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Iteru [2.4K]
2 years ago
9

Someone help me please

Mathematics
2 answers:
GenaCL600 [577]2 years ago
4 0

Answer:

3x + 5y = 15

Find gradient:

coordinates taken: (5, 0), (0, 3)

\rightarrow \sf \dfrac{y_2-y_1}{x_2-x_1}

\rightarrow \sf \dfrac{3-0}{0-5}

\rightarrow \sf -\dfrac{3}{5}

  • Here the y-intercept: 3
  • gradient: -3/5

equation:

\sf \bold{ y = m(x) + b}

\rightarrow \sf y = -\dfrac{3}{5} x + 3

\rightarrow  \sf y = -\dfrac{3}{5} x + \dfrac{15}{5}

\rightarrow  \sf 5y =-3x+15

\rightarrow  \sf 3x+5y =15

viva [34]2 years ago
3 0

Take two points

  • (0,3)
  • (5,0)
  • x intercept=a=5
  • y intercept=b=3

So

equation of line in intercept form

\\ \rm\rightarrowtail \dfrac{x}{a}+\dfrac{y}{b}=1

\\ \rm\rightarrowtail \dfrac{x}{5}+\dfrac{y}{3}=1

  • Multiply each by 15

\\ \rm\rightarrowtail 3x+5y=15

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Answer:

The general solution of the given differential equation

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Step-by-step explanation:

Step(I):-

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The auxiliary equation

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        we will use formula  ( a + b)² = a² + 2 a b + b²    

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                      m^{2} =-9\\m= - 3i and m=3i

            m² + 9 = 0      

      m² = -9\\m= -3i and m=3i

The complex roots are   0± 3 i ,0 ± 3 i

<em>Step(ii)</em>:-

The complementary function  

              y = e^{\alpha x } ( c_{1} + c_{2} x ) cos\beta x + (  c_{3} +c_{4} x) sin\beta x

The general solution of the given differential equation

          y = e^{0 x } ( c_{1} + c_{2} x ) cos3 x + (  c_{3} +c_{4} x) sin3 x

The general solution of the given differential equation

   y =  ( c_{1} + c_{2} x ) cos3 x + (  c_{3} +c_{4} x) sin3 x    



             



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