−2,9〉 F

3. 〈x,y,z〉=〈−1,−7,3〉+t〈4,5,6〉 T

4. 〈x,y,z〉=〈3,−2,9〉+t〈−1,−7,3〉 F

5. 〈x,y,z〉=〈3,−2,9〉+t〈4,5,6〉 F

Step-by-step explanation:

The vector AB is the vectorial difference between point A and B, that is:

\overrightarrow {AB} = \vec B - \vec A

Given that \vec A = (-1,-7,3) and \vec B = (3,-2,9), the vector AB is:

\overrightarrow {AB} = (3,-2,9)-(-1,-7,3)

\overrightarrow{AB} = (3-(-1),-2-(-7),9-3)

\overrightarrow {AB} = (4,5,6)

The vectorial equation of the line is represented by:

\langle x, y, z\rangle = \vec A + t \cdot \overrightarrow {AB}

Where t is the parametric variable, dimensionless. Given that \vec A = (-1,-7,3) and \overrightarrow {AB} = (4,5,6)

\langle x,y,z \rangle = \langle -1,-7,3 \rangle + t\cdot \langle 4,5,6 \rangle

Finally, the list of questions are now checked:

1. 〈x,y,z〉=〈−1,−7,3〉+t〈−4,−5,−6〉 F

2. 〈x,y,z〉=〈−1,−7,3〉+t〈3,−2,9〉 F

3. 〈x,y,z〉=〈−1,−7,3〉+t〈4,5,6〉 T

4. 〈x,y,z〉=〈3,−2,9〉+t〈−1,−7,3〉 F

5. 〈x,y,z〉=〈3,−2,9〉+t〈4,5,6〉 F

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Step-by-step explanation:

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