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andrezito [222]
2 years ago
7

Sketch the curve represented by the parametric equations. Use arrows to indicate the direction of the curve as t increases. Find

a rectangular-coordinate equation for the curve by eliminating the parameter. Express the vector v with initial point P and terminal point Q in component form. (Assume that each point lies on the gridlines.)

Mathematics
1 answer:
Paul [167]2 years ago
8 0

The <em>lower right</em> image represents the image of the <em>parametric</em> formulas, whose <em>rectangular</em> formula is \frac{x^{2}}{16} + \frac{y^{2}}{25} = 1. The vector in <em>component</em> form is \vec v = (8, -4).

<h3>How to analyze parametric equations and vectors</h3>

<em>Parametric</em> formulas are <em>vectorial</em> expressions in terms of a parameter (t). Planar parametric expression are of the form \vec r(t) = (x(t), y(t)). Ellipses centered at the origin are described by the following expression:

\vec r (t) = (a\cdot \cos t, b \cdot \sin t)     (1)

Where a, b are the lengths of the <em>major</em> and <em>minor</em> semiaxes.

By direct observation of the given <em>parametric</em> equations, we conclude that the ellipse of the lower <em>right</em> image represents the two equations.

The <em>rectangular</em> equation of the ellipse is found by eliminating the parameter:

\cos ^{2}t + \sin ^{2}t = 1  

(\frac{x}{4})^{2} + \left(\frac{y}{5} \right)^{2} = 1

\frac{x^{2}}{16} + \frac{y^{2}}{25} = 1

According to the geometry, vectors can be generated from two points, one of them as the <em>initial</em> point. A vector can be defined as a subtraction between two vectors with <em>initial</em> points at the origin:

\vec v = B(x, y) - A(x, y)     (2)

Where:

  • A(x, y) - Initial point
  • B(x, y) - Final point

If we know that A(x, y) = (1, 8) and B(x, y) = (9, 4), then the equation of the vector is:

\vec v = (9, 4) - (1, 8)

\vec v = (9 - 1, 4 - 8)

\vec v = (8, -4)

To learn more on parametric equations: brainly.com/question/12718642

#SPJ1

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Read 2 more answers
If x&gt;2, then x^2-x-6/x^2-4=
Viefleur [7K]
Consider the expression \frac{x^{2} -x-6}{ x^{2} -4}

To factorize the expression in the denominator we use difference of squares: x^{2} -4=x^{2} - 2^{2} =(x-2)(x+2)

To factorize x^{2} -x-6 we use the following method:

x^{2} -x-6=(x-a)(x-b)

where a, b are 2 numbers such that a+b= -1, the coefficient of x,

and a*b= -6, the constant.

such 2 numbers can be easily checked to be -3 and 2

(-3*2=6, -3+2=-1)

So x^{2} -x-6=(x-a)(x-b)=(x+3)(x-2)

&#10; \frac{x^{2} -x-6}{ x^{2} -4}= \frac{(x+3)(x-2)}{(x-2)(x+2)}= \frac{x+3}{x+2}


\frac{x+3}{x+2}= \frac{x+2+1}{x+2}= \frac{x+2}{x+2}+ \frac{1}{x+2}=1+ \frac{1}{x+2}

for x>2

\frac{1}{x+2}\ \textless \  \frac{1}{2+2}= \frac{1}{4}

thus

for x>2, 

1+ \frac{1}{x+2}\ \textless \ 1+ \frac{1}{4}= \frac{5}{4}


Answer: 

for x>2

\frac{x^{2} -x-6}{ x^{2} -4} =  \frac{x+3}{x+2} \ \textless \  \frac{5}{4}, (but the expression is never 0)
8 0
3 years ago
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