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3 years ago

Brainlist to whoever shows me how to solve this step by step.

Mathematics

1 answer:

Andrews [41]3 years ago

4 0

Answer:

x ∈ {-4, 2}

Step-by-step explanation:

The x-coefficient (+2) is the sum of the constants in the binomial factors of the equation, and the constant term (-8) is their product.

Since you're familiar with the divisors of 8, you know that the two factors of -8 that total +2 are -2 and +4. These are the constants in the binomial factors:

f(x) = x^2 +2x -8

f(x) = (x -2)(x +4)

The roots of f(x) are the values of x that make these factors be zero. They are x=2 and x=-4.

The roots of f(x) are -4 and 2.

_____

Above is the solution by factoring. We can also "complete the square" to find the roots. Anytime we're looking for roots, we want the values of x that make f(x) = 0.

x^2 +2x -8 = 0

x^2 +2x = 8 . . . . . add the opposite of the constant

x^2 +2x +1 = 8 +1 . . . . . add the square of half the x-coefficient: (2/2)² = 1

(x +1)² = 9 . . . . . . . . . . . write as squares

x +1 = ±√9 = ±3 . . . . . . take the square root

x = -1 ± 3 . . . . . . . . . . . subtract 1

The roots are x=-4, x=2.

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3 years ago

How much is 7 cups in pints<br> Write your answer as a whole or mixed number in simplest form

katrin [286]

The answer is: 3 1/2

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3 years ago

Which answer choice do I choose?

Ratling [72]

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3 years ago

Use the method of undetermined coefficients to solve the given nonhomogeneous system. x' = −1 5 −1 1 x + sin(t) −2 cos(t)

AlekseyPX

It looks like the system is

$x' = \begin{bmatrix} -1 & 5 \\ -1 & 1 \end{bmatrix} x + \begin{bmatrix} \sin(t) \\ -2 \cos(t) \end{bmatrix}$

Compute the eigenvalues of the coefficient matrix.

$\begin{vmatrix} -1 - \lambda & 5 \\ -1 & 1 - \lambda \end{vmatrix} = \lambda^2 + 4 = 0 \implies \lambda = \pm2i$

For $\lambda = 2i$ , the corresponding eigenvector is $\eta=\begin{bmatrix}\eta_1&\eta_2\end{bmatrix}^\top$ such that

$\begin{bmatrix} -1 - 2i & 5 \\ -1 & 1 - 2i \end{bmatrix} \begin{bmatrix} \eta_1 \\ \eta_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$

Notice that the first row is 1 + 2i times the second row, so

$(1+2i) \eta_1 - 5\eta_2 = 0$

Let $\eta_1 = 1-2i$ ; then $\eta_2=1$ , so that

$\begin{bmatrix} -1 & 5 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 - 2i \\ 1 \end{bmatrix} = 2i \begin{bmatrix} 1 - 2i \\ 1 \end{bmatrix}$

The eigenvector corresponding to $\lambda=-2i$ is the complex conjugate of $\eta$ .

So, the characteristic solution to the homogeneous system is

$x = C_1 e^{2it} \begin{bmatrix} 1 - 2i \\ 1 \end{bmatrix} + C_2 e^{-2it} \begin{bmatrix} 1 + 2i \\ 1 \end{bmatrix}$

The characteristic solution contains $\cos(2t)$ and $\sin(2t)$ , both of which are linearly independent to $\cos(t)$ and $\sin(t)$ . So for the nonhomogeneous part, we consider the ansatz particular solution

$x = \cos(t) \begin{bmatrix} a \\ b \end{bmatrix} + \sin(t) \begin{bmatrix} c \\ d \end{bmatrix}$

Differentiating this and substituting into the ODE system gives

$-\sin(t) \begin{bmatrix} a \\ b \end{bmatrix} + \cos(t) \begin{bmatrix} c \\ d \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ -1 & 1 \end{bmatrix} \left(\cos(t) \begin{bmatrix} a \\ b \end{bmatrix} + \sin(t) \begin{bmatrix} c \\ d \end{bmatrix}\right) + \begin{bmatrix} \sin(t) \\ -2 \cos(t) \end{bmatrix}$

$\implies \begin{cases}a - 5c + d = 1 \\ b - c + d = 0 \\ 5a - b + c = 0 \\ a - b + d = -2 \end{cases} \implies a=\dfrac{11}{41}, b=\dfrac{38}{41}, c=-\dfrac{17}{41}, d=-\dfrac{55}{41}$

Then the general solution to the system is

$x = C_1 e^{2it} \begin{bmatrix} 1 - 2i \\ 1 \end{bmatrix} + C_2 e^{-2it} \begin{bmatrix} 1 + 2i \\ 1 \end{bmatrix} + \dfrac1{41} \cos(t) \begin{bmatrix} 11 \\ 38 \end{bmatrix} - \dfrac1{41} \sin(t) \begin{bmatrix} 17 \\ 55 \end{bmatrix}$

7 0

2 years ago

Just please help help help help

ludmilkaskok [199]

Answer:

C.{(-3,10),(-2,5),(1,2),(-2,-4)}

7 0

3 years ago