Answer:
As shown in picture, shaded space area:
A = outer rectangle area - inner rectangle area
= 11 x 8 - 9 x 6
= 88 - 54
= 34 (yd2)
Hope this helps!
:)
You would move each coordinate 6 left and 1 up
A(2,5)
B(0,2)
C(-1,3)
hope this helps!
Answer:
boom ganag
Step-by-step explanation:
Believe it or not, :), the width of the ribbon needs to be known to get the right answer. If the ribbon had an infinitely small width the ribbon needed would just be the perimeter.
P=2(L+W)
P=2(12+15)
P=54 in
However an infinitely small width is not something that has any practical application. For example, if the width of the ribbon were 1/2 in.
P=2((L+1)+W)
P=2(16+12)
P=56 in So the perimeter increased by 2 in when the ribbon was only 1/2 in wide.
The "infinitely small" width and the perimeter of 54 in is only a decent approximation if the ribbon were a very thin string.
Option C
The radius of circle is 13 units
<em><u>Solution:</u></em>
<em><u>The equation of circle is given by formula:</u></em>
Where, the center being at the point (h, k) and the radius being "r"
<em><u>Given that circle has its center at the origin</u></em>
(h, k) = (0, 0)
<em><u>(5, -12) is a point on the circle</u></em>
(x, y) = (5, 12)
<em><u>Substituting in equation we get,</u></em>
Thus radius of circle is 13 units
