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Mandarinka [93]
3 years ago
8

2.

Mathematics
1 answer:
9966 [12]3 years ago
6 0

Answer:

(b) -m, m + 3

Step-by-step explanation:

x² − 3x − m(m + 3) = 0

x² − 3x = m(m + 3)

x² − 3x + 9/4 = m(m + 3) + 9/4

(x − 3/2)² = m(m + 3) + 9/4

(x − 3/2)² = m² + 3m + 9/4

(x − 3/2)² = (m + 3/2)²

x − 3/2 = ±(m + 3/2)

x − 3/2 = m + 3/2, -m − 3/2

x = m + 3, -m

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Find parametric equations and symmetric equations for the line. (Use the parameter t.) The line through (4, −5, 2) and parallel
Nataliya [291]

Answer:

Step-by-step explanation:

From the given information, the symmetric equations for the line pass through(4, -5, 2) i.e (x_o, y_o, z_o) and are parallel to \dfrac{x+5}{1} = \dfrac{y}{2}= \dfrac{z-3}{1}

The parallel vector to the line i + zj+k = ai + bj + ck

Hence, the equation for the line is :

x = x_o + at \\ \\ x = y_o + bt \\ \\ x = z_o + ct

x = 4 + t

y = -5 + 2t

z = 2 + t

Thus, x, y, z = ( 4+t, -5+2t, 2+t )

The symmetric equation can now be as follows:

\begin  {vmatrix} x = 4+ t   \\ \\  \dfrac{x-4}{1} = t  \begin {vmatirx} \end {vmatrix}\begin {vmatrix} y = - 5+2t  \\ \\ \dfrac{y+5}{2}  =t      \end {vmatrix}\begin {vmatrix} z =2+t  \\ \\ \dfrac{z-2}{1}  =t      \end {vmatrix}

∴

\dfrac{x-4}{1}= \dfrac{y+5}{2}=\dfrac{z-2}{1}

8 0
3 years ago
4x-9 , 3x-8 Add<br> cause i'm too tired
Dovator [93]

Answer:

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Step-by-step explanation:

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8 0
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Read 2 more answers
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Check whether the function yequalsStartFraction cosine 2 x Over x EndFraction is a solution of x y prime plus yequalsnegative 2
Jobisdone [24]

The question is:

Check whether the function:

y = [cos(2x)]/x

is a solution of

xy' + y = -2sin(2x)

with the initial condition y(π/4) = 0

Answer:

To check if the function y = [cos(2x)]/x is a solution of the differential equation xy' + y = -2sin(2x), we need to substitute the value of y and the value of the derivative of y on the left hand side of the differential equation and see if we obtain the right hand side of the equation.

Let us do that.

y = [cos(2x)]/x

y' = (-1/x²) [cos(2x)] - (2/x) [sin(2x)]

Now,

xy' + y = x{(-1/x²) [cos(2x)] - (2/x) [sin(2x)]} + ([cos(2x)]/x

= (-1/x)cos(2x) - 2sin(2x) + (1/x)cos(2x)

= -2sin(2x)

Which is the right hand side of the differential equation.

Hence, y is a solution to the differential equation.

6 0
4 years ago
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