Answer:
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Step-by-step explanation:y
Answer:
The horizontal displacement is 11 units, the vertical displacement is 9 units, and the projection angle is 39.3 degrees.
Step-by-step explanation:
We can start using the definition of displacement in one dimension between any 2 points which is the difference between them, so we have

And apply it to get the horizontal and vertical displacements.
Once we have found them, we can use trigonometric functions to find the projection angle with respect the horizontal.
Linear displacements.
Using the definition of displacement, we can write the horizontal displacement as

So we can use the given points
on the displacement formula

In the same manner we can look at the y components of those points to find the vertical displacement

Thus the horizontal displacement is 11 units and the vertical displacement is 9 units.
Projection angle.
The projection angle with respect the horizontal is the angle that is made between the line that connects the points P1 and P2 and the horizontal, so we can use the linear displacements previously found to write

Solving for the angle we get

Replacing values

Which give us

So the projection angle is 39.3 degrees.
Answer:
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Step-by-step explanation:
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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.
Answer:
1222.2
explanation:
53.95 x 12 = 647.40
17.95 x 12 = 215.40
29.95 x 12 = 359.40
647.40 + 359.40 + 215.40 = 1222.2