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1 year ago

18

Mathematics

1 answer:

goldfiish [28.3K]1 year ago

8 0

Answer:

Step-by-step explanation:

S.A for open-faced box= lb + 2(bh+lh)

= (7-2h)(7-2h) + 2[h(7-2h)+h(7-2h)]

= (7-2h)² + 2[2h(7-2h)]

= (7-2h)² + 4h(7-2h) in²

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storchak [24]

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3 years ago

Write the given second order equation as its equivalent system of first order equations. u′′−5u′−4u=1.5sin(3t),u(1)=1,u′(1)=2.5

lakkis [162]

Answer:

hi your question options is not available but attached to the answer is a complete question with the question options that you seek answer to

Answer: v = 5v + 4u + 1.5sin(3t),

1.5sin(3t)

Step-by-step explanation:

u" - 5u' - 4u = 1.5sin(3t) where u'(1) = 2.5 u(1) = 1

v represents the "velocity function" i.e v = u'(t)

As v = u'(t)

u' = v

since u' = v

v' = u"

v' = 5u' + 4u + 1.5sin(3t) ( given that u" - 5u' - 4u = 1.5sin(3t) )

= 5v + 4u + 1.5sin(3t) ( noting that v = u' )

so v' = 5v + 4u + 1.5sin(3t)

d/dt $\left[\begin{array}{ccc}u&\\v&\\\end{array}\right]$ = $\left[\begin{array}{ccc}0&1&\\4&5&\\\end{array}\right]$ $\left[\begin{array}{ccc}u&\\v&\\\end{array}\right]$ + $\left[\begin{array}{ccc}0&\\1.5sin(3t)&\\\end{array}\right]$

Given that u(1) = 1 and u'(1) = 2.5

since v = u'

v(1) = 2.5

note: the initial value for the vector valued function is given as

$\left[\begin{array}{ccc}u(1)&\\v(1)\\\end{array}\right]$ = $\left[\begin{array}{ccc}1\\2.5\\\end{array}\right]$

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ss7ja [257]

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Step-by-step explanation: please see attachment for explanation. Thanks.!