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marusya05 [52]
3 years ago
11

an observer on a cliff 1200 feet above sea level sights two ships due East. The angle of depression to the shops are 48° and 33°

. What is the distance between the ships?

Mathematics
1 answer:
ziro4ka [17]3 years ago
4 0
See the attached figure to better understand the problem

we know that
in the triangle ABC
tan 48°=AB/AC--------> AC=AB/tan48°------> 1200/tan 48°------> 1080.48 ft

in the triangle ABD
tan 33°=AB/AD---------> AD=AB/tan 33°-----> 1200/tan 33°------> 1847.88 ft

<span>the distance between the ships is AD-AC---> 1847.88-1080.48----> 767.4 ft

the answer is 
</span>the distance between the ships is 767.4 ft<span>

</span>

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Answer:

y = -2x + 6

Step-by-step explanation:

y=mx + c

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y= -2x + c

To find c, just sub one of the coordinates into the eqn:

By using the coordinates (1,4)

4 = -2(1) + c

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3 years ago
Simplify 6m plus 2m minus 5
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Answer:

8m-5

Step-by-step explanation:

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8m-5

That is as simple as it gets.

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3 years ago
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8 0
2 years ago
If 1st and 4tg terms of G.p are 500 and 32 respectively it's second term is ?
bixtya [17]

Answer:

T_{2} = 200

Step-by-step explanation:

Given

Geometry Progression

T_1 = 500

T_4 = 32

Required

Calculate the second term

First, we need to write out the formula to calculate the nth term of a GP

T_n = ar^{n-1}

For first term: Tn = 500 and n = 1

500 = ar^{1-1}

500 = ar^{0}

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For fought term: Tn = 32 and n = 4

32 = ar^{4-1}

32 = ar^3

Substitute 500 for a

32 = 500 * r^3

Make r^3 the subject

r^3 = \frac{32}{500}

r^3 = 0.064

Take cube roots

\sqrt[3]{r^3} = \sqrt[3]{0.064}

r  = \sqrt[3]{0.064}

r = 0.4

Using:  T_n = ar^{n-1}

n = 2     r = 0.4     and a = 500

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T_{2} = 200

<em>Hence, the second term is 200</em>

5 0
3 years ago
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