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3 years ago

Convert the equation to slope- intercept and standard form. y-4=5 + 1/2x

Mathematics

1 answer:

Vitek1552 [10]3 years ago

4 0

Answer:

note :

The equation of a linear function in point-slope form is y – y1 = m(x – x1)

The point is A (x1 , y1)

Step-by-step explanation:

in this exercice : y - 4 = 5 +(1/2)x

y - 4 = 0.5 ( x - 5/0.5)

y - 4 = 0.5( x - 10).....

note : the standard forme is : y = ax +b

in this exercice : y = (1/2)x+9

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Which of the following is an example of a variable expense?

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Answer:

Step-by-step explanation:

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3 years ago

Which of the following is the standard equation of the ellipse with vertices at (1,0) and (27,0) and an eccentricity of 5/13?

Dovator [93]

The equation of the elipse is given by:

$\frac{(x - 14)^2}{169} + \frac{y^2}{144} = 1$

The equation of an elipse of center $(x_0, y_0)$ is given by:

$\frac{(x - x_0)^2}{a^2} + \frac{(y - y_0)^2}{b^2} = 0$

Values a and b are found according to the <u>vertices and the eccentricity</u>.

It has vertices at (1,0) and (27,0), thus:

$x_0 = \frac{27 + 1}{2} = 14$

$y_0 = \frac{0 + 0}{2} = 0$

$a = \frac{27 - 1}{2} = 13$

$a^2 = 169$

It has eccentricity of $\frac{5}{13}$ , thus:

$\frac{5}{13} = \frac{c}{a}$

$\frac{5}{13} = \frac{c}{13}$

$c = 13$

Thus, b is given according to the following equation:

$c^2 = a^2 - b^2$

$b^2 = a^2 - c^2$

$b^2 = 169 - 25$

$b = \sqrt{144}$

$b = 12$

The equation of the elipse is:

$\frac{(x - 14)^2}{169} + \frac{y^2}{144} = 1$

A similar problem is given at brainly.com/question/21405803

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