darren has 3/4 of a gallon of soil.he needs to plant 5 small plants.if he splits the soil evenly between the plants.how much soi
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Answer:
The answer is below
Step-by-step explanation:
A cell tower is located 3 miles east and 4 miles north of the center of a small town. The cell tower has a coverage radius of 3 miles.
a) Start by drawing a diagram of this situation. Your diagram might include a coordinate plane, the cell tower, and a circle representing the cell tower's coverage boundary.
a) A house is located 5 miles north of the center of the town and is to the east of the cell tower. If the house lies on the boundary of the cell tower's coverage, how far east of the center of the town is the house?
Answer:
Let the center of the town represent the origin (0,0). Also 1 unit = 1 mile. Since the cell tower is located 3 miles east and 4 miles north of the center of a small town, it is represented by A(3, 4)
The cell tower has a coverage of radius 3 miles. This can be represented by a circle with equation:
(x - a)² + (y - b)² = r². where (a,b) is the center of the circle and r is the radius. Hence:
(x - 3)² + (y - 4)² = 3²
(x - 3)² + (y - 4)² = 9
The diagram is drawn using geogebra.
b) The house is 5 miles north. It can be represented by y = 5 line.
To find the distance east of the house we have to substitute y = 5 and solve for x, hence:
(x - 3)² + (y - 4)² = 9
(x - 3)² + (5 - 4)² = 9
(x - 3)² + 1 = 9
(x - 3)² = 8
(x - 3) = √8
x - 3 = ±2.83
x = 3 ± 2.83
x = 5.83 or 1.83
Since it is to the east to the cell tower, hence x = 5.83.
Therefore the house is located 5.83 miles to the east of the cell tower
Answer:
Step-by-step explanation:
Given:
°
From the triangle, using the theorem that center angle by an arc is twice the angle it subtend at the circumference.
Also, the diameter of the circle is BD. As per the theorem that says that angle subtended by the diameter at the circumference is always 90°,
From the Δ BCD, which is a right angled triangle,
Now, using the theorem that angle between the tangent and a chord is equal to the angle subtended by the same chord at the circumference.
Here, chords CD and BC subtend angles 40 and 50 at the circumference as shown in the diagram by angles and EF is a tangent to the circle at point C.
Therefore,
Again, using the same theorem as above,
Hence, all the angles are as follows: