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Naddik [55]
2 years ago
7

HELP PLEASE!!!! math questions:

Mathematics
2 answers:
Doss [256]2 years ago
6 0

Answer:

1)Rate of change, intersect, one, graph, intersect  (i'm not really sure but this is what i think the answers are for 1)

Step-by-step explanation:

den301095 [7]2 years ago
3 0
The answer is in you brain sorry ;$
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Which of the following functions are solutions of the differential equation y'' + y = 3 sin x? (select all that apply)
DiKsa [7]

Answer:

C

Step-by-step explanation:

We must compute the derivatives and check if the equation is satisfied.

A. y'=(3\sin x)'=3\cos x. Differentiate again to get y''=(3\cos x)'=-3\sin x, then y''+y=-3\sin x+3\sin x=0\neq 3\sin x so this choice of y doesn't solve the equation.

B. y'=3\sin x+3x\cos x-5\cosx +5x\sin x=(5x+3)\sin x+(3x-5)\cos x and y''=5\sin x+(5x+3)\cos x+3\cos x-(3x-5)\sin x=(10-3x)\sin x+(5x+6)\cos x, then y''+y=10\sin x+6\cos x\neq 3\sin x so y is not a solution

C. y'=\frac{-3}{2}\cos x+\frac{3}{2}x\sin x hence y''=\frac{3}{2}\sin x+\frac{3}{2}\sin x+\frac{3}{2}x\cos x=3\sin x+\frac{3}{2}x\cos x. Then y''+y=3\sin x+\frac{3}{2}x\cos x-\frac{3}{2}x\cos x=3\sin x so y is a solution.

D.y'=-3\sin x and y''=-3\cos x, then y''+y=0 thus y isn't a solution

E. y'=\frac{3}{2}\sin x+\frac{3}{2}x\cos x hence y''=\frac{3}{2}\cos x+\frac{3}{2}\cos x-\frac{3}{2}x\sin x=3\cos x-\frac{3}{2}x\cos x. Then y''+y=3\cos x-\frac{3}{2}x\cos x-\frac{3}{2}x\cos x=3\cos x\neq 3\sin x then y is not a solution.

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Answer:  y= (x+2)² − 5

Step-by-step explanation:

The way I got this answer is by completing the square. The first step though, when looking at this equation, is to see if we can factor it. The way to check is to look at the coefficient for x² which is 1, and the constant, in this case -1. If we multiply those together, we get  −. Now we look at the middle term, 4x. We need to find any numbers that multiply to equal  −  1x² and add to  4x. There aren't any, which means it is not factorable.

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