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Lynna [10]
3 years ago
7

Can someone please help me

Mathematics
1 answer:
lions [1.4K]3 years ago
8 0

Problem 1

<h3>Answer:  Choice B) x = -2</h3>

------------------

Explanation:

Add the equations straight down.

The x terms add to 5x+(-3x) = 2x

The y terms cancel out since 3y+(-3y) = 0y = 0

The right hand sides add to -7+3 = -3

After adding the equations you should get 2x = -4 which solves to x = -2

=====================================================

Problem 2

<h3>Answer:  y = 4</h3>

------------------

Explanation:

Since 5x = 0, this means x = 0 after dividing both sides by 5.

Use this to find y

y = -4x+4

y = -4(0)+4

y = 4

If you want the value of z, then

x = -3y+z+6

0 = -3(4)+z+6

0 = z-6

6 = z

z = 6

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Consider the initial value problem:
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Step-by-step explanation:

The given initial value problem is;

y''-5y'+6y=-5\sin(2t)---(1)\\y(0)=-5,y'(0)=2

Let

u(t)=y(t)----(2)\\v(t)=y'(t)----(3)

Differentiating both sides of equation (1) with respect to t, we obtain:

u'(t)=y'(t)---(4)\\ \\\implies u'(t)=v(t)---(5)

Differentiating both sides of equation (2) with respect to t gives:

v'(t)=y''(t)----(6)

From equation (1),

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Putting t=0 into equation (2) yields

u(0)=y(0)\\\implies u(0)=-5

Also putting t=0 into equation (3)

u'(0)=y'(0)\\u'(0)=2

The system of first order equations is:

\left \{ {{u'=v} \atop {v'=5v-6u-5\sin(2t)}} \right. \\u(0)=-5;u'(0)=2

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