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Pani-rosa [81]
3 years ago
11

Susan and Mark are standing at different places on a beach and watching a bird. The angles of elevation they make are 20º and 50

º, respectively.
If Susan and Mark are 7 kilometers apart and the bird is between them, the bird is at a height of kilometers [Blank] from the ground.
Mathematics
2 answers:
sukhopar [10]3 years ago
8 0
The picture is in the attachment.
x- horizontal distance between Susan and a bird
7 - x - horizontal distance between Math and a bird; 
h / x = tan 20° = 0.364
h / (7 - x) = tan 50°= 1.192
----------------------------------
h = 0.364 x
0.364 x / (7 - x ) = 1.192
0.364 x = 1.192 ( 7 - x ) 
0.364 x = 8.344 - 1.192 x
1.556 x = 8.344
x = 8.344 : 1.556 = 5.362 km
h = 0.364 · 5.362 
h = 1.952 km 
.. the bird is at height of 1.952 km from the ground. 
Download docx
san4es73 [151]3 years ago
5 0

Answer:

the answer would be 1.95



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Lelechka [254]

Answer:

1) \frac{(-2)^{-5}}{(-2)^{-10}}=-32

2) 2^{-1}.2^{-4} = \frac{1}{32}

3) (-\frac{1}{2} )^3.(-\frac{1}{2} )^2=-\frac{1}{32}

4) \frac{2}{2^{-4}} = 32

Step-by-step explanation:

1) \frac{(-2)^{-5}}{(-2)^{-10}}

Solving using exponent rule: a^{-m}=\frac{1}{a^m}

\frac{(-2)^{-5}}{(-2)^{-10}}\\=(-2)^{-5+10}\\=(-2)^{5}\\=-32

So, \frac{(-2)^{-5}}{(-2)^{-10}}=-32

2) 2^{-1}.2^{-4}

Using the exponent rule: a^m.a^n=a^{m+n}

We have:

2^{-1}.2^{-4}\\=2^{-1-4}\\=2^{-5}

We also know that: a^{-m}=\frac{1}{a^m}

Using this rule:

2^{-5}\\=\frac{1}{2^5}\\=\frac{1}{32}

So, 2^{-1}.2^{-4} = \frac{1}{32}

3) (-\frac{1}{2} )^3.(-\frac{1}{2} )^2

Solving:

(-\frac{1}{2} )^3.(-\frac{1}{2} )^2\\=(-\frac{1}{8} ).(\frac{1}{4} )\\=-\frac{1}{32}

So, (-\frac{1}{2} )^3.(-\frac{1}{2} )^2=-\frac{1}{32}

4) \frac{2}{2^{-4}}

We know that: a^{-m}=\frac{1}{a^m}

\frac{2}{2^{-4}}\\=2\times 2^4\\=2(16)\\=32

So, \frac{2}{2^{-4}} = 32

3 0
3 years ago
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