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GaryK [48]
4 years ago
10

A model rocket is launched from a roof into a large field. The path of the rocket can be modeled by the equation y=-0.8x^2 + 12x

+25.8 where x is the horizontal distance, in meters, from the starting point on the roof and y is the height, in meters, of the rocket above the ground. How far horizontally from its starting point will the rocket land?
a. 25.80 m
b. 37.00 m
c. 17.24 m
d. 16.91 m
Mathematics
1 answer:
svet-max [94.6K]4 years ago
8 0
The rocket will land about 16.85 meters horizontally from the starting point. I would select choice D. I think you have a typo in your problem.

To find this answer, we need to find the point on the graph where this parabola hits the x-axis (or ground). 

Using the quadratic equation, we will get two answer about -1.85 and 16.85. The first value doesn't apply in the context of this story. The second answer would be the distance.
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Add the square of half the x-coefficient to both sides.

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a garden is 15 m × 10 m . It has a 2.6. m wide flower bed all around outside it. There is also a 2.5 m wide grassy path all arou
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Answer:

flower bed = 157.04 m^{2}

grassy path = 202 m^{2}

Step-by-step explanation:

If you draw this out, it's basically a rectangle inside a rectangle inside a rectangle. I'm going to call those rectangles A, B, and C. with A being the smallest and C being the biggest.

first we need to find the area of A, B, and C.

A will be easy because we're already given that its 15x10

Area A = 150 m^{2}

To find area B, we need to add the 2.6 pathway to all the sides, so it'll be:

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Area B = 307.04 m^{2}

Then we do the same for area C but with 2.5 m.

length: 20.2+2.5(2) = 25.2 m

width: 15.2+2.5(2) = 20.2 m

Area C = 509.04 m^{2}

To find the area of just the flower bed, we need to subtract area A from area B, then to find grassy path we need to subtract area B from area C.

flower bed: 307.04-150 = 157.04 m^{2}

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Answer:

\boxed{ \bold{ \huge{ \boxed{ \sf{128°}}}}}

Step-by-step explanation:

Sum of angles of triangle add up to 180°

Let the third angle be x

Create an equation and solve for x

\sf{x + 12° + 40° = 180°}

\dashrightarrow{ \sf{x + 52° = 180°}}

\dashrightarrow{ \sf{x  = 180° - 52°}}

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