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Kazeer [188]
2 years ago
14

Evaluate using the values given : 3 pm; use m = 4, and p = 3

Mathematics
1 answer:
Goryan [66]2 years ago
5 0

Answer:

36

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

Step-by-step explanation:

<u>Step 1: Define</u>

3pm

m = 4

p = 3

<u>Step 2: Evaluate</u>

  1. Substitute:                   3(3)(4)
  2. Multiply:                       9(4)
  3. Multiply:                       36
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April carries 5 suitcases to the car.Each suitcase weighs 6 1/3 pounds.How many pounds does April carry in all?
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The answer to your question is 31 2/3.
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3 years ago
Consider the following differential equation to be solved by undetermined coefficients. y(4) − 2y''' + y'' = ex + 1 Write the gi
kompoz [17]

Answer:

The general solution is

y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))

     + \frac{x^2}{2}

Step-by-step explanation:

Step :1:-

Given differential equation  y(4) − 2y''' + y'' = e^x + 1

The differential operator form of the given differential equation

(D^4 -2D^3+D^2)y = e^x+1

comparing f(D)y = e^ x+1

The auxiliary equation (A.E) f(m) = 0

                         m^4 -2m^3+m^2 = 0

                         m^2(m^2 -2m+1) = 0

(m^2 -2m+1) this is the expansion of (a-b)^2

                        m^2 =0 and (m-1)^2 =0

The roots are m=0,0 and m =1,1

complementary function is y_{c} = (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x

<u>Step 2</u>:-

The particular equation is    \frac{1}{f(D)} Q

P.I = \frac{1}{D^2(D-1)^2} e^x+1

P.I = \frac{1}{D^2(D-1)^2} e^x+\frac{1}{D^2(D-1)^2}e^{0x}

P.I = I_{1} +I_{2}

\frac{1}{D^2} (\frac{x^2}{2!} )e^x + \frac{1}{D^{2} } e^{0x}

\frac{1}{D} means integration

\frac{1}{D^2} (\frac{x^2}{2!} )e^x = \frac{1}{2D} \int\limits {x^2e^x} \, dx

applying in integration u v formula

\int\limits {uv} \, dx = u\int\limits {v} \, dx - \int\limits ({u^{l}\int\limits{v} \, dx  } )\, dx

I_{1} = \frac{1}{D^2(D-1)^2} e^x

\frac{1}{2D} (e^x(x^2)-e^x(2x)+e^x(2))

\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))

I_{2}= \frac{1}{D^2(D-1)^2}e^{0x}

\frac{1}{D} \int\limits {1} \, dx= \frac{1}{D} x

again integration  \frac{1}{D} x = \frac{x^2}{2!}

The general solution is y = y_{C} +y_{P}

         y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))

      + \frac{x^2}{2!}

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The required graph is attached.

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Answer:

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