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Alenkasestr [34]
3 years ago
6

Meanwhile, at the ballpark Juan was practicing his hitting while talking to the girls standing around watching. He started telli

ng them how he hit the ball over the 50 foot light tower yesterday. The girls were unsure that he really did that. The formula for this hit is:
H(x)=-16x^2+60x+4 where h is the height of the ball and x is the number of seconds the ball is in the air. How high did Juan actually hit the ball?
Mathematics
1 answer:
Naddik [55]3 years ago
7 0

Answer:

5.2 seconds

Step-by-step explanation:

We want to determine the value of x when h(x) = 0

:

-16x^2 +83x + 4 = 0

:

use quadratic formula to determine the value for x

:

x = ( -83 + sqrt(83^2 - 4*(-16)*4) ) / (2*(-16) = −0.047753182

:

x = ( -83 - sqrt(83^2 - 4*(-16)*4) ) / (2*(-16) = 5.235253182

:

we want the positive value for x since time runs forward

:

the ball hung in the air for approx 5.2 seconds

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\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\
% left side templates
\begin{array}{llll}
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\\ \quad \\
y=&{{  A}}({{  B}}x+{{  C}})+{{  D}}
\\ \quad \\
f(x)=&{{  A}}\sqrt{{{  B}}x+{{  C}}}+{{  D}}
\\ \quad \\
f(x)=&{{  A}}(\mathbb{R})^{{{  B}}x+{{  C}}}+{{  D}}
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f(x)=&{{  A}} sin\left({{ B }}x+{{  C}}  \right)+{{  D}}
\end{array}\\\\
--------------------\\\\

\bf \bullet \textit{ stretches or shrinks horizontally by  } {{  A}}\cdot {{  B}}\\\\
\bullet \textit{ flips it upside-down if }{{  A}}\textit{ is negative}\\
\left. \qquad   \right.  \textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if }{{  B}}\textit{ is negative}\\
\left. \qquad   \right.  \textit{reflection over the y-axis}

\bf \bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\
\left. \qquad  \right. if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\\\
\left. \qquad  \right.  if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\\\
\bullet \textit{ vertical shift by }{{  D}}\\
\left. \qquad  \right. if\ {{  D}}\textit{ is negative, downwards}\\\\
\left. \qquad  \right. if\ {{  D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{  B}}}

with that template in mind, let's see

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a shift down by 3 units, means a vertical shift downwards, so D needs to drop by 3 units.

\bf h(x)=\stackrel{A}{-1}(\stackrel{B}{1}x\stackrel{C}{+0})\boxed{\stackrel{D}{+5-3}}\implies h(x)=-1(1x+0)+2
\\\\\\
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