3y(2y+5) solve the problem
1 answer:
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This question is solved applying the formula of the area of the rectangle, and finding it's width. To do this, we solve a quadratic equation, and we get that the cardboard has a width of 1.5 feet.
Area of a rectangle:
The area of rectangle of length l and width w is given by:
w(2w + 3) = 9
From this, we get that:
Solving a quadratic equation:
Given a second order polynomial expressed by the following equation:
.
This polynomial has roots such that
, given by the following formulas:
In this question:
Thus a quadratic equation with
Then
Width is a positive measure, thus, the width of the cardboard is of 1.5 feet.
Another similar problem can be found at brainly.com/question/16995958
Answer:
(a) 7.5 seconds
(b) The horizontal distance the package travel during its descent is 1737.8 ft
Step-by-step explanation:
(a) The given function for the height of the object is s = -16·t² + v₀·t + s₀
The initial height of the object s₀ = 900 feet
The initial vertical velocity of the object v₀ = 0 m/s
The time it takes the package to strike the ground is found as follows;
0 = -16·t² + 0×t + 900
900 = 16·t²
t² = 900/16 = 62.25
t = √62.25 = 7.5 seconds
(b) Given that the horizontal velocity of the package is given as 158 miles/hour, we have
158 miles/hour = 231.7126 ft/s
The horizontal distance the package covers in the 7.5 second of vertical flight = 231.7126 ft/s × 7.5 s = 1737.8445 feet = 1737.8 ft to one decimal place
The horizontal distance the package travel during its descent = 1737.8 ft.