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Amiraneli [1.4K]
4 years ago
7

Type the correct answer in each box. A circle is centered at the point (5, -4) and passes through the point (-3, 2). The equatio

n of this circle is (x + )2 + (y + )2 = .
Mathematics
1 answer:
Galina-37 [17]4 years ago
4 0

Answer:

(x+ \boxed{-5})^2+(y+\boxed4)^2=\boxed{100}

Step-by-step explanation:

Given:

Center of circle is at (5, -4).

A point on the circle is (x_1,y_1)=(-3, 2)

Equation of a circle with center (h,k) and radius 'r' is given as:

(x-h)^2+(y-k)^2=r^2

Here, (h,k)=(5,-4)

Radius of a circle is equal to the distance of point on the circle from the center of the circle and is given using the distance formula for square of the distance as:

r^2=(h-x_1)^2+(k-y_1)^2

Using distance formula for the points (5, -4) and (-3, 2), we get

r^2=(5-(-3))^2+(-4-2)^2\\r^2=(5+3)^2+(-6)^2\\r^2=8^2+6^2\\r^2=64+36=100

Therefore, the equation of the circle is:

(x-5)^2+(y-(-4))^2=100\\(x-5)^2+(y+4)^2=100

Now, rewriting it in the form asked in the question, we get

(x+ \boxed{-5})^2+(y+\boxed4)^2=\boxed{100}

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a. F = ma

F/a = ma/a

m = F/a

b. F = ma

-30 = -6a

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3 years ago
Find the first partial derivatives of the function f(x,y,z)=4xsin(y−z)
Amanda [17]

Answer:

f_x(x,y,z)=4\sin (y-z)

f_x(x,y,z)=4x\cos (y-z)

f_z(x,y,z)=-4x\cos (y-z)

Step-by-step explanation:

The given function is

f(x,y,z)=4x\sin (y-z)

We need to find first partial derivatives of the function.

Differentiate partially w.r.t. x and y, z are constants.

f_x(x,y,z)=4(1)\sin (y-z)

f_x(x,y,z)=4\sin (y-z)

Differentiate partially w.r.t. y and x, z are constants.

f_y(x,y,z)=4x\cos (y-z)\dfrac{\partial}{\partial y}(y-z)

f_y(x,y,z)=4x\cos (y-z)

Differentiate partially w.r.t. z and x, y are constants.

f_z(x,y,z)=4x\cos (y-z)\dfrac{\partial}{\partial z}(y-z)

f_z(x,y,z)=4x\cos (y-z)(-1)

f_z(x,y,z)=-4x\cos (y-z)

Therefore, the first partial derivatives of the function are f_x(x,y,z)=4\sin (y-z), f_x(x,y,z)=4x\cos (y-z)\text{ and }f_z(x,y,z)=-4x\cos (y-z).

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4 years ago
Suppose that X is a random variable with mean 30 and standard deviation 4. Also suppose that Y is a random variable with mean 50
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Answer:

a) Var[z] = 1600

    D[z] = 40

b) Var[z] = 2304

    D[z] = 48

c) Var[z] = 80

    D[z] = 8.94

d) Var[z] = 80

    D[z] = 8.94

e) Var[z] = 320

    D[z] = 17.88

Step-by-step explanation:

In general

V([x+y] = V[x] + V[y] +2Cov[xy]

how in this problem Cov[XY] = 0, then

V[x+y] = V[x] + V[y]

Also we must use this properti of the variance  

V[ax+b] = a^{2}V[x]

and remember that

standard desviation = \sqrt{Var[x]}

a) z = 35-10x

   Var[z] = 10^{2} Var[x] = 100*16 = 1600

   D[z] = \sqrt{1600} = 40

b) z = 12x -5

   Var[z] = 12^{2} Var[x] = 144*16 = 2304

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c) z = x + y

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d) z = x - y

   Var[z] =  Var[x-y] = Var[x] + Var[y] = 16 + 64 = 80

   D[z] = \sqrt{80} = 8.94

e) z = -2x + 2y

   Var[z] = 4Var[x] + 4Var[y] = 4*16 + 4*64 = 320

  D[z] = \sqrt{320} = 17.88

   

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