Answer:
Perpendicular lines are lines that intersect at right angles. If you multiply the slopes of two perpendicular lines in the plane, you get −1 . That is, the slopes of perpendicular lines are opposite reciprocals .
Hope this help:)
She can arrange them in the following;
1 row of 36
2 rows of 18 (doesnt count because it says other and this has already been mentioned)
3 rows of 12
4 rows of 9 (doesnt count because it says other and this has already been mentioned)
6 rows of 6
Answer:
i) 
\frac{3}{5} + (- \frac{1}{2}) = \frac{6}{10} - \frac{5}{10} = \frac{1}{10}
ii)
4^{3} + 8^{2} + \sqrt{9}
iii) ![(\frac{4}{5})^{2}. \sqrt[3]{8} \leqx^{3} - 3x + 6 \leq \sqrt{\frac{1}{3}} \Rightarrow \hspace{0.2cm}](https://tex.z-dn.net/?f=%28%5Cfrac%7B4%7D%7B5%7D%29%5E%7B2%7D.%20%5Csqrt%5B3%5D%7B8%7D%20%5Cleqx%5E%7B3%7D%20-%203x%20%2B%206%20%5Cleq%20%5Csqrt%7B%5Cfrac%7B1%7D%7B3%7D%7D%20%5CRightarrow%20%5Chspace%7B0.2cm%7D)
(\frac{4}{5})^{2}. \sqrt[3]{8} \leqx^{3} - 3x + 6 \leq \sqrt{\frac{1}{3}}
Step-by-step explanation:
i) \frac{3}{5} + (- \frac{1}{2}) = \frac{6}{10} - \frac{5}{10} = \frac{1}{10}
ii) 4^{3} + 8^{2} + \sqrt{9}
iii) (\frac{4}{5})^{2}. \sqrt[3]{8} \leqx^{3} - 3x + 6 \leq \sqrt{\frac{1}{3}}
and ends at 
, so we have an arc length of
