I believe it's the first 3 boxes
Answer:-2x^2(2-1)
Step-by-step explanation:
-4x^2 + 2x^2=-2x^2(2 - 1)
Answer:
7: 45.2 (rounded from 45.21)
8: 114.5 (rounded from 114.45)
9: 115.9 (rounded from 115.86)
10: 105.6 (rounded from 105.56)
and of course the unit is cm3
Step-by-step explanation:
(a)

Substitute <em>x</em> = 3 tan(<em>t</em> ) and d<em>x</em> = 3 sec²(<em>t </em>) d<em>t</em> :


(b) The series

converges by comparison to the convergent <em>p</em>-series,

(c) The series

converges absolutely, since

That is, ∑ (-1)ⁿ (<em>n</em> ² + 9)/<em>e</em>ⁿ converges absolutely because ∑ |(-1)ⁿ (<em>n</em> ² + 9)/<em>e</em>ⁿ| = ∑ (<em>n</em> ² + 9)/<em>e</em>ⁿ in turn converges by comparison to a geometric series.