Answer:
12. 1 second, 35 ft; 2 seconds, 32 ft
13. (t, y) = (1.4 seconds, 37.6 feet)
14. 37.6 ft; the vertex is the highest point
Step-by-step explanation:
12. You have properly answered question 12.
After 1 second, the height is 35 feet; after 2 seconds, it is 32 feet.
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13. My method of choice is to plot the graph on a graphing calculator and let it show me the coordinates of the vertex when I highlight that point. (See attached.) The vertex is ...
(t, y) = (1.4, 37.6)
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14. The graph is a graph of height when the object is launched with a vertical velocity of 45 ft/s. So, the maximum of the graph will correspond to the maximum height of the object. The vertex is that maximum point, and its y-coordinate is that maximum height.
The maximum height is 37.6 feet.
Can u explain it a little more clearly...
Answer:
Hi, please provide a photo or options to pick from. I will answer then.
Step-by-step explanation:
The answer is A. for this, you have to set up a system of equations. the first one will be the area equation. since you know area=length x width, your equation will be LxW=50. the next equation is L=2W, since the length is two times the width. then, plug in 2W for the L in the other equation and you get 2W^2=50. divide by 2 and get W^2=25. square root both sides and you get W=5. plug back into the other equation to find L=10. Then, add the sides of the rectangle for the perimeter. 10+5+10+5=30.
Let h represent the height of the trapezoid, the perpendicular distance between AB and DC. Then the area of the trapezoid is
Area = (1/2)(AB + DC)·h
We are given a relationship between AB and DC, so we can write
Area = (1/2)(AB + AB/4)·h = (5/8)AB·h
The given dimensions let us determine the area of ∆BCE to be
Area ∆BCE = (1/2)(5 cm)(12 cm) = 30 cm²
The total area of the trapezoid is also the sum of the areas ...
Area = Area ∆BCE + Area ∆ABE + Area ∆DCE
Since AE = 1/3(AD), the perpendicular distance from E to AB will be h/3. The areas of the two smaller triangles can be computed as
Area ∆ABE = (1/2)(AB)·h/3 = (1/6)AB·h
Area ∆DCE = (1/2)(DC)·(2/3)h = (1/2)(AB/4)·(2/3)h = (1/12)AB·h
Putting all of the above into the equation for the total area of the trapezoid, we have
Area = (5/8)AB·h = 30 cm² + (1/6)AB·h + (1/12)AB·h
(5/8 -1/6 -1/12)AB·h = 30 cm²
AB·h = (30 cm²)/(3/8) = 80 cm²
Then the area of the trapezoid is
Area = (5/8)AB·h = (5/8)·80 cm² = 50 cm²
