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
Answer:
Minimum value of function
is 63 occurs at point (3,6).
Step-by-step explanation:
To minimize :

Subject to constraints:

Eq (1) is in blue in figure attached and region satisfying (1) is on left of blue line
Eq (2) is in green in figure attached and region satisfying (2) is below the green line
Considering
, corresponding coordinates point to draw line are (0,9) and (9,0).
Eq (3) makes line in orange in figure attached and region satisfying (3) is above the orange line
Feasible region is in triangle ABC with common points A(0,9), B(3,9) and C(3,6)
Now calculate the value of function to be minimized at each of these points.

at A(0,9)

at B(3,9)

at C(3,6)

Minimum value of function
is 63 occurs at point C (3,6).
Answer:
C
Step-by-step explanation:
Answered by Gauthmath
Answer:
106
Step-by-step explanation:
C.
Solve for x by simplifying both sides of the equation, then isolating the variable.
x = 77, -77