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nirvana33 [79]
3 years ago
7

Arrange the functions in descending order, starting with the function that eventually has the greatest value and ending with the

function that eventually has the least value.
4x+9
9^x+1
x^2+3
4x
9^x
x^2
Mathematics
1 answer:
Marina86 [1]3 years ago
8 0
I'll just assign a value of x and solve it so that I'll be able to arrange them in order from greatest to least.

I am assuming that x = 2.

4x + 9 ⇒ 4(2) + 9 = 8 + 9 = 17
9^x + 1 ⇒ 9² + 1= 81 + 1 = 82
x² + 3 ⇒ 2² + 3 = 4 + 3 = 7
4x ⇒ 4(2) = 8
9^x ⇒ 9² = 81
x² ⇒ 2² = 4

based on my assumption, the order from greatest value to least value is:
1) 9^x + 1
2) 9^x
3) 4x + 9
4) 4x
5) x² + 3
6) x²
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the missing is 4.98

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Laura and Martin obtain a 30-year, $180,000 conventional mortgage at 10.0% on a house selling for $210,000. Their monthly mortga
Westkost [7]

Answer:

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Step-by-step explanation:

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4 0
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),

  • 0
  • 1
  •  4
  • 5
  • 0
  • 1.5sin(3t)
  • 1
  •  2.5

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)

<em>u' = v</em>

since <em>u' = v </em>

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]

7 0
3 years ago
HELP ASAP WILL GIVE BRAINLIST
Basile [38]

The answer to your question is 0                                                                                                                



6 0
3 years ago
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