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avanturin [10]
3 years ago
9

Directions:Find the solution of the follow equations and check your work

Mathematics
1 answer:
Mekhanik [1.2K]3 years ago
7 0

y=28×9

=252

x=334-150

=184

h=254×100

=25400

Step-by-step explanation:

to check the equations substitute the values by the variables

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Step-by-step explanation:

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<em>Now let's take the area of the square which is 3ft long and 3ft wide</em>

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A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by th
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Answer:

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Step-by-step explanation:

The height of the rocket, after x seconds, is given by the following equation:

y = -16x^2 + 177x + 98

It hits the ground when y = 0, so we have to find x for which y = 0, which is a quadratic equation.

Finding the roots of a quadratic equation:

Given a second order polynomial expressed by the following equation:

ax^{2} + bx + c, a\neq0.

This polynomial has roots x_{1}, x_{2} such that ax^{2} + bx + c = a(x - x_{1})*(x - x_{2}), given by the following formulas:

x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}

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In this question:

y = -16x^2 + 177x + 98

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x_{2} = \frac{-177 - \sqrt{37601}}{2*(-16)} = 11.59

Since time is a positive measure, the rocket hits the ground at a time of 11.59 seconds.

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