In the figure, lines a and b are parallel.
1 answer:
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The maximum height of the projectile will be 226 feet that is, as per question, the option c.
A quadratic equation , y = ax²+bx+c (equation 1)
given, axis of symmetry is
x =
As per the question:
The path of the projectile is modeled using the equation :
h(t) = -16t²+48t+190 (equation 2)
h(t) = height after t time.
comparing equation 1 and equation 2, we get
a = -16 and b = 48
further, t =
=
= 1.5sec
Substituting the found value in equation 2, we get
h(1.5) = -16(1.5)²+48(1.5)+190
h(1.5) = -36+72+190
h(1.5) = 226ft.
Thus, the maximum height of the projectile at 1.5 sec is, 226 feet.
Note that the full question is:
A projectile with an initial velocity of 48 feet per second is launched from a building 190 feet tall. The path of the projectile is modeled using the equation h(t) = –16t2 + 48t + 190.
What is the maximum height of the projectile?
A. 82 feet
B. 190 feet
C. 226 feet
D. 250 feet
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Answer: They can order from 0 to 449 bars, but no more.
Step-by-step explanation:
Given: The school band is ordering health bars to sell for a fundraiser. The company that sells the bars charges $0.40 per bar plus $20.00 for shipping regardless of the size of the order. The band must spend less than $200.00 on bars for the fundraiser.
The inequality below relates x, the number of bars that could be ordered, with the shipping costs and their spending requirements.
To solve for x, subtract 20 on both sides, we get
Divide 0.40 on both sides, we get
......... → the number of bars that could be ordered, with the shipping costs and their spending requirements must less than 450.
Thus, they can order from 0 to 449 bars, but no more.