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kykrilka [37]
3 years ago
13

10d - 6d = 4 find what d equals

Mathematics
2 answers:
nevsk [136]3 years ago
8 0

Answer:

  • d = 1

Step-by-step explanation:

  • 10d - 6d = 4
  • 4d = 4
  • d = 4/4
  • d = 1
Nutka1998 [239]3 years ago
5 0

Answer:

\huge\boxed{\sf d = 1}

Step-by-step explanation:

10d - 6d = 4

Take d common

d (10 - 6) = 4

d (4) = 4

4d = 4

Divide both sides by 4

d = 1

\rule[225]{225}{2}

Hope this helped!

<h3>~AH1807</h3>
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The second-order equation as its equivalent system of first-order equations is

\left[\begin{array}{ccc}u(1)\\\\v(1)\end{array}\right] = \left[\begin{array}{ccc}7.5\\\\9\end{array}\right]

An equivalent system that has the identical answer is known as an equivalent structure. Given a gadget of two equations, we can produce an equal system by way of replacing one equation by means of the sum of the 2 equations, or by way of changing an equation by means of a couple of of itself.

Systems of linear equations are equivalent if and handiest in the event that they have an equal set of solutions. In other phrases, two systems are equal if and only if each answer of one in all of them is likewise a solution of the opposite.

In the structures sciences, a machine equivalent system is the conduct of a parameter or thing of a machine in a way just like a parameter or component of a distinctive system. Similarity means that mathematically the parameters and additives will be indistinguishable from each different.

Taking v = u, we have:

u" + 4u' + 6u = 4sin(3t)

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So the system of equations is:

u' = 0u + 1v

v' = -6u - 4v + 4sin(3t)

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u(1) = 7.5

v(1) = u'(1) = 9

So the initial condition matrix is:

\left[\begin{array}{ccc}u(1)\\\\v(1)\end{array}\right] = \left[\begin{array}{ccc}7.5\\\\9\end{array}\right]

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