HELP ASAP!!! Characterize the slope of the line in the graph. A. Zero B. Undefined C. Positive D. Negative

Mathematics

1 answer:

Alexxandr [17]3 years ago

5 0

Answer:

Undefined

Explanation:

When a line is vertical like that, the slope is undefined. If it was horizontal, then it would be zero. Positive slopes go in an upward right direction. Negative slopes go in a downward left direction.

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What is the perimeter of a triangle with vertices of (5,1) (-3,3) (-7,-3)

timurjin [86]

$Vertices:\\
A(5,1)\\B(-3,3)\\C(-7,-3)\\\\ You\ must\ find\ length\ of\ AB,\ BC,\ AC.\\\\
A(5,1)\ \ \ \ \ \ B(-3,3)\\\\Distance\ between\ A \ and\ B:\\AB=\sqrt{(x_B-x_A)^2+(y_A-y_B)^2}\\\\AB=\sqrt{(-3-5)^2+(3-1)^2}\\AB=\sqrt{8^2+2^2}\\AB=\sqrt{64+4}\\AB=\sqrt{68}
$ $
 B(-3,3)\ \ \ \ C(-7,-3)\\\\Distance\ between\ B \ and\ C:\\BC=\sqrt{(x_C-x_B)^2+(y_C-y_B)^2}\\\\BC=\sqrt{(-7-(-3))^2+(-3-3)^2}\\BC=\sqrt{(-4)^2+(-6)^2}\\BC=\sqrt{16+36}\\BC=\sqrt{52}$ $A(5,1)\ \ \ \ \ \ C(-7,-3)\\\\Distance\ between\ A \ and\ C:\\AC=\sqrt{(x_C-x_A)^2+(y_C-y_A)^2}\\\\AC=\sqrt{(-7-5)^2+(-3-1)^2}\\AC=\sqrt{(-12)^2+(-4)^2}\\AC=\sqrt{144+16}\\AC=\sqrt{160}$ $Perimeter\ of\ triangle=\ AB+AC+BC=\\\sqrt{68}+\sqrt{52}+\sqrt{160}=2\sqrt{17}+2\sqrt{13}+2\sqrt{40}$