Acell tower is located 2 miles east and 5 miles north of a small town. the cell tower has a coverage radius of 4 miles. a house is located 7 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 towers coverage, how far east of thr center of the town is the house?

The attached figure represents the explanation of the problem.

The distance between the house and the point B = 7 - 5 = 2

The distance between the house and the cell tower = 4

While the angle B is a right angle.

So, The distance between the house and the cell tower represents the hypotenuse

x and the distance between the house and the cell tower will be the legs

So, x =

The east distance between Town and House = 2 + x = 2 + 3.464 = 5.464 ≈ 5.5 miles

So, The house will be located 5.5 miles east of the town

Here we have a problem with circles, we want to find the location of the house given that it is inside the coverage area of the tower.

We will see that the house is 5.46 miles east of the center of the town.

First, we need to define the coordinate axis where we are working on.

Let's define the center of the small town as the (0, 0), north as the positive y-axis, and east as the positive x-axis.

Now, we know that the cell tower is located 2 miles east and 5 miles north of the small town, and it has a radius coverage of 4 miles.

Then the coverage area is defined by:

(x - 2mi)^2 + (y - 5mi)^2 ≤ (4mi)^2

Where:

(x - a)^2 + (y - b)^2 ≤ R^2

Represents all the area inside a circle of radius R and centered on the point (a, b).

Now we know that the house is located at 7 miles north of the town center, then:

y = 7mi

And we also know that the house lies on the boundary of the cell tower's coverage area, then we can write:

(x - 2mi)^2 + (7mi - 5mi)^2 = (4mi)^2

(x - 2mi)^2 + 4mi^2 = 16^2

(x - 2mi)^2 = 12 mi^2

x - 2mi = √(12 mi^2)

x = 2mi + √(12 mi^2) = 5.46 mi

This means that the town is exactly 5.46 miles to the east of the town.

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(n^3 - n)

= n(n-1)(n+1)

since it is divisible by 3 and 2 ,because it is a multiple 3 consecutive number

if, n-1 is odd then n+1 is also odd, then it is a multiple of 9

but if, n-1 is even then n+1 is also even, then it is a multiple of 4

so, largest multiple = max(9*2, 4*3) = max(18, 12) = 18

step-by-step explanation:

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