Solve this A tourist starts off from town A and travels for 50km on a bearing of N80°W to town B. At town B ,he continues for another 40kmbon a bearing of N20°E toC. How far is C from A?

The point C is 12.68 km away from the point A on a bearing of S23.23°W.

Step-by-step explanation:

Given that AB is 50 km and BC is 40 km as shown in the figure.

From the figure, the length of x-component of AC = |AB sin 80° - BC cos 20°|

=|50 sin 80° - 40 cos 20°|=11.65 km

The length of y-component of AC = |AB cos 80° - BC sin 20°|

=|50 cos 80° - 40 sin 20°|= 5 km

tan= 5/11.65

=23.23°

AC= km

Hence, the point C is 12.68 km away from the point A on a bearing of S23.23°W.

the formula of a slope:

from a table we have

substitute:

the slope of f(x) is -1.

the graph of funcfion g(x) is a horizontal line. the slope of a horizontal line is equal 0. therefore the slope of g(x) is 0.

-1 < 0

a reflection across the x-axis , followed by the by a reflection across the y-axis , and then a scale dilation of 0.5 .