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jenyasd209 [6]
1 year ago
15

What is the solution set of x2 + 5x- 5 = 0? oss-355 S +385 o sus S* o {stay -S-345 , 30 O S=5+375 5+3/5 2 0 --375 -5 +385 2

Mathematics
1 answer:
jolli1 [7]1 year ago
6 0

Given the quadratic equation:

x^2+5x-5=0

Using the general solution for quadratic equations:

ax^2+bx+c=0\Rightarrow x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

From the problem, we identify:

\begin{gathered} a=1 \\ b=5 \\ c=-5 \end{gathered}

Then:

x=\frac{-5\pm\sqrt{25+20}}{2}=\frac{-5\pm3\sqrt{5}}{2}

And the solution set is:

\lbrack\frac{-5-3\sqrt{5}}{2},\frac{-5+3\sqrt{5}}{2}]

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Answer:

Kennan will be from home approximately an hour and 48 minutes.

Step-by-step explanation:

We must know that total time (t_{T}) that Keenan will be from home is the sum of run (t_{R}), hang out (t_{H}) and walk times (t_{W}), measured in hours:

t_{T} = t_{R}+t_{H}+t_{W}

If Keenan runs and walks at constant speed, then equation above can be expanded:

t_{T} = \frac{x_{R}}{v_{R}}+t_{H}+ \frac{x_{W}}{v_{W}}

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v_{R}, v_{W} - Run and walk speeds, measured in miles per hour.

Given that x_{R}=x_{W} = 2.5\,mi, v_{R} = 6\,\frac{mi}{h}, v_{W} = 4\,\frac{mi}{h} and t_{H} = 0.75\,h, the total time is:

t_{T} = \frac{2.5\,mi}{6\,\frac{mi}{h} } + 0.75\,h+\frac{2.5\,mi}{4\,\frac{mi}{h} }

t_{T} = 1.792\,h (1\,h\,48\,m)

Kennan will be from home approximately an hour and 48 minutes.

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