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3 years ago

Write an equation of a line in point-slope form that is perpendicular to the line y=1/2x-7 and goes through the point (4,-3)

Mathematics

1 answer:

photoshop1234 [79]3 years ago

5 0

Answer:

y = - 2x + 5

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

y = $\frac{1}{2}$ x - 7 ← is in slope- intercept form

with slope m = $\frac{1}{2}$

Given a line with slope m then the slope of a line perpendicular to it is

$m_{perpendicular}$ = - $\frac{1}{m}$ = - $\frac{1}{\frac{1}{2} }$ = - 2, hence

y = - 2x + c ← is the partial equation of the perpendicular line

To find c substitute (4, - 3) into the partial equation

- 3 = - 8 + c ⇒ c = - 3 + 8 = 5

y = - 2x + 5 ← equation of perpendicular line

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Suppose a change of coordinates T:R2→R2 from the uv-plane to the xy-plane is given by x=e−2ucos(5v), y=e−2usin(5v). Find the abs

anzhelika [568]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The solution is $\frac{\delta (x,y)}{\delta (u, v)} | = 10e^{-4u}$

Step-by-step explanation:

From the question we are told that

$x = e^{-2a} cos (5v)$

and $y = e^{-2a} sin(5v)$

Generally the absolute value of the determinant of the Jacobian for this change of coordinates is mathematically evaluated as

$| \frac{\delta (x,y)}{\delta (u, v)} | = | \ det \left[\begin{array}{ccc}{\frac{\delta x}{\delta u} }&{\frac{\delta x}{\delta v} }\\\frac{\delta y}{\delta u}&\frac{\delta y}{\delta v}\end{array}\right] |$

$= |\ det\ \left[\begin{array}{ccc}{-2e^{-2u} cos(5v)}&{-5e^{-2u} sin(5v)}\\{-2e^{-2u} sin(5v)}&{-2e^{-2u} cos(5v)}\end{array}\right] |$

$Let \ a = -2e^{-2u} cos(5v), \\ b=-2e^{-2u} sin(5v),\\c =-2e^{-2u} sin(5v),\\d=-2e^{-2u} cos(5v)$

$\frac{\delta (x,y)}{\delta (u, v)} | = |det \left[\begin{array}{ccc}a&b\\c&d\\\end{array}\right] |$

=> $\frac{\delta (x,y)}{\delta (u, v)} | = | a * b - c* d |$

substituting for a, b, c,d

=> $\frac{\delta (x,y)}{\delta (u, v)} | = | -10 (e^{-2u})^2 cos^2 (5v) - 10 e^{-4u} sin^2(5v)|$

=> $\frac{\delta (x,y)}{\delta (u, v)} | = | -10 e^{-4u} (cos^2 (5v) + sin^2 (5v))|$

=> $\frac{\delta (x,y)}{\delta (u, v)} | = 10e^{-4u}$

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Nataliya [291]

<h3>Answer: C) 4/8 = 0.5</h3>

=========================================================

Explanation:

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sin(angle) = opposite/hypotenuse

sin(J) = KL/LJ

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