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Svetradugi [14.3K]
3 years ago
11

What is the value of y that satisfies the equation below? y= 12 3A3 C 36 B4 D 63

Mathematics
1 answer:
denpristay [2]3 years ago
5 0

Answer:

Step-by-step explanation:

B i think

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You are driving along the coast at 45 mph. How many minutes should it take you to get to Ventura?
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45 miles per hour is 3/4 mile per minute. Divide 4.8 by 0.75.
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3 years ago
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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Answer: I don’t think I understand can you set this up differently so I can answer this please

Step-by-step explanation:

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Answer:

hmmm i think you need to dod

Step-by-step explanation:

yeah  

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An acute triangle has a side measuring 10 cm and 16 cm. The legnth of the third side is unknown. Which best describes the range
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