Answer: The missing length is 40 m.
Step-by-step explanation:
30^2 + b^2 = 50^2
900 + b^2 = 2500
-900 -900
b^2 = 1600
b= 40
AA Theorem: To show two triangles are similar, it is sufficient to show that two angles of one triangle are congruent (equal) to two angles of the other triangle. All three triangles have a right angle and they all share the common angle at point S.
Flagpole is 32 meters from point S: so 1.5/8 = x/32 where x is FG
cross-multiply: 32 • 1.5 = 8x
48 = 8x
x = 6 meters
building is a total of 240 meters from point S, so: 1.5/8 = x / 240 where x is BD
cross-multiply: 240 • 1.5 = 8x
360 = 8x
<span>x = 45 meters</span>
Answer:
a
Step-by-step explanation:
Answer:
see below
Step-by-step explanation:
The recursive function tells you the first term is 1. That is the meaning of ...

It also tells you that to find the next term, you multiply this term by 1/2:

And, to find the next term, you multiply this term by 1/2:

Then the first three terms are ...
1, 1/2, 1/4
The slope of the tangent line to the curve at (8, 2) is given by the derivative
at that point. By the chain rule,

Differentiate the given parametric equations with respect to
:


Then

We have
and
when
, so the slope at the given point is
.
The normal line to the same point is perpendicular to the tangent line, so its slope is +4. Then using the point-slope formula for a line, the normal line has equation

Alternatively, we can eliminate the parameter and express
explicitly in terms of
:

Then the slope of the tangent line is

At
, the slope is again
, so the normal has slope +4, and so on.