PLEASE SOMEONE HELP ME

Three cell phones towers can be modeled by the points X(6,0), Y(8,4), and Z(3,9). Determine the location of another cell phone t

Marat540 [252]

Answer:

The location of the new cell phone tower is $(h,k) = (3,4)$ , and the equation of the circle is $x^{2}+y^{2} -6\cdot x - 8\cdot y = 0$ .

Step-by-step explanation:

The location of the cell phone tower coincides with the location of a circunference passing through the three cell phone towers. By Analytical Geometry, the equation of the circle is represented by the following general formula:

$x^{2} + y^{2}+A\cdot x + B\cdot y +C = 0$ (1)

Where:

$x$ - Independent variable.

$y$ - Dependent variable.

$A$ , $B$ , $C$ - Circunference constants.

Given the number of variable, we need the location of three distinct points:

$(x_{1},y_{1}) = (6,0)$

$36 +6\cdot A + C = 0$

$(x_{2},y_{2}) = (8,4)$

$80 + 8\cdot A + 4\cdot B + C = 0$

$(x_{3},y_{3}) = (3,9)$

$90 + 3\cdot A + 9\cdot B + C = 0$

Then, we have the following system of linear equations:

$6\cdot A + C = -36$ (2)

$8\cdot A +4\cdot B + C = -80$ (3)

$3\cdot A + 9\cdot B + C = -90$ (4)

The solution of this system is:

$A = -6$ , $B = -8$ , $C = 0$

By comparing the general form with the standard form of the equation of the circunference is:

$A = -2\cdot h$ (5)

$B = -2\cdot k$ (6)

$C = h^{2}+k^{2}-r^{2}$ (7)

Where:

$h$ , $k$ - Coordinates of the center of the circle.

$r$ - Radius of the circle.

If we know that $A = -6$ , $B = -8$ and $C = 0$ , then coordinates of the center of the circle and its radius are, respectively:

$h = -\frac{A}{2}$

$k = -\frac{B}{2}$

$r = \sqrt{h^{2}+k^{2}-C}$

$h = 3$ , $k = 4$ , $r = 5$

The location of the new cell phone tower is $(h,k) = (3,4)$ , and the equation of the circle is $x^{2}+y^{2} -6\cdot x - 8\cdot y = 0$ .

3 0

2 years ago

I need answers to all please

valina [46]

Answer:

Step-by-step explanation:

2a.x=10+5

b. x=11-7

C. x=5*2

4 0

3 years ago

Read 2 more answers

The graphs of f(x)=2x+4 and g(x)=10−4x intersect at (1,6) . What is the solution of the equation 2x+4=10−4x ? Enter your

nordsb [41]

X = 1, since they intersect at (1,6) and are thus equal at that point

8 0

3 years ago

Read 2 more answers

Lucy will rent a car for the weekend. She can choose one of two plans. The first plan has an initial fee of $53 and costs an add

Alchen [17]

Answer:

The amount driven would be 275, and the cost for both plans will be 77.75$

Step-by-step explanation:

make both equations

y=0.09x+53

y=0.13x+42

set them equal to each other and solve for x to get the distance

take that number and put it in one equation and solve for y to get the price of the plan for that value

6 0

2 years ago

Find gradient xe^y + 4 ln y = x² at (1, 1)

cricket20 [7]

$xe^y+4\ln y=x^2$

Differentiate both sides with respect to x, assuming y = y(x).

$\dfrac{\mathrm d(xe^y+4\ln y)}{\mathrm dx}=\dfrac{\mathrm d(x^2)}{\mathrm dx}$

$\dfrac{\mathrm d(xe^y)}{\mathrm dx}+\dfrac{\mathrm d(4\ln y)}{\mathrm dx}=2x$

$\dfrac{\mathrm d(x)}{\mathrm dx}e^y+x\dfrac{\mathrm d(e^y)}{\mathrm dx}+\dfrac4y\dfrac{\mathrm dy}{\mathrm dx}=2x$

$e^y+xe^y\dfrac{\mathrm dy}{\mathrm dx}+\dfrac4y\dfrac{\mathrm dy}{\mathrm dx}=2x$

Solve for dy/dx :

$e^y+\left(xe^y+\dfrac4y\right)\dfrac{\mathrm dy}{\mathrm dx}=2x$

$\left(xe^y+\dfrac4y\right)\dfrac{\mathrm dy}{\mathrm dx}=2x-e^y$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2x-e^y}{xe^y+\frac4y}$

If y ≠ 0, we can write

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2xy-ye^y}{xye^y+4}$

At the point (1, 1), the derivative is

$\dfrac{\mathrm dy}{\mathrm dx}\bigg|_{x=1,y=1}=\boxed{\dfrac{2-e}{e+4}}$

4 0

3 years ago