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Zepler [3.9K]
3 years ago
9

PLSSSS HELP!! The radius of a circle is 8 feet. What is the length of a 135° arc?​

Mathematics
1 answer:
dangina [55]3 years ago
7 0

Answer:

Use https://www.bartleby.com/  it has helped me out a lot . Its actual tutors/teachers who answer your questions in about 30 minutes !!!!!!!!

Step-by-step explanation:

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Subtract 1.05 from a certain number. Multiply the difference by 0.8, add 2.84 to the product then divide the sum by 0.01 and get
Dima020 [189]
The answer to your question would be 6.25 and the equation you need to solve is: (0.8x+2)/0.01=700. 

                         Hope this helps!
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2 years ago
4+ -1 2/3 please help
kogti [31]

The sum would end us as: 2.33333333

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3 years ago
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If ab = 5 and a² + b² = 25, then (a + b)² =​
prisoha [69]

Answer:

100

Step-by-step explanation:

because 5+5=10

the 10^2=10*10=100

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3 years ago
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an aircraft carrier leads port heading due west at 28 knots. A helicopter is 175 nautical miles from the carrier at an angle of
emmainna [20.7K]
Let’s say it takes time t hours for the interception.
In that time the carrier travels 28t nm west and the helicopter 130t nm.
Now we use the sine rule to find the angle x which lies between the distance 175 and 130t.
sin(x)/28t=sin35/130t. So sin(x)=28sin35/130=14sin35/65=0.1235 and x=7.0964 degrees.
Therefore the helicopter should use a bearing of 35+7.0964=42.0964 degrees north of east or 47.9 degrees east of north approx.
8 0
3 years ago
An airplane pilot can see the top of a traffic control tower at a 20 degree angle of depression. the airplane is 5,000 feet away
daser333 [38]

The given question describes a right triangle with with one of the angles as 20 degrees and the side adjacent to the angle 20 degrees is of length 5,000 feet. We are looking for the length of the side opposite the angle 20 degrees.

Let the required length be x, then

\tan{20^o}=\cfrac{opp}{hyp}=\cfrac{x}{5,000}\\ \\ \Rightarrow x=5,000\tan{20^o}=1,819.85

Therefore, the height of the airplane above the tower is 1,819.85 feet.

8 0
3 years ago
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