Number of real solution of the equation, (x-3) to the power 2x²-7x+3 = (x²-6x+9) to the power x²-x-6. (A)2 (B)1 (C)0 (D)3​

Graph of liner equation.

⇒ 3x + 4y = 20. - - - - - (1).

⇒ x - 2y = 0. - - - - - (2).

As we know that,

From equation (1), we get.

⇒ 3x + 4y = 20. - - - - - (1).

Taking y - axis it means x = 0.

Put the value of x = 0 in the equation, we get.

⇒ 3(0) + 4y = 20.

⇒ 4y = 20.

⇒ y = 5.

Their Co-ordinates = (0,5).

Taking x - axis it means y = 0.

Put the value of y = 0 in the equation, we get.

⇒ 3x + 4(0) = 20.

⇒ 3x = 20.

⇒ x = 6.67.

Their Co-ordinates = (6.67,0).

From equation (2), we get.

⇒ x - 2y = 0. - - - - - (2).

Taking y - axis it means x = 0.

Put the value of x = 0 in the equation, we get.

⇒ (0) - 2y = 0.

⇒ - 2y = 0.

⇒ y = 0.

Their Co-ordinates = (0,0).

Taking x - axis it means y = 0.

Put the value of y = 0 in the equation, we get.

⇒ x - 2(0) = 0.

⇒ x = 0.

Their Co-ordinates = (0,0).

Both curves intersects at a point = (4,2).

As we know that,

Area of triangle = 1/2 x base x height.

Base = 20/3.

Height = 2.

Using this formula in this equation, we get.

Area of triangle = 1/2 x (20/3) x 2.

Area of triangle = 20/3.

Area of triangle = 6.67 sq. units.

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Given,

x−y+1=0....(1)

3x+2y−12=0....(2)

put x=0 in (1)

0−y+1=0⇒y=1

(0,1).....(a)

put y=0 in (2)

3x+2(0)−12=0→x=4

(4,0).....(b)

put x=0 in (2)

3(0)+2y−12=0→y=6

(0,6)......(c)

put y=0 in (1)

x−0+1=0⇒x=−1

(−1,0)...(d)

From figure, the vertices of triangle are (−1,0),(2,3),(4,0)

Answer:

(5(8

1/3

+27

1/3

)

3

)

1/4

= (5(2+3)

3

)

1/4

⇒ (5(5)

3

)

1/4

⇒ (5×125)

1/4

⇒

4

625

⇒ 5

∴ (5(8

1/3

+27

1/3

)

3

)

1/4

=5

Step-by-step explanation:

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Answer:

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Answer:

The secant line is the red line to the right that passes through two points on the curve. The tangent line is the green line that just grazes the curve at a point.

Explanation:

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Answer:

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