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2 years ago

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Find the least common denominator

a1" title="\frac{3}{9} \frac{8}{4}" alt="\frac{3}{9} \frac{8}{4}" align="absmiddle" class="latex-formula">

1 answer:

MA_775_DIABLO [31]2 years ago

8 0

Answer:

36 is your least common denominator. Hope this helps

Step-by-step explanation:

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What value of c makes the statement true?

Vanyuwa [196]

$c-(-17.6)=-4 \\\\ c+17.6=-4 \\\\ c=-4-17.6 \\\\ \boxed{c=-21.6\to \ variant \ A}$

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Given data:

The given cube .

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Step-by-step explanation:

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3 years ago

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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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3 years ago

How can you tell if a system of equations has infinite solutions?

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To tell if an equation has infinite solutions, the equation will be equal to each other.

For example,

x = x
x + 1 = x + 1
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And so on...

And in special cases,

0x = 0
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They have infinite solutions because there's no constant to determine the variable.

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