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2 years ago

Write the equation in standard form for a circle that has a diameter with endpoints (8,0) and (-8,0)

Mathematics

1 answer:

riadik2000 [5.3K]2 years ago

6 0

Answer:

x²+y² = 64

Step-by-step explanation:

The standard form of writing the equation of a circle is expressed as;

(x-a)²+(y-b)² = r² where;

(a, b) is the centre of the circle

r is the radius of the circle

Given the diameter of the circle with endpoints (8,0) and (-8, 0)

d = √(x2-x1)²+(y2-y1)²

d = √(-8-8)²+(0-0)²

d = √(-16)²

d = √256

d = 16

radius r = 16/2

r = 8

The centre of the circle will be the midpoint of the coordinates

M = (8-8/2, 0+0/2)

M = (0/2, 0/2)

M = (0,0)

Hence the centre is at (0,0)

Get the required equation

Recall that (x-a)²+(y-b)² = r²

Substitute the centre and the radius

(x-0)²+(y-0)² = 8²

x²+y² = 64

This gives the required equation

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stellarik [79]

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2 years ago

1. Write the form of the partial fraction decomposition of the rational expression. Do not solve for the constants.

german

Answer:

Step-by-step explanation:

To write the form of the partial fraction decomposition of the rational expression:

We have:

$\mathbf{\dfrac{8x-4}{x(x^2+1)^2}= \dfrac{A}{x}+\dfrac{Bx+C}{x^2+1}+\dfrac{Dx+E}{(x^2+1)^2}}$

Using partial fraction decomposition to find the definite integral of:

$\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}dx$

By using the long division method; we have:

$x^2-8x-20 | \dfrac{2x}{2x^3-16x^2-39x+20 }$

$- 2x^3 -16x^2-40x$

$x+ 20$

So;

$\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}= 2x+\dfrac{x+20}{x^2-8x-20}$

By using partial fraction decomposition:

$\dfrac{x+20}{(x-10)(x+2)}= \dfrac{A}{x-10}+\dfrac{B}{x+2}$

$= \dfrac{A(x+2)+B(x-10)}{(x-10)(x+2)}$

x + 20 = A(x + 2) + B(x - 10)

x + 20 = (A + B)x + (2A - 10B)

Now; we have to relate like terms on both sides; we have:

A + B = 1 ; 2A - 10 B = 20

By solvong the expressions above; we have:

$A = \dfrac{5}{2}$ $B = \dfrac{3}{2}$

Now;

$\dfrac{x+20}{(x-10)(x+2)} = \dfrac{5}{2(x-10)} + \dfrac{3}{2(x+2)}$

Thus;

$\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}= 2x + \dfrac{5}{2(x-10)}+ \dfrac{3}{2(x+2)}$

Now; the integral is:

$\int \dfrac{2x^3-16x^2-39x+20}{x^2-8x-20} \ dx = \int \begin {bmatrix} 2x + \dfrac{5}{2(x-10)}+ \dfrac{3}{2(x+2)} \end {bmatrix} \ dx$

$\mathbf{\int \dfrac{2x^3-16x^2-39x+20}{x^2-8x-20} \ dx = x^2 + \dfrac{5}{2}In | x-10|\dfrac{3}{2} In |x+2|+C}$

3. Due to the fact that the maximum words this text box can contain are 5000 words, we decided to write the solution for question 3 and upload it in an image format.

Please check to the attached image below for the solution to question number 3.

4 0

2 years ago

A red bus leaves a theme park every 24 minutes and a blue bus leaves the park every 20 minutes. they both leave the park at noon

earnstyle [38]

Answer:

2:00 PM

Step-by-step explanation:

20 = 2² × 5

4 0

2 years ago

Find the volume of a cylinder with a radius of 4.8 cm and a height of 14.6 cm. [Use it =

juin [17]

Answer:

1056.24 cm^3

Step-by-step explanation:

The volume of a cylinder is $\pi r^2h$ . The problem says to use $\pi$ =3.14

Plug in your numbers $3.14*4.8^2*14.6=1056.24$

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1 year ago

Is point R collinear with points Q and S?

ZanzabumX [31]

Answer:

can't hwlp you without an image

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2 years ago

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