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bcox32314
02.03.2020 •
Mathematics
EXAMPLE 1 Show that every member of the family of functions y = 1 + cet 1 − cet is a solution of the differential equation y' = 1 2 (y2 − 1). SOLUTION We use the Quotient Rule to differentiate the expression for y: y' = (1 − cet) − (1 + cet)(−cet) (1 − cet)2 = − c2e2t + cet + c2e2t (1 − cet)2 = (1 − cet)2 . The right side of the differential equation becomes 1 2 (y2 − 1) = 1 2 2 − 1 = 1 2 (1 + cet)2 − 2 (1 − cet)2 = 1 2 (1 − cet)2 = (1 − cet)2 . Therefore, for every value of c, the given function is a solution of the differential equation.
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Ответ:
3
Step-by-step explanation:
Im assuming you meant 8ab/2bc
So adding all the numbers you'll have
8x3x12/2x12x4
You first reduce the 12 in the fractions making it
8x3/2x4
and then you reduce the 8 and 2 making it 4
so you'll have
4x3/4
One last time you reduce the two fours making the
3