we have the number
![6-\sqrt[]{-40}](https://tex.z-dn.net/?f=6-%5Csqrt%5B%5D%7B-40%7D)
Remember that
i^2=-1
so
substitute
![6-\sqrt[]{(i^2)40}](https://tex.z-dn.net/?f=6-%5Csqrt%5B%5D%7B%28i%5E2%2940%7D)
![6-2i\sqrt[]{10}](https://tex.z-dn.net/?f=6-2i%5Csqrt%5B%5D%7B10%7D)
therefore
the real part is 6
the imaginary part is -2√10
If Ms. Callahan has 24 feet of fencing, and she is building a pen, the PERIMETER of the pen must be 24 feet. The perimeter is basically the distance around a figure. The perimeter of a rectangle is equal to length plus width plus length plus width, AKA l+w+l+w, or P=2l+2w. In a rectangle, two pairs of sides are of equal length--so the two lengths and the two widths must be equal.
So, the formula is P=2l+2w. P, the perimeter, is 24, so 24=2l+2w. Let's try some values for l and see what we get for w. If the length is 1, l=1. 24=(2*1)+2w. 24=2+2w. 22=2w. w=11. So if length is 1 foot, width is 11 feet.
What if l=2? 24=(2*2)+2w. 24=4+2w. 2w=20. w=10. If l=2, w=10. And l=3? 24=(2*3)+2w. 24=6+2w. 18=2w. w=9. If l=3, w=9. Do you see a pattern? Every time we add 1 to l, we subtract 1 from w. So if l=4, w=8. If l=5, w=7. If l=6, w=6. Here, we start getting similar answers: if l=7, w=5. If l=8, w=4. Since we already know these values work, it doesn't matter whether we call it length or width. So, our answers are below.
Answer: Ms Callahan can make a pen with a length of 1 foot and a width of 11 feet, a length of 2 feet and a width of 10 feet, a length of 3 feet and a width of 9 feet, a length of 4 feet and a width of 8 feet, a length of 5 feet and a width of 7 feet, or a length of 6 feet and a width of 6 feet.
Ok what is the question lol
