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OleMash [197]
3 years ago
11

1. Suppose that you are watching the tree warden trim branches from a large tree in your yard. He has climbed up 15 meters and h

is assistant is holding a rope that will be used to guide the branch when it falls. To the nearest meter, how long is the rope?
2. For XYZ, what is the measure of the smallest angle to the nearest tenth of a degree? (ZY = 8) (YX = 3.5) (ZX = x )
Mathematics
1 answer:
RSB [31]3 years ago
7 0
In item# 1, we are only given with the height of the tree a=15 meters and we are asked to solve for the hypotenuse. Since we are not given with angle created between the rope the ground level we will assign it as ∠A = ∅
Solving for the length of the rope, we have
sin A = 15/Length of rope
Length of rope = 15 meters / sin A

In item#2, assuming that it is a right triangle we have:
8 ²= x²+3.5²
x=7.2

The smallest angle:
sin Z =3.5/8
Z = 25.95°
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The order pair below is a solution to which equation (24,6)
Illusion [34]
It would be much easier for me to help you pick the correct equation
if you would let me see the list of choices.

Since I can't see any of the choices, I'll just make up a few equations
that have the ordered pair  (24, 6)  as a solution:

         Y  =  0.25 x

         Y  =  x  -  18

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7 0
3 years ago
The numbers of teams remaining in each round of a single-elimination tennis tournament represent a geometric sequence where an i
Anit [1.1K]

Answer:

a_n = 128\bigg(\dfrac{1}{2}\bigg)^{n-1}

Step-by-step explanation:

We are given the following in the question:

The numbers of teams remaining in each round follows a geometric sequence.

Let a be the first the of the geometric sequence and r be the common ration.

The n^{th} term of geometric sequence is given by:

a_n = ar^{n-1}

a_4 = 16 = ar^3\\a_6 = 4 = ar^5

Dividing the two equations, we get,

\dfrac{16}{4} = \dfrac{ar^3}{ar^5}\\\\4}=\dfrac{1}{r^2}\\\\\Rightarrow r^2 = \dfrac{1}{4}\\\Rightarrow r = \dfrac{1}{2}

the first term can be calculated as:

16=a(\dfrac{1}{2})^3\\\\a = 16\times 6\\a = 128

Thus, the required geometric sequence is

a_n = 128\bigg(\dfrac{1}{2}\bigg)^{n-1}

4 0
3 years ago
Marisol has 5 times as many books as jerry. Jerry has 15 books. How many more books does marisol have
ICE Princess25 [194]

Answer:

5 times more books

Step-by-step explanation:

In total Marisol has 75 books.  She has 5 times as many books and Jerry has 15.  5*15=75

6 0
2 years ago
Can someone explain to me why are we adding 2kpi when we are doing zeros for sin and cos, but adding kpi when doing zeros for tg
makvit [3.9K]

on the first exercise, you got a solution angle of π/18, that's a good solution for the I Quadrant only, however, on a circle, we have angles that go from 0 to 2π, however we can always keep on going around and continute to 2π + π/2 or 3π or 4π, or 115π/3 or 1,000,000π/18 and so on, and we're really just going around the circle many times over, getting a larger and larger angle, same circular motion.

π/18 on that exercise works for the I Quadrant, however if we continue and go around say 2π, we'll find that 2π/3 + π/18 is a coterminal angle with π/18, and thus that angle has also the same sine value.

π/18 + 2kπ/3 , where k = integer, is a way to say, all angles around the circle that look like this have the same sine, namely

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so using the "k" as some sequence multiplier, is a generic notational way to say, "all these angles".

you'll also find that "n" is used as well for the same notation, say for example

2π/3  + 2πn.

8 0
3 years ago
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