Recall the following property of integrals: if f ( t ) ≥ g ( t ) for all t ≥ 0 , then for all x ≥ 0 , ∫ x 0 f ( t ) d t ≥ ∫ x 0 g ( t ) d t 4 the inequality e t ≥ 1 holds for t ≥ 0 because e > 1. use (4) to prove that e x ≥ 1 + x for x ≥ 0 then prove, by successive integration, the following inequalities (for x ≥ 0 ): e x ≥ 1 + x + 1 2 x 2 e x ≥ 1 + x + 1 2 x 2 + 1 6 x 3.

answer: my brother can figure it out, imma go get him so i can answer : ) edit: my brother is sleeping, sorry man

step-by-step explanation:

step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v

step-by-step explanation: