The area of a square poster is 59 in.squared 2. Find the length of one side of the poster to the nearest tenth of an inch.

Mathematics

2 answers:

mafiozo [28]3 years ago

6 0

The answer would be 7.4 , because if you divide 59 by 2 that's the perimeter so divide that by 4 and round it. That should be the answer.

amm18123 years ago

6 0

Length of each side of square is 7.6 inch.

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Given the circle with the equation (x - 3)2 + y2 = 49, determine the location of each point with respect to the graph of the cir

Bas_tet [7]

To find out if a point is inside, on, or outside a circle, we need to substitute the ordered pair into the equation of the circle:
(x-xc)^2+(y-yc)^2=r^2
where (xc,yc) is the centre of the circle, and r=radius of the circle.

If the left-hand side [(x-xc)^2+(y-yc)^2] is less than r^2, then point (x,y) is INSIDE the circle. If the left-hand side is equal to r^2, the point is ON the circle.
Finally, if the left-hand side is greater than r^2, the point is OUTSIDE the circle.

For the given problem, we have xc=3, yc=0, or centre at (3,0), r=sqrt(49)=7
(x-xc)^2+(y-yc)^2=r^2 => (x-3)^2+y^2=7^2

A. (-1,1),
(x-3)^2+y^2=7^2 => (-1-3)^2+1^2=16+1=17 <49 [inside circle]

B. (10,0)
(x-3)^2+y^2=7^2 => (10-3)^2+0^2=49+0=49 [on circle]

C. (4,-8)
(x-3)^2+y^2=7^2 => (4-3)^2+(-8)^2=1+64=65 > 49 [outside circle]