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bixtya [17]
3 years ago
5

How do You do solving system of equations by solving for a variable

Mathematics
1 answer:
In-s [12.5K]3 years ago
6 0
Well, you have to do it with the substitution method now it's not easy, it takes some time to get used to it. I'm not really sure how to explain it but if you go on Khan Academy it will show you how to do it.
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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
I need your help please I don’t understand this
Step2247 [10]
The answer is y = -5/7x + 4/7! The first step is to subtract the 5x so it’s on the other side of the equal sign. So now the equation is 7y = -5x + 4. Then divide each of them for 7
3 0
3 years ago
Read 2 more answers
What is the simplest form of 3 square root 27 a^3b^7?
4vir4ik [10]
\sqrt[3]{27a^3b^7} = \sqrt[3]{3^3 \cdot a^3 \cdot b^6 \cdot b^} = 3ab^2(\sqrt[3]{b} )
5 0
3 years ago
can you help me evaluate this: 7m + 14m - 6n - 5n + 2m. why do you add 6n with 5n instead of subtract?
Mademuasel [1]

Step-by-step explanation:

The reason is because it is not just 6 it is - 6 so when there are two negative numbers they don't subtract they add if it was 6n then it was to be subtracted

21m - 11n + 2m

23m - 11n

6 0
3 years ago
491,852 nearest hundred thousand
r-ruslan [8.4K]

500,000

Rounded to the hundred thousands

6 0
3 years ago
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