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Licemer1 [7]
11 months ago
14

What are the solutions to this quadratic equation x2+6x-5=0

Mathematics
2 answers:
marishachu [46]11 months ago
8 0

Answer:

See Below.

Explanation:

Quadratic Form:

a {x}^{2}  + bx + c

Quadratic Formula:

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

Based on the information given from the question, we can deduce:

a = 1

b = 6

c = -5

Now we can substitute all these values into the formula to find x.

x =   \frac{ - 6 +  -  \sqrt{ {6}^{2}  - 4(1)( - 5)} }{2(1)}  \\  =   - 3 +  \sqrt{14}   \: or \:  - 3 -  \sqrt{14}

ale4655 [162]11 months ago
6 0
Thus, the solutions of the quadratic equation x 2 + 5x + 6 = 0 are x = -2 and x = -3.
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The number of minutes advertisement should use is found.

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(MISSING PART OF THE QUESTION: AVERAGE WAITING TIME = 2.5 MINUTES)

<h3 /><h3>Step 1</h3>

For such problems, we can use probability density function, in which probability is found out by taking integral of a function across an interval.

Probability Density Function is given by:

f(t)=\left \{ {{0 ,\-t

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f(t)=\frac{e^{-t/\mu}}{\mu}\\

Where Average waiting time = μ = 2.5

The function f(t) becomes

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<h3>Step 2</h3>

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The probability that a costumer has to wait for more than x minutes is:

\int\limits^\infty_x {f(t)} \, dt= \int\limits^\infty_x {}0.4e^{-0.4t}dt

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<h3>Step 3</h3>

Solve the equation for x

\int\limits^{\infty}_x {0.4e^{-0.4t}} \, dt =0.01\\\\\frac{0.4e^{-0.4t}}{-0.4}=0.01\\\\-e^{-0.4t} |^\infty_x =0.01\\\\e^{-0.4x}=0.01

Take natural log on both sides

ln (e^{-0.4x})=ln(0.01)\\-0.4x=ln(0.01)\\-0.4x=-4.61\\x= 11.53

<h3>Results</h3>

The costumer has to wait x = 11.53 mins ≅ 12 mins to get a free hamburger

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