C
6x^2-13X-5=0
(3X+1)(2X-5)
3x-1=0
X=1/3. 2x-5=0. X= 5/2
<span><span>x3-6x2+11x-6=0</span> </span>Three solutions were found :<span> x = 3 x = 2<span> x = 1</span></span>
So in order for us to know the area of the square that is not covered by the circle, we need to find first both the areas of the square and the circle.
So for the area of the square it is A = sxs. And for the circle is A = pi*r^2.
Let us find the area of the square first given that the side is 3 inches.
So A = 3*3
A = 9 square inches.
Next is the area of the circle. Since the center of the circle is the same with the center of the square, the radius would be 1.5.
SO, A = (3.14)(1.5)^2
A = 2.25 (3.14)
A = 7.065 square inches.
Next, we deduct the area of circle from area of square and the result would be 1.935 <span>in². So the answer for this would be option B.
Hope this answer helps.</span>
<em>Answer</em><em>:</em>
<em>Range</em><em>=</em><em>{</em><em> </em><em>-</em><em>5</em><em>,</em><em>-</em><em>1</em><em>,</em><em>3</em><em>,</em><em>7</em><em>)</em><em>. </em><em> </em><em> </em><em> </em><em>[</em><em> </em><em>option</em><em> </em><em>C</em><em>]</em>
<em>Please</em><em> </em><em>see</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>picture</em><em> </em><em>for</em><em> </em><em>full</em><em> </em><em>sol</em><em>ution</em><em>.</em><em>.</em>
<em>Hope</em><em> </em><em>it</em><em> </em><em>helps</em><em>.</em><em>.</em>
<em>Good</em><em> </em><em>luck</em><em> </em><em>on</em><em> </em><em>your</em><em> </em><em>assignment</em><em>.</em><em>.</em>
Thank you for posting your question here at brianly. I hope the answer will help you. Below is the solution:
or segment from (4,0) to (-4,0), y is always zero because straight segment.
<span>in that case x*y=x*0=0 </span>
<span>and x+y=x+0=x </span>
<span>if you choose x=4-t where t goes from 0 to 8 you should get </span>
<span>r(t)=0i+(4-t)j </span>
