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Troyanec [42]
2 years ago
5

The map shows the location J of Jose's house, the location of Tyrell's house, and the

Mathematics
1 answer:
sergij07 [2.7K]2 years ago
7 0

Using distance between two points, it is found that Jose traveled 25 units of length to football practice.

The distance between two points (x_1, y_1) and (x_2, y_2) is given by:

D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

First, Jose goes from his house, at (-6,8), to Tyrell's house, at (6,8), thus:

D_1 = \sqrt{(6 - (-6))^2 + (8 - 8)^2} = \sqrt{12^2} = 12

Then, he goes from Tyrell's house, at (6,8), to the football field, at (6, -5), thus:

D_2 = \sqrt{(6 - 6)^2 + (-5 - 8)^2} = \sqrt{13^2} = 13

The total distance is:

D = D_1 + D_2 = 12 + 13 = 25

Jose traveled 25 units of length to football practice.

A similar problem is given at brainly.com/question/22532602

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Assignment: \bold{Solve \ Equation: \ 3x^2+15x=18}

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Answer: \boxed{\bold{x=1,\:x=-6}}

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Explanation: \downarrow\downarrow\downarrow

<><><><><><><>

[ Step One ] Subtract 18 From Both Sides

\bold{3x^2+15x-18=18-18}

[ Step Two ] Simplify

\bold{3x^2+15x-18=0}

[ Step Three ] Solve With Quadratic Formula

Note: \bold{For\:a\:quadratic\:equation\:of\:the\:form\: ax^2+bx+c=0}

\bold{the \ solutions \ are \ x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}}

\bold{a=3,\:b=15,\:c=-18:\quad x_{1,\:2}=\frac{-15\pm \sqrt{15^2-4\cdot \:3\left(-18\right)}}{2\cdot \:3}}

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\bold{\frac{-15+\sqrt{15^2-4\cdot \:3\left(-18\right)}}{2\cdot \:3}: \ 1}

\bold{\frac{-15-\sqrt{15^2-4\cdot \:3\left(-18\right)}}{2\cdot \:3}: \ -6}

[ Step Four ] Combine Solutions

\bold{x=1,\:x=-6}

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\bold{\rightarrow Mordancy \leftarrow}

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