Given that,
The camera sights the stadium at a 7 degree angle of depression.
The altitude of the blimp i slide 300 m.
To find,
The line of sight distance from the camera to the stadium.
Solution,
If we consider a right angled triangle. Let x is its hypotenuse i.e. the line of sight distance from the camera to the stadium. Using trigonometry :

So, the line of sight is at a distance of 2461.65 m from the camera to the stadium.
Answer:
4
Step-by-step explanation:
What we have here in the picture is a graph of a segment. The goal is to find the segment's length, which we can do by using the distance formula:

Note that the distance between two points is equal to the length of the line segment that joins them.
So now we already have answers for two of the boxes.
The difference in the x-coordinates gives us the horizontal distance between the two points, while the difference in the y-coordinates gives us the vertical distance between the points.
So, the other two boxes should contain:
Step-by-step explanation:
the introduction of a fraction tells us that we are dealing with multiplications, and therefore a geometric sequence (where every new term is created by multiplying the previous term by a constant factor, the ratio r).
I think your teacher made a mistake, or you made one when typing the question in here.
there is no factor r that creates
15×r = 9
and
9×r = 5/27
it would mean that
15 × r² = 5/27
r² = 5/27 / 15 = 5/27 × 1/15 = 5/405 = 1/81
r = 1/9
but 15 × 1/9 = 5 × 1/3 = 5/3 is NOT 9
and 9 × 1/9 = 9/9 = 1 is NOT 5/27
so, this can't be right.
on the other hand
15 × r = 9
r = 9/15 = 3/5
and then
9 × 3/5 = 27/5
so, either the sequence should have been
15, 5/3, 5/27
or (and I suspect this to be true)
15, 9, 27/5
under that assumption we have
s1 = 15
r = 3/5
sn = sn-1 × r = s1 × r^(n-1) = 15 × (3/5)^(n-1)
s10 = 15 × (3/5)⁹ = 15 × 19683/1953125 =
= 3 × 19683/390625 = 59049/390625 =
= 0.15116544 ≈ 0.151