<u>Answer:
</u>
The fraction of the bleachers filled with the home team is
<u>Solution:
</u>
Given that,
The bleachers at the football game are
In those bleachers
of the fans are rooting for the home team
So, the fans that are filled with the home team is \left(\left(\frac{7}{8}\right) \times\left(\frac{1}{2}\right)\right)
Hence, the required fraction is \left(\left(\frac{7}{8}\right) \times\left(\frac{1}{2}\right)\right)
Removing the brackets we get,
\frac{7 \times 1}{8 \times 2}
=
The required fraction is
D2 - d + 4d = d2 - 8d
d2 + 3d = d2 - 8d
d2 - d2 = - 8d - 3d
0 = - 11d
Then AE < BC. or am i wrong
(6^-2)^2 = 1/1296 = 0.0007716049382716 hope this is what you were looking for. the math is right tho for (6^-2)^2 most calculators can do this easy just do 6^-2 then after just do ^2