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3 years ago

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

1 answer:

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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3 years ago

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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3 years ago

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mote1985 [20]

X = A/y
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3 years ago

The historical walking tour of a town covers a total distance of 5 miles. The map shows that the path of the tour travels 3 inch

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Since the problem is given all the total walking distance, the first thing we are going to do is find the total distance in the map:
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We can conclude that each inch the map represents 0.5 miles of walking distance.

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3 years ago

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Margarita [4]

Answer:

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