Y = C1 cos 2x + c2 sin 2x is a two-parameter family of solutions of the second-order DE y'" + 4y = 0. If possible, find a solution of the differential equation that satisfies the given side conditions. The conditions specified at two different points are called boundary conditions. 0, y(T) 0 40. y(0) 39. y(0) 0, y(T/4) 3 42. y(0) = 1, y'(m) - 41. y'(0) 0, y'(7/6) 0 44. y'(T/2) 1, y'(m) = 0 43. y(0) 0, y(T) = 2

d (5 x (-20)) * (100 x (0.7))

step-by-step explanation:

5÷1/100 * -.7 *-20

copy dot flip

5*100 * -.7 *-20

rearrange

5*-20   * 100 *-.7

-100   *-70

7000

answer: the answer is one of

step-by-step explanation:   we are given to write a proper fraction where the whole is divided into an even number of equal parts less than ten, and an odd number of those parts exist.

an even number less than 10 may be one of 2, 4, 6 or 8. so, the whole may be one of these four evens.

now, an odd number of these parts must exist, so the numerator must be an odd number, one of 1, 3, 5 or 7. it cannot be 9 because then the fraction will become improper.

thus, our fraction may be one of the following

2 and 1.

step-by-step explanation:

plug in each answer. 0 doesn't work, 1 does. so does 2. 4 doesn't:

0 = 1 is not true.

2 = 2 is true.

4 = 4 is true.

8 = 16 is not true.

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