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3 years ago

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Mathematics

1 answer:

Softa [21]3 years ago

8 0

$180^o-112^o=68^o$

$\left\{\begin{array}{ccc}2x+3y=68&|\cdot(-15)\\30x+11y=112\end{array}\right\\\underline{+\left\{\begin{array}{ccc}-30x-45y=-1020\\30x+11y=112\end{array}\right}\\.\ \ \ \ \ \ \ -34y=-908\ \ \ |:(-34)\\\\.\ \ \ \ \ \ \ \ \ \ \ y=\dfrac{908}{34}\to y=\dfrac{454}{17}\\\\2x+3\cdot\dfrac{454}{17}=68\ \ \ \ |\cdot17\\\\34x+1362=1156\ \ \ \ |-1362\\34x=-206\ \ \ \ |:34\\\\x=-\dfrac{206}{34}\to x=-\dfrac{103}{17}\\\\ \left\{\begin{array}{ccc}x=-\dfrac{103}{17}\\\\y=\dfrac{454}{17}\end{array}\right$

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<u><em>The slope of a line that includes the points (4, -2) and (5, 0) is 2</em></u>

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