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OLEGan [10]
2 years ago
10

Solve the system of equations.

D20x-400%7D%7D%20%5Cright.%20%5C%5C" id="TexFormula1" title="\left \{ {{y=220x-160} \atop {y=20x-400}} \right. \\" alt="\left \{ {{y=220x-160} \atop {y=20x-400}} \right. \\" align="absmiddle" class="latex-formula">

Mathematics
1 answer:
Leni [432]2 years ago
4 0

\begin{cases} y=220x-160\\ y=20x^2-400 \end{cases}\qquad \qquad \stackrel{\textit{substituting on the 2nd equation}}{\stackrel{\textit{\Large y}}{220x-160}~~ = ~~20x^2-400} \\\\\\ -160=20x^2-220x-400\implies 0=20x^2-220x-240 \\\\\\ 0=20(x^2-11x-12)\implies 0=x^2-11x-12 \\\\\\ 0=(x-12)(x+1)\implies x= \begin{cases} 12\\ -1 \end{cases}

\dotfill\\\\ x=12\qquad \qquad \stackrel{\textit{substituting on the 1st equation}}{y=220(12)-160}\implies y=2480 \\\\\\ x=-1\qquad \qquad \stackrel{\textit{substituting on the 1st equation}}{y=220(-1)-160}\implies y=-380 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill (12~~,~~2480)\qquad (-1~~,~~-380)~\hfill

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d = r * t

t = d/r

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240 = 12(r_boat + current)

240/ (r_boat - current) = 24       Multiply both sides by r_boat - r_current

240 = 24*(r_boat - current)    

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r_boat + current  = 2 (r_boat - current)             Remove the brackets.

r_boat  + current = 2*r_boat - 2* current          add 2 currents to both sides

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3 currents = r_boat.

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5 = current

The rate of the current = 5 miles / hour

3 currents = r_boat

3*5 = rate_boat

15 = rate of the boat

8 0
3 years ago
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