By applying the <em>quadratic</em> formula and discriminant of the <em>quadratic</em> formula, we find that the <em>maximum</em> height of the ball is equal to 75.926 meters.
<h3>How to determine the maximum height of the ball</h3>
Herein we have a <em>quadratic</em> equation that models the height of a ball in time and the <em>maximum</em> height represents the vertex of the parabola, hence we must use the <em>quadratic</em> formula for the following expression:
- 4.8 · t² + 19.9 · t + (55.3 - h) = 0
The height of the ball is a maximum when the discriminant is equal to zero:
19.9² - 4 · (- 4.8) · (55.3 - h) = 0
396.01 + 19.2 · (55.3 - h) = 0
19.2 · (55.3 - h) = -396.01
55.3 - h = -20.626
h = 55.3 + 20.626
h = 75.926 m
By applying the <em>quadratic</em> formula and discriminant of the <em>quadratic</em> formula, we find that the <em>maximum</em> height of the ball is equal to 75.926 meters.
To learn more on quadratic equations: brainly.com/question/17177510
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Answer:
Graph
Step-by-step explanation:
Answer:
The answer is "
".
Step-by-step explanation:
![\bold{\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]}](https://tex.z-dn.net/?f=%5Cbold%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%262%5C%5C3%264%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%265%5C%5C%2019%268%5Cend%7Barray%7D%5Cright%5D%7D)
Solve the L.H.S part:
![\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right]\\\\\\\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%262%5C%5C3%264%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%2B2c%26b%2B2d%5C%5C3a%2B4c%263b%2B4d%5Cend%7Barray%7D%5Cright%5D)
After calculating the L.H.S part compare the value with R.H.S:
![\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]= \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]} \\\\](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%2B2c%26b%2B2d%5C%5C3a%2B4c%263b%2B4d%5Cend%7Barray%7D%5Cright%5D%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%265%5C%5C%2019%268%5Cend%7Barray%7D%5Cright%5D%7D%20%5C%5C%5C%5C)

In equation (i) multiply by 3 and subtract by equation (iii):

put the value of c in equation (i):

In equation (ii) multiply by 3 then subtract by equation (iv):

put the value of d in equation (iv):

The final answer is "
".
Answer:
$0.31
Step-by-step explanation:
1.55/5=0.31
So if each ear of corn cost $0.31 than 5 times .31 should equal $1.55
5x.31=1.55
Yes!