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kakasveta [241]
1 year ago
13

A rocket is launched from the ground at an initial velocity of 39.2 meters per second. Which equation can be used to model the h

eight of the rocket after t seconds?
s(t) = –4.9t2 + 39.2
s(t) = –4.9t2 + 39.2t
s(t) = –4.9t2 + 39.2t + 39.2
s(t) = –4.9t2 + 39.2t – 39.2
Mathematics
1 answer:
Naya [18.7K]1 year ago
8 0

The quadratic equation that models the height of the rocket after t seconds is:

s(t) = -4.9t² + 39.2t.

<h3>What is the quadratic function for a projectile's height?</h3>

Considering the height in meters, the equation is given by:

s(t) = -4.9t² + v(0)t + h(0)

In which:

  • v(0) is the initial velocity, in m/s.
  • h(0) is the initial height, in m.

In this problem, we have that v(0) = 39.2, h(0) = 0, hence the equation is:

s(t) = -4.9t² + 39.2t.

More can be learned about quadratic functions at brainly.com/question/24737967

#SPJ1

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Remark

If the lines are parallel, there are no solutions to the system of equations. Start with the equation you know the most about.

x + 6y = 7                Subtract x from both sides

x - x + 6y = 7 - x      Combine

6y =  - x + 7             Switch and divide by 6

y = -x / 6 + 7/6

The general equation for a line is y = mx + b where m is the slope of the line.

m = - 1/6

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Now look at the second equation

10ay - 5x = 32               Add 5x to both sides

10ay = 5x + 32              Divide by 10a

y = (5/10a)x + 32/(10a)  

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Now you must make

5/10a = - 1/6                     Cross Multiply

5* 6 = - 10a * 1

30 = - 10a                        Divide by - 10

a = 30 / - 10

a = - 3

So these two equations will have no solution when a = - 3



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3 years ago
What is 14.9 divided by 0.02
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for a given speed, the distance traveled varies directly with the time. Kate's school is 5 miles away from her home and it takes
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Answer: It will take Josh 4 minutes to get to school.

Step-by-step explanation:

  1. Divide the distance by the time.
  2. So, 10 minutes/ 5 miles
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  5. Multiply 2 miles by 2 minutes
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A certain bridge arch is in the shape of half an ellipse 106 feet wide and 33.9 feet high. At what horizontal distance from the
nata0808 [166]

Answer:

The horizontal distance from the center is 49.3883 feet

Step-by-step explanation:

The equation of an ellipse is equal to:

\frac{x^2}{a^{2} } +\frac{y^2}{b^{2} } =1

Where a is the half of the wide, b is the high of the ellipse, x is the horizontal distance from the center and y is the height of the ellipse at that distance.

Then, replacing a by 106/2 and b by 33.9, we get:

\frac{x^2}{53^{2} } +\frac{y^2}{33.9^{2} } =1\\\frac{x^2}{2809} +\frac{y^2}{1149.21} =1

Therefore, the horizontal distances from the center of the arch where the height is equal to 12.3 feet is calculated replacing y by 12.3 and solving for x as:

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So, the horizontal distance from the center is 49.3883 feet

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