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Aleonysh [2.5K]
3 years ago
12

Se lanza un objeto desde una plataforma.

Mathematics
1 answer:
jasenka [17]3 years ago
7 0

Answer:

6 seconds

Step-by-step explanation:

<u><em>The question in English is</em></u>

An object is thrown from a platform.

Its height (in meters), x seconds after launch, is modeled by:

h(x)=-5x^2+20x+60

How many seconds after the launch does the object reach the ground?

Let

x ----> the time in seconds

h(x) ---> the height of the object

we have

h(x)=-5x^2+20x+60

we know that

When the object hit the ground the height is equal to zero

so

For h(x)=0

we have

-5x^2+20x+60=60

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

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

in this problem we have

-5x^2+20x+60=60

so

a=-5\\b=20\\c=60

substitute in the formula

x=\frac{-20\pm\sqrt{20^{2}-4(-5)(60)}} {2(-5)}

x=\frac{-20\pm\sqrt{1,600}} {-10}

x=\frac{-20\pm40} {-10}

x=\frac{-20+40} {-10}=-2

x=\frac{-20-40} {-10}=6

The solution is x=6 sec

The he object reach the ground at x=6 seconds

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3 years ago
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Answer:

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3 0
3 years ago
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What value of x is in the solution set of 2(4+2x)&gt;5x+5
Serggg [28]

Answer:

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Step-by-step explanation:

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deff fn [24]
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