Find the center and radius of the sphere whose equation is given by x2+y2+z2−2x−4y+8z+17=0x2+y2+z2−2x−4y+8z+17=0.

1 answer:

4 0

We know that
the equation of a sphere is
(x-h)²+(y-k)²+(z-l)²=r²
where (h,k,l) is the center and r is the radius

we have
x²+y²+z²<span>−2x−4y+8z+17=0
</span>

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(x²+2x)+(y²-4y)+(z²+8z)=-17
<span>Complete the square. Remember to balance the equation by adding the same constants to each side
</span>(x²+2x+1)+(y²-4y+4)+(z²+8z+16)=-17+1+4+16
(x²+2x+1)+(y²-4y+4)+(z²+8z+16)=4

Rewrite as perfect squares

(x+1)²+(y-2²)+(z+4)²=4

(x+1)²+(y-2²)+(z+4)²=2²

the center is the point (-1,2,-4) and the radius is 2 units

You might be interested in

Can you help me?

zalisa [80]

Answer:

ig this is free pionts, sry whats the question???

Step-by-step explanation:

4 0

3 years ago

HELP ME PLEASE THIS MEANS A LOT TO ME

nevsk [136]

Answer:

I cannot see the screen. Write it out please and I will answer.

Step-by-step explanation:

7 0

3 years ago

Pumping stations deliver oil at the rate modeled by the function D, given by d of t equals the quotient of 5 times t and the qua

goblinko [34]

<h2>Hello!</h2>

The answer is: There is a total of 5.797 gallons pumped during the given period.

To solve this equation, we need to integrate the function at the given period (from t=0 to t=4)

The given function is:

$D(t)=\frac{5t}{1+3t}$

So, the integral will be:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dx$

So, integrating we have:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dt=5\int\limits^4_0 {\frac{t}{1+3t}} \ dx$

Performing a change of variable, we have:

$1+t=u\\du=1+3t=3dt\\x=\frac{u-1}{3}$

Then, substituting, we have:

$\frac{5}{3}*\frac{1}{3}\int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du\\\\\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u}{u} -\frac{1}{u } \ du$