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3 years ago

Which system of equations has a solution of approximately (1.8,-0.9)

Mathematics

2 answers:

Solnce55 [7]3 years ago

3 0

Solve the equation systems or set the values x = 1.8 and y = -0.9 on the equation and check the equality.

6x - 5y = 15

L = 6(1.8) - 5(-0.9) = 10.8 + 4.5 = 15.3

R = 15

L ≈ R

x + 2y = 0

L = 1.8 + 2(-0.9) = 1.8 - 1.8 = 0

L = R

4x + 5y = 8

L = 4(1.8) + 5(-0.9) = 7.2 - 4.5 = 2.7

R = 8

L ≠ R

x - 2y = 4

L = 1.8 - 2(-0.9) = 1.8 + 1.8 = 3.6

R = 4

L ≈ R

4x + 5y = 8

L = 4(1.8) + 5(-0.9) = 7.2 - 4.5 = 2.7

R = 8

L ≠ R

6x - 5y = 15

L = 6(1.8) - 5(-0.9) = 10.8 + 4.5 = 15.3

R = 15

L ≈ R

x - 2y = 4

L = 1.8 - 2(-0.9) = 1.8 + 1.8 = 3.6

R = 4

L ≈ R

We have two probability solutions: A and D.

But I think, Your answer is A.

Andrews [41]3 years ago

3 0

6x - 5y = 15

L = 6(1.8) - 5(-0.9) = 10.8 + 4.5 = 15.3

R = 15

L ≈ R

x + 2y = 0

L = 1.8 + 2(-0.9) = 1.8 - 1.8 = 0

L = R

4x + 5y = 8

L = 4(1.8) + 5(-0.9) = 7.2 - 4.5 = 2.7

R = 8

L ≠ R

x - 2y = 4

L = 1.8 - 2(-0.9) = 1.8 + 1.8 = 3.6

R = 4

L ≈ R

4x + 5y = 8

L = 4(1.8) + 5(-0.9) = 7.2 - 4.5 = 2.7

R = 8

L ≠ R

6x - 5y = 15

L = 6(1.8) - 5(-0.9) = 10.8 + 4.5 = 15.3

R = 15

L ≈ R

x - 2y = 4

L = 1.8 - 2(-0.9) = 1.8 + 1.8 = 3.6

R = 4

L ≈ R

Yeah so your answer is gonna be

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2 years ago

Which of the following statements is true?

sdas [7]

Answer:

Option The product of $(-x+4)(x^2+3x-3)$ is

$-x^3+x^2+15x-12$ is correct.

That is $(-x+4)(x^2+3x-3)=-x^3+x^2+15x-12$

Step-by-step explanation:

Given expression is $(-x+4)(x^2+3x-3)$

To find the product of the given expression :

$(-x+4)(x^2+3x-3)$

( By using the distributive property each term in the factor is multiplied by each term in the another factor )

$=[-x(x^2)+(-x)(3x)+(-x)(-3)]+[4(x^2)+4(3x)+4(-3)]$

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$=-x^3+x^2+15x-12$

Therefore $(-x+4)(x^2+3x-3)=-x^3+x^2+15x-12$

Therefore Option The product of

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2 years ago