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aniked [119]
4 years ago
12

Solve the inequality 3(x – 2) + 1 ≥ x + 2(x + 2).

Mathematics
1 answer:
Olin [163]4 years ago
6 0

Answer:

no solution

Step-by-step explanation:

3(x – 2) + 1 ≥ x + 2(x + 2).

Distribute

3x – 6 + 1 ≥ x + 2x + 4

Combine like terms

3x -5 ≥ 3x +4

Subtract 3x from each side

-5 ≥ 4

This is never true so there is no solution

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Is the given point interior , exterior, or on the circle k (x+2)2 + (y-3)2 =18 P (8,4)
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One way would be to find the distance from the point to the center of the circle and compare it to the radius

for
(x-h)^2+(y-k)^2=r^2
the center is (h,k) and the radius is r

and the distance formula is
distance between (x_1,y_1) and (x_2,y_2) is
D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}


r=radius
D=distance form (8,4) to center

if r>D, then (8,4) is inside the circle
if r=D, then (8,4) is on the circle
if r<D, then (8,4) is outside the circle


so
(x+2)^2+(y-3)^2=18
(x-(-2))^2+(y-3)^2=(\sqrt{18})^2
(x-(-2))^2+(y-3)^2=(3\sqrt{2})^2

the radius is 3\sqrt{2}
center is (-2,3)

find distance between (8,4) and (-2,3)

D=\sqrt{(8-(-2))^2+(4-3)^2}
D=\sqrt{(8+2)^2+(1)^2}
D=\sqrt{10^2+1}
D=\sqrt{100+1}
D=\sqrt{101}




r=3\sqrt{2}≈4.2
D=\sqrt{101}≈10.04

do r<D

(8,4) is outside the circle

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