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galina1969 [7]
3 years ago
9

!!!!!!!please help!!!!!!!!

Mathematics
1 answer:
GuDViN [60]3 years ago
4 0

Answer:

The first one is No. The second one I do not know unfortunately.

Step-by-step explanation:

I am in the middle of this lesson in my math class as well and I know that in order to make a triangle, you need the two smaller sides to add up to become larger than the biggest side. In the case of the first question, the two smaller sides, 6 and 3 equal 9, which is not bigger than the biggest side which is 10. I hope you can find the answer to the second one, but for now, the first one is No.

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C. Solving this equation results in the statement –48 = 48. Because this is a false statement, the equation has no solution.

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One of the tallest buildings in a country is topped by a high antenna. The angle of elevation from the position of a surveyor on
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Answer:

a. distance of the surveyor to the base of the building = 2051.90 ft

b. height of the building = 1384 ft

c. Angle of elevation from the surveyor to the top of the antenna = 38.31°

d. Height of antenna  =  237.08 ft

Step-by-step explanation:

​The picture above is a illustration of the described event.

a = the height of the flag

b = the height of the building

c = distance of the surveyor from the base of the building

the angle of elevation from the position of the surveyor on the ground to the top of the building = 34°  

distance from her position to the top of the building  = 2475 ft

distance from her position to the top of the flag  = 2615 ft

​(a) How far away from the base of the building is the surveyor​ located?​

using the SOHCAHTOA principle

cos 34° = c/2475

c =  0.8290375726  × 2475

c = 2051.8679921

c = 2051.90 ft

(b) How tall is the​ building

The height of the building = b

sin 34° = opposite /hypotenuse

0.5591929035 = b/2475

b =  0.5591929035  × 2475

b =  1384.0024361

b =  1384.00 ft

​(c) What is the angle of elevation from the surveyor to the top of the​ antenna?

let the angle = ∅

cos ∅ = adjacent/hypotenuse

cos ∅ = 2051.90/2615

cos ∅ =  0.784665392

∅ = cos-1  0.784665392

∅ =   38.310258303

∅ =  38.31°

​(d) How tall is the​ antenna?

height of the antenna = a

sin 38.31° = opposite/hypotenuse

sin 38.31° = (a + b)/2615

sin 38.31° × 2615 = (a + b)

(a + b) =  0.6199159917  × 2615

(a + b) =  1621.0803182

(a + b) = 1621. 08 ft

Height of antenna = 1621. 08 - 1384.00  =  237.08031822 ft

Height of antenna  =  237.08 ft

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"isotherms are drawn at regular intervals; on this map, the interval between successive isotherms is ______ fahrenheit degrees."
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