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lapo4ka [179]
3 years ago
9

The algebraic expression 6x2 + 9 x + 3 represents the area of a rectangle. What is the area of the rectangle when x = 3m?​

Mathematics
1 answer:
ioda3 years ago
8 0

The area of the rectangle calculate using the algebraic model given is 84 m²

<u>Given the algebraic representation </u><u>for</u><u> </u><u>the area of a rectangle</u> :

  • Area of rectangle = 6x² + 9x + 3

<u>The Area of the rectangle when x = 3 can be calculated thus</u> :

Substitute the value x = 3 into the equation :

x = 3

Area of rectangle = 6(3)² + 9(3) + 3

Area of rectangle = 54 + 27 + 3 = 84 m²

Therefore, the area of the rectangle is 84 m²

Learn more :brainly.com/question/18112348

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Consider the following higher-order differential equation. d3u dt3 + d2u dt2 − 2u = 0 Find all the roots of the auxiliary equati
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Answer:

Therefore the auxiliary solution is

y=C_1e^{t}+e^{-t}[C_2sin \ t+C_3 cos \ t]

Step-by-step explanation:

To find auxiliary equation we have to put u=e^{mt} in the given differential equation.

The degree of the differential equation is 3.

Therefore the number of root of the differential equation is 3.

Let  \lambda_1, \lambda_2 and \lambda_3 be three roots of the auxiliary equation.

  • If three roots are real and equal.

Then y= e^{\lambda_1t} (C_1+C_2t+C_3t^2)

  • If three roots are real and distinct.

Then y=C_1e^{\lambda_1t}+C_2e^{\lambda_2t}+C_3e^{\lambda_3x}

  • If two roots imaginary and one root real , \lambda_1= a+ib \ and \ \lambda_2= a-ib  

Then y=C_1e^{\lambda_3t}+e^{at}(C_2sin \ bt+C_3cos \ bt)

Now, u=e^{mt},u'=me^{mt},u"=m^2e^{mt}  \ and \ u'"= m^3e^{mt}

Given differential equation is

\frac{d^3u}{dt^3} +\frac{d^2u}{dt^2}-2u=0

The auxiliary equation is

(m^3+m^2-2)e^{mt}=0

\Rightarrow (m^3+m^2-2)=0

\Rightarrow m^3-m^2+2m^2-2m+2m-2=0

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

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

\Rightarrow m= 1,\frac{-2\pm\sqrt{2^2-4.2.1} }{2.1}

\Rightarrow m= 1,\frac{-2\pm\sqrt{-4} }{2.1}

\Rightarrow m= 1,-1\pm i

Here \lambda_1= -1+i  , \lambda_2=- 1-i \ and \ \lambda_3=1

Therefore ,

y=C_1e^{1.t}+e^{-1.t}[C_2sin \ (1.t)+C_3 cos \ (1.t)]

\Rightarrow y=C_1e^{t}+e^{-t}[C_2sin \ t+C_3 cos \ t]

5 0
4 years ago
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