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TEA [102]
2 years ago
6

The hypotenuse of right triangle ABC, line segment AC, measures 13 cm. The length of line segment BC is 5 cm.

Mathematics
1 answer:
slavikrds [6]2 years ago
7 0
Cos ( ∠C ) = 5/13 = 0.38462
m ∠C = cos^(-1) 0.38642 = 67.4°
m ∠A = 90° - 67.4° = 22.6°
m ∠C - m ∠A = 67.4° - 22.6° = 44.8°
Answer:
B ) 44.8° 
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Remember the cost cannot be a number lesser than zero. so the domain must be such that the total cost C(m) is not negative. Also the range will always be 0.5 and above provided our domain is greater than or equal to zero.

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A fishing boat lies 200 m due south of a large tree on the shoreline and 300 m southwest of the dock. The shoreline runs East to
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Tree...........................dock
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use Pythagorean Theorem to satisfy...a^2 + b^2 = c^2

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2 years ago
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Erik's disabled sailboat is floating stationary 3 miles east and 2 miles north of Kingston. A ferry leaves Kingston heading towa
sergij07 [2.7K]

Answer:

Step-by-step explanation:

Let's suppose that Ballard is an origin with coordinates (0;0)

Ballard is 8 miles south and 1 mile west of Edmonds ⇒ Edmonds is 1 mile east and 8 miles north of Ballard.

Thus, coordinates of Edmonds (0+1; 0+8) (1;8)

Edmonds is 6 miles due east of Kingston. So, is 6 miles due of Edmonds

Thus, Kingston coordinates (1-6;8) (-5:8)

Sailboat is 3 miles east and 2 miles north of kingston

So, coordinates of sailboat are (-5+3; 8+2) (-2;10)

a)

Ferry leaves Kingston towards Edmonds at mph and Edmonds is 6 miles due east of Kingston.

Initial ferry coordinates as calculated above are equal to Kingston coordinates (-5;8)

After 20 minutes the ferry turns to south.

This distance travelled (20min) is d=9mph*20/60h=3 miles

So, ferry travelled 3 miles toward east in 20 min

Then, coordinates became (-5+3;8+0) (-2;8)

Thus, we have a line connecting two points (-5;8) and (-2;8)

Line connecting them has an equation:

y-8 = (8-8)/(-2+5) * (x+5)

y-8=0

y=8 - This is the equation for the first 20 minutes of travel

Then, ferry turns due south (-2;8) and has a vertiacl line

The equation of verical line is x=a, so the equation will be x=2

b) The sailboat has a radar scope that will detect any object within 3 miles of the sailboat.

region looks like a circular disc with a center (-2;10)

(x+2)^2+(y-10)^2=3^2

(x+2)^2+(y-10)^2 <9 (interior of the circular disc)

(x+2)^2+(y-10)^2 >9 (exterior)

The equation of the line joining the Kingston and Edmonds is y=8

the point of intersection:

(x+2)^2+(8-10)^2=9

(x+2)^2=5

x is approximately 0.24; -4.24

(0.24;8) (-4.24;8) - intersection points

c) The ferry exits the radar during the trid due south long x=-2

The points of intersection of the circle and the line x=-2:

(-2+2)^2+(y-10)^2=9

0+(y-10)^2=0

y is 7.13

(-2;7)

d) Takes south turn at (-2;8), then ferry travels 0.8 miles up to point (-2;7) where it exists the radar zone

Speed is 9mph

The time taken to cover 1 mile = t=1/9hr=1/9 *60=6.667min

So, it exists after 6.67 minutes

e) Ferry enters at (-4.24;8) and takes turn due to south at (-2;8)

The distance travelled is = [-4.24+2]=2.24 miles

The time taken to cover these miles is = t=2.24/9 *60=14.93 min

After turning to south ferry remains in radar for 6.7 min

So, it remains in radar zone for 14.93+6.7=21.63min

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3/2x +7/4x = ...........

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