Center: <span><span>(5,4)</span><span>
</span></span>Radius: <span>2<span>√<span>10
Hope this helped☺☺</span></span></span>
Triangle QRS is plotted in Quadrant I. After a 180° rotation about the origin, which quadrant would its image be? quadrant III
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:
x > 5
Step-by-step explanation:
14 - 3x < -1
-3x < -15
x > 5
