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Lesechka [4]
3 years ago
11

How do you solve 2x^2 - 7x -5= 0 in quadratic form?

Mathematics
2 answers:
lesantik [10]3 years ago
7 0

Answer:

Once the equation is in standard form, factor the quadratic expression. 2x2 + 7x + 3 = 0 (2x + 1)(x + 3) = 0. Using the Zero Product Property set ...

2x2 + 7x = -3

2x2 + 7x + 3 = 0

Once the equation is in standard form, factor the quadratic expression.

2x2 + 7x + 3 = 0

(2x + 1)(x + 3) = 0

Using the Zero Product Property set each factor equal to 0 and solve for x.

2x + 1 = 0  

2x + 1 - 1 = 0 - 1 x + 3 = 0

2x = -1 x + 3 - 3 = 0 - 3

2x 2 = -1 2 x = -3

x = -1 2  

The solutions to the equation are -1 2 and -3.

Ray Of Light [21]3 years ago
5 0

Answer:

x^1 = 7-square root89/4

x^2 = 7+square root89/4

Step-by-step explanation:

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Step-by-step explanation:

6 0
2 years ago
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If nelson invests $5000 for 4 years and earns $600, what is the simple interest rate?
klio [65]

Answer: Simple interest rate = 3%.


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Simple interest formula.

I = P R T.

Where I is the earned interest, P is the invested amount , R is the rate of interest and T is the time in years.

Plugging values in formula, we get

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4 0
3 years ago
What degree of rotation is represented on this matrix
Korvikt [17]

Answer:

Option B is correct

the degree of rotation is, -90^{\circ}

Step-by-step explanation:

A rotation matrix is a matrix that is used to perform a rotation in Euclidean space.

To find the degree of rotation using a standard rotation matrix i.e,

R = \begin{bmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}

Given the matrix: \begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}

Now, equate the given matrix with standard matrix we have;

\begin{bmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} =  \begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}

On comparing we get;

\cos \theta = 0       and -\sin \theta =1  

As,we know:

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\cos \theta = \cos(90^{\circ}) = \cos( -90^{\circ})

we get;

\theta = -90^{\circ}

and

\sin \theta =- \sin (90^{\circ}) = \sin ( -90^{\circ})

we get;

\theta = -90^{\circ}

Therefore, the degree of rotation is, -90^{\circ}

7 0
3 years ago
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