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Amanda [17]
2 years ago
6

8. Which of the following does NOT deseribe a

Mathematics
1 answer:
topjm [15]2 years ago
7 0

Answer:

B. a decimal that does not terminate and does

not repeat.

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Really need help on this , thanks so much! (:
Mazyrski [523]
-2y>12x-42
y<21-6x
so a is true
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3 years ago
If /|| m, what is the value of X?
Elis [28]

Answer:

43° is correct answer

Step-by-step explanation:

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Keith receives $25000 salary for working as a account. If keith spends 40% of his salary on expenses each year, then how much mo
Scrat [10]

Answer:

$25000 * 40% = 100.000

Step-by-step explanation:

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3 years ago
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5 is what percent of 23
Eddi Din [679]
5 = what percent of 23

5 = x% of 23

5 = (x/100)*23

5 = 23x/100

5*100 = 23x

23x = 5*100

x = 5*100/23

x = 21.739

5 is ≈ 21.739% of 23 
8 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
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