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telo118 [61]
2 years ago
12

Tell whether the equation has one, zero, or infinitely many solutions.9(x − 3) + 15 = 9x − 11.

Mathematics
2 answers:
romanna [79]2 years ago
7 0

Answer:

No solution

Step-by-step explanation:

9(x-3)+15= 9x-11

9x-27+15=9x-11

-15 on both sides

9x-27=9x-26

add 26 on both sides

9x -1=9x

add 1 on both sides

9x=9x+1

divide 9 on both sides

x=x+1/9

No solution

V125BC [204]2 years ago
5 0

Answer:

Zero solutions; there are no variables and thus there are no solutions

Step-by-step explanation:

We start out with the equation 9 (x - 3) + 15 = 9x - 11

Which simplifies to 9x - 27 + 15 = 9x - 11

The 9x terms cancel each other out, and we're left with 15 - 27 = -11

And we end up with -12 = -11, which is not true.

This tells us that the equation has zero solutions, because no matter what value we substitute for x, the equation will not be true.

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Q.6. The equation of the ellipse whose centre is at the origin and the x-axis, the major axis, which passes
azamat

<h3>Answer:</h3>

Equation of the ellipse = 3x² + 5y² = 32

<h3>Step-by-step explanation:</h3>

<h2>Given:</h2>

  • The centre of the ellipse is at the origin and the X axis is the major axis

  • It passes through the points (-3, 1) and (2, -2)

<h2>To Find:</h2>

  • The equation of the ellipse

<h2>Solution:</h2>

The equation of an ellipse is given by,

\sf \dfrac{x^2}{a^2} +\dfrac{y^2}{b^2} =1

Given that the ellipse passes through the point (-3, 1)

Hence,

\sf \dfrac{(-3)^2}{a^2} +\dfrac{1^2}{b^2} =1

Cross multiplying we get,

  • 9b² + a² = 1 ²× a²b²
  • a²b² = 9b² + a²

Multiply by 4 on both sides,

  • 4a²b² = 36b² + 4a²------(1)

Also by given the ellipse passes through the point (2, -2)

Substituting this,

\sf \dfrac{2^2}{a^2} +\dfrac{(-2)^2}{b^2} =1

Cross multiply,

  • 4b² + 4a² = 1 × a²b²
  • a²b² = 4b² + 4a²-------(2)

Subtracting equations 2 and 1,

  • 3a²b² = 32b²
  • 3a² = 32
  • a² = 32/3----(3)

Substituting in 2,

  • 32/3 × b² = 4b² + 4 × 32/3
  • 32/3 b² = 4b² + 128/3
  • 32/3 b² = (12b² + 128)/3
  • 32b² = 12b² + 128
  • 20b² = 128
  • b² = 128/20 = 32/5

Substituting the values in the equation for ellipse,

\sf \dfrac{x^2}{32/3} +\dfrac{y^2}{32/5} =1

\sf \dfrac{3x^2}{32} +\dfrac{5y^2}{32} =1

Multiplying whole equation by 32 we get,

3x² + 5y² = 32

<h3>Hence equation of the ellipse is 3x² + 5y² = 32</h3>
8 0
3 years ago
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