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olganol [36]
3 years ago
6

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.

Mathematics
1 answer:
Vedmedyk [2.9K]3 years ago
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




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The answer is a = \frac{4}{-3}

Step-by-step explanation:

<em>1. Convert the mixed fraction to an improper fraction</em>

To find the numerator, multiply the denominator by the whole number and add the numerator to it.

The denominator remains the same.

So, 2\frac{2}{3} will be \frac{8}{3}

<em>2. Now the equation is,</em>

\frac{3}{2} a - \frac{4}{3} a = \frac{10}{3} + \frac{8}{3} a

<em>3. Take LCM on both sides. </em>

For the left side, multiply the first fraction by \frac{3}{3} and multiply the second fraction by \frac{2}{2}

\frac{3*3}{2*3} a - \frac{4*3}{3*3} a = \frac{10+8a}{3}

<em>4. Solve by making a the subject</em>

\frac{9a-8a}{6} = \frac{10+8a}{3}

\frac{a}{6} = \frac{10+8a}{3}

\frac{3a}{6} =10+8a

\frac{a}{2} = 10 + 8a

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a = 20 + 16a

a-16a = 20

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a = \frac{20}{-15}

a = \frac{4}{-3}

Therefore, the answer is a = \frac{4}{-3}

Keyword: Equations

Learn more about equations at

  • brainly.com/question/10666510
  • brainly.com/question/4460262
  • brainly.com/question/8955867

#LearnwithBrainly

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Now we assume that 1/x is rational (we want to prove that this implies that x will be also rational and because we know that x is irrational assuming that 1/x is rational will lead to an incongruence), then:

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