write an equation that is perpendicular to x - 3y =3 and passes through the point (5,-9) in slope-intercept form.

Mathematics

1 answer:

Aleksandr [31]3 years ago

6 0

9514 1404 393

Answer:

y = -3x +6

Step-by-step explanation:

The perpendicular line will have the coefficients of x and y swapped (and one of them negated), and will have a constant appropriate to the point the line needs to go through.

The equation will look like ...

x -3y = 3 . . . . . . . . . . . . . given line

3x +y = constant . . . . . . perpendicular line

where "constant" can be found by substituting for x and y.

3x +y = 3(5) +(-9)

3x +y = 6 . . . . equation in standard form

To put this in slope-intercept form subtract 3x:

y = -3x +6

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The circumference of the ellipse approximate. Which equation is the result of solving the formula of the circumference for b?

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Answer:

$b = \sqrt{\frac{C^{2} }{2(\pi )^{2} } - a^{2}}$

Step-by-step explanation:

Given - The circumference of the ellipse approximated by $C = 2\pi \sqrt{\frac{a^{2} + b^{2} }{2} }$ where 2a and 2b are the lengths of 2 the axes of the ellipse.

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Squaring Both sides, we get

$[\frac{C}{2\pi }]^{2} = [\sqrt{\frac{a^{2} + b^{2} }{2} }]^{2} \\\frac{C^{2} }{(2\pi)^{2} } = {\frac{a^{2} + b^{2} }{2} }\\2\frac{C^{2} }{4(\pi)^{2} } = {{a^{2} + b^{2} }$