the net is pretty much the net of a long box, kinda like the one in the example in the picture below. Due to that, we can pretty much assume the two sides sticking up and down, are just two small 6x3 rectangles, namely, they have a height of 3, reason why we assume that, if that if we fold the other sides to make out the box, those two sides sticking out, must be 3m to neatly snugfit.
now, if we close the box as it stands, the sides(laterals) will be on the left-right sides two 3x15 rectangles, and on the front-back sides, two 6x15 rectangles.
we're excluding the top and bottom sides, because those are not "laterals", or sides of the box.
Your going to divide them then take the same score and double it then subtract the total and the doubles number
X=7 and x= -2. You can use calculator find the answer
Short Answer: AB
Argument
Take F to be the center. The line segment that IS a diameter or radius must either go through F (that would make the line a diameter) or to be a radius the segment mus end in F and touch the circumference once. See below.
The diameter must not only go through F, it must touch the circumference in two places. EFB is a diameter. So is AFC
The radius must have 1 endpoint at the center and one endpoint on the circumference. DF is a radius.
So what isn't? Answer: AB isn't. It neither goes through F nor is F one of the end points.
Answer: AB
Answer:
72
Step-by-step explanation:
<em>here's</em><em> your</em><em> solution</em>
<em>=</em><em>></em><em> </em><em>x=</em><em> </em><em>8</em>
<em>=</em><em>></em><em> </em><em>y=</em><em> </em><em>-</em><em>3</em>
<em>=</em><em>></em><em> </em><em>putting</em><em> the</em><em> value</em>
<em>=</em><em>></em><em> </em><em> </em><em>8</em><em>*</em><em>6</em><em> </em><em>-</em><em> </em><em>(</em><em>8</em><em>*</em><em>-</em><em>3</em><em>)</em>
<em>=</em><em>></em><em> </em><em> </em><em>4</em><em>8</em><em> </em><em>+</em><em> </em><em>2</em><em>4</em>
<em>=</em><em>></em><em> </em><em>7</em><em>2</em>
<em>hope</em><em> it</em><em> helps</em>
