What is the slope of the line passing through (-3, 5) and (5, -3)?
2 answers:
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Answer:
Step-by-step explanation:
The scenario is represented in the attached photo. Triangle ABC is formed. AB represents her distance from her base camp. We would determine BC by applying the law of Cosines which is expressed as
a² = b² + c² - 2abCosA
Where a,b and c are the length of each side of the triangle and B is the angle corresponding to b. It becomes
AB² = AC² + BC² - 2(AC × BC)CosC
AB² = 42² + 28² - 2(42 × 28)Cos58
AB² = 1764 + 784 - 2(1176Cos58)
AB² = 2548 - 1246.37 = 1301.63
AB = √1301.63
AB = 36.08 km
To find the bearing, we would determine angle B by applying sine rule
AB/SinC = AC/SinB
36.08/Sin58 = 42/SinB
Cross multiplying, it becomes
36.08SinB = 42Sin58
SinB = 42Sin58/36.08 = 0.987
B = Sin^-1(0.987)
B = 81°
Therefore, her bearing from the base camp is
360 - 81 = 279°
Answer:
is outside the circle of radius of
centered at
.
Step-by-step explanation:
Let and
denote the center and the radius of this circle, respectively. Let
be a point in the plane.
Let denote the Euclidean distance between point
and point
.
In other words, if is at
while
is at
, then
.
Point would be inside this circle if
. (In other words, the distance between
and the center of this circle is smaller than the radius of this circle.)
Point would be on this circle if
. (In other words, the distance between
and the center of this circle is exactly equal to the radius of this circle.)
Point would be outside this circle if
. (In other words, the distance between
and the center of this circle exceeds the radius of this circle.)
Calculate the actual distance between and
:
.
On the other hand, notice that the radius of this circle, , is smaller than
. Therefore, point
would be outside this circle.