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Feliz [49]
3 years ago
11

Identify the graph of the equation. What is the angle of rotation for the equation?xy=-2.5

Mathematics
2 answers:
AveGali [126]3 years ago
7 0

Answer:

The correct option is b

Step-by-step explanation:

The given equation is

xy=-2.5

It can be written as

xy+2.5=0              .... (1)

The general forms of conic is

Ax^2+Bxy+Cy^2+Dx+Ey+F=0             .... (2)

From (1) and (2), we get

A=0,B=1,C=0,D=0,E=0,E=2.5

B^2-4AC=1-4(0)(0)=1>0

Since the value of B²- 4AC > 0, then it is hyperbola.

The formula form angle of rotation is

\tan 2\theta=\frac{B}{A-C}

\tan 2\theta=\frac{1}{0-0}

\tan 2\theta=\infty

\tan 2\theta=\tan (90^{\circ})

2\theta=90^{\circ}

\theta=45^{\circ}

The angle of rotation is 45°. Therefore the correct option is b.

Maurinko [17]3 years ago
3 0

Answer:

It is B. hyperbola, 45 degrees.

SteIt is p-by-step explanation:

If we rotate the standard form  x^2 - y^2 = 1 through 45 degrees we get xy = 1/2.

xy = -2.5 comes from x^2 - y^2 = -5  being rotated 45 degrees.

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Match each equation with its solution set. Tiles a2 − 9a + 14 = 0 a2 + 9a + 14 = 0 a2 + 3a − 10 = 0 a2 + 5a − 14 = 0 a2 − 5a − 1
sattari [20]
We have that

N 1)
a²<span> − 9a + 14 = 0 
</span>

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² − 9a)=-14

Complete the square  Remember to balance the equation by adding the same constants to each side 

(a² − 9a+20.25)=-14+20.25

Rewrite as perfect squares

(a-4.5)²=6.25--------> (a-4.5)=(+/-)√6.25

a1=4.5+√6.25-----> a1=7

a2=4.5-√6.25-----> a2=2

the solution problem N 1 is the pair {7, 2}


N 2) 

a²<span> + 9a + 14 = 0
</span>

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² + 9a)=-14

Complete the square  Remember to balance the equation by adding the same constants to each side 

(a² +9a+20.25)=-14+20.25

Rewrite as perfect squares

(a+4.5)²=6.25--------> (a+4.5)=(+/-)√6.25

a1=-4.5+√6.25-----> a1=-2

a2=-4.5-√6.25-----> a2=-7

the solution problem N 2 is the pair {-2,-7}

N 3) 

a² + 3a − 10 = 0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² + 3a)=10

Complete the square  Remember to balance the equation by adding the same constants to each side 

(a² + 3a+2.25)=10+2.25

Rewrite as perfect squares

(a+1.5)²=12.25------> (a+1.5)=(+/-)√12.25

a1=-1.5+√12.25-----> a1=2

a2=-1.5-√12.25-----> a2=-5

the solution problem N 3 is the pair {2, -5}


N 4)

a²<span> + 5a − 14 = 0
</span>

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² + 5a) =14

Complete the square  Remember to balance the equation by adding the same constants to each side 

(a² + 5a+6.25) =14+6.25

Rewrite as perfect squares

(a+2.5)² =20.25-------> (a+2.5)=(+/-)√20.25

a1=-2.5+√20.25-----> a1=2

a2=-2.5-√20.25-----> a2=-7

the solution problem N 4 is the pair {2, -7}


N 5) 

a² − 5a − 14 = 0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² − 5a)=14

Complete the square  Remember to balance the equation by adding the same constants to each side 

(a² − 5a+6.25)=14+6.25

Rewrite as perfect squares

(a-2.5)²=2025--------> (a-2.5)=(+/-)√20.25

a1=2.5+√20.25-----> a1=7

a2=2.5-√20.25-----> a2=-2

the solution problem N 5 is the pair {7, -2}

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