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SVETLANKA909090 [29]
3 years ago
7

Third-degree, with zeros of -3,-1 and 2, and passes through the point (4,7)

Mathematics
1 answer:
Leokris [45]3 years ago
7 0
<h2>Required Expression is \frac{\textbf{1}}{\textbf{10}}(\textbf{x}^{\textbf{3}}\textbf{+}\textbf{2x}^{\textbf{2}}\textbf{-}\textbf{5x}\textbf{-}\textbf{6})</h2>

Step-by-step explanation:

       Assuming that the question is to find a third degree expression f(x) that has zeros -3,-1,2 and the equation y=f(x) passes through (4,7),

       If the roots/zeroes of a n^{th} order expression are given as r_{1},r_{2},r_{3}..... r_{n}, the expression is given by f(x)=c(x-r_{1})(x-r_{2})(x-r_{3}).....(x-r_{n}).

       Since we know the three roots of the third degree expression, the function is

f(x)=c(x-(-3))(x-(-1))(x-2)=c(x+3)(x+1)(x-2)=c(x^{3}+2x^{2}-5x-6)

       Also, y=f(x) passes through (4,7), so

7=c(4^{3}+2(4)^{2}-5(4)-6)\\7=c(64+32-20-6)=70c\\c=\frac{1}{10}

∴Required expression is \frac{1}{10}(x^{3}+2x^{2}-5x-6)

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x = c1 cos(t) + c2 sin(t) is a two-parameter family of solutions of the second-order DE x'' + x = 0. Find a solution of the seco
igomit [66]

Answer:

x=-cos(t)+2sin(t)

Step-by-step explanation:

The problem is very simple, since they give us the solution from the start. However I will show you how they came to that solution:

A differential equation of the form:

a_n y^n +a_n_-_1y^{n-1}+...+a_1y'+a_oy=0

Will have a characteristic equation of the form:

a_n r^n +a_n_-_1r^{n-1}+...+a_1r+a_o=0

Where solutions r_1,r_2...,r_n are the roots from which the general solution can be found.

For real roots the solution is given by:

y(t)=c_1e^{r_1t} +c_2e^{r_2t}

For real repeated roots the solution is given by:

y(t)=c_1e^{rt} +c_2te^{rt}

For complex roots the solution is given by:

y(t)=c_1e^{\lambda t} cos(\mu t)+c_2e^{\lambda t} sin(\mu t)

Where:

r_1_,_2=\lambda \pm \mu i

Let's find the solution for x''+x=0 using the previous information:

The characteristic equation is:

r^{2} +1=0

So, the roots are given by:

r_1_,_2=0\pm \sqrt{-1} =\pm i

Therefore, the solution is:

x(t)=c_1cos(t)+c_2sin(t)

As you can see, is the same solution provided by the problem.

Moving on, let's find the derivative of x(t) in order to find the constants c_1 and c_2:

x'(t)=-c_1sin(t)+c_2cos(t)

Evaluating the initial conditions:

x(0)=-1\\\\-1=c_1cos(0)+c_2sin(0)\\\\-1=c_1

And

x'(0)=2\\\\2=-c_1sin(0)+c_2cos(0)\\\\2=c_2

Now we have found the value of the constants, the solution of the second-order IVP is:

x=-cos(t)+2sin(t)

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Answer:

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Measures of the opposite angles of a rhombus are equal.

Therefore,

m \angle \: 1 = 120 \degree

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