Lemma 3 (burkholder-davis-gundy). let ? be a nondecreasingfunction on [0, oo], finite and convex on [0, oo) with d(0) 0, ? (00) and ? (2) cd(a) for all ? > 0 and sonne ce (0.00). then there exists a constant b (0,00) such that for every sequence {zn, n> 1} of non- negative measurable functions on (2, f, p) ? (a-) oo,

there are many solutions.

step-by-step explanation:

step 1: simplify both sides of the equation.

4x+6=4x^2+10x−4

step 2: subtract 4x^2+10x-4 from both sides.

4x+6−(4x^2+10x−4)=4x^2+10x−4−(4x^2+10x−4)

−4x^2−6x+10=0

step 3: factor left side of equation.

−2(2x+5)(x−1)=0

step 4: set factors equal to 0.

x=-5/2 or x=1

ca = ec

cpctc makes it so that corresponding parts of the triangle you proved to be equal are equal, the other options aren't part of the part of the triangles you proved to be equal.

step-by-step explanation:

well if you round, things become much more simple. round 5.95 to 6, 4.25 to 4, and 1.05 to 1. now, add 6+4+1 and that would equal 11. so the items together would equal to about $11.