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Yuki888 [10]
3 years ago
9

Determine what type of model best fits the given situation:

Mathematics
1 answer:
kondaur [170]3 years ago
8 0

Answer:

Ok, i will suppose the situation that:

The rocket has a constant speed S (So the acceleration of the rocket is equal in magnitude, but opposite in direction, to the gravitational acceleration)

Here we can remember that:

Velocity = distance/time.

Then if the distance is the height of the rocket, we can write this as:

H = velocity*time

h = S*t

This is a linear model that represents the height of the rocket as a function of time.

Case 2:

The rocket is fired with an initial velocity v0, but no acceleration:

In this case the only acceleration acting on the rocket is the gravitatonal acceleration pulling the rocket down, so the acceleration is:

a = -g

To get the velocity as a function of time, we should integrate:

a = -g*t + v0

To get the height as a function of time, we integrate again:

h(t) = (-g/2)*t^2 + v0*t + p0

Where p0 is the initial position of te rocket, but the rocket starts at the ground, so p0 = 0m.

The height as a function of time is:

h(t) = (-g/2)*t^2 + v0*t

This is a quadratic equation.

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7. Use the quadratic formula to find the solution(s). x² + 2x - 8 = 0​
morpeh [17]

Hey there!

<u>Use the quadratic formula to find the solution(s). x² + 2x - 8 = 0</u>

  • Answer :

x = -4 or x = 2 ✅

  • Explanation :

<em><u>Quadratic</u></em><em><u> </u></em><em><u>formula </u></em><em><u>:</u></em><em><u> </u></em>ax² + bx + c = 0 where a ≠ 0

The number of real-number solutions <em>(roots)</em> is determined by the discriminant (b² - 4ac) :

  • If b² - 4ac > 0 , There are 2 real-number solutions

  • If b² - 4ac = 0 , There is 1 real-number solution.

  • If b² - 4ac < 0 , There is no real-number solution.

The <em><u>roots</u></em> of the equation are determined by the following calculation:

x =  \frac{ - b \pm  \sqrt{ {b}^{2} - 4ac } }{2a}

Here, we have :

  • a = 1
  • b = 2
  • c = -8

1) <u>Calculate </u><u>the </u><u>discrim</u><u>i</u><u>n</u><u>ant</u><u> </u><u>:</u>

b² - 4ac ⇔ 2² - 4(1)(-8) ⇔ 4 - (-32) ⇔ 36

b² - 4ac = 36 > 0 ; The equation admits two real-number solutions

2) <u>Calculate </u><u>the </u><u>roots </u><u>of </u><u>the </u><u>equation</u><u>:</u>

▪️ (1)

x_1 =  \frac{ - b -  \sqrt{ {b}^{2}  - 4ac} }{2a}  \\  \\ x_1 =  \frac{ - 2 -  \sqrt{36} }{2(1) }  \\  \\ x_1 =  \frac{ - 2 - 6}{2}   \\ \\ x_1 =  \frac{ - 8}{2}  \\  \\ \blue{\boxed{\red{x_1 = -4}}}

▪️ (2)

x_2 =  \frac{ - b  +   \sqrt{ {b}^{2} - 4ac } }{2a}  \\  \\ x_2 =  \frac{ - 2 +  \sqrt{36} }{2(1)}  \\  \\ x_2 =  \frac{ - 2 + 6}{2}  \\  \\ x_2 =  \frac{4}{2}  \\  \\ \red{\boxed{\blue{x_2 = 2}}}

>> Therefore, your answers are x = -4 or x = 2.

Learn more about <u>quadratic equations</u>:

brainly.com/question/27638369

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y^1/2/xz^2 option 3

Step-by-step explanation:

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The answer would be 9
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