Cathy bought 2 jars of sauce for $3. What is the constant of proportionality ?

sketch graph of a polynomial function with degree 5, lead coefficient less than 0, a real zero at 4, 6, and -2 with multiplicity

labwork [276]

What’s your answer choice

4 0

2 years ago

Find parametric equations for the path of a particle that moves along the circle x2 + (y − 1)2 = 16 in the manner described. (En

ArbitrLikvidat [17]

Answer:

a) $x = 4\cdot \cos t$ , $y = 1 + 4\cdot \sin t$ , b) $x = 4\cdot \cos t$ , $y = 1 + 4\cdot \sin t$ , c) $x = 4\cdot \cos \left(t+\frac{\pi}{2} \right)$ , $y = 1 + 4\cdot \sin \left(t + \frac{\pi}{2} \right)$ .

Step-by-step explanation:

The equation of the circle is:

$x^{2} + (y-1)^{2} = 16$

After some algebraic and trigonometric handling:

$\frac{x^{2}}{16} + \frac{(y-1)^{2}}{16} = 1$

$\frac{x^{2}}{16} + \frac{(y-1)^{2}}{16} = \cos^{2} t + \sin^{2} t$

Where:

$\frac{x}{4} = \cos t$

$\frac{y-1}{4} = \sin t$

Finally,

$x = 4\cdot \cos t$

$y = 1 + 4\cdot \sin t$

a) $x = 4\cdot \cos t$ , $y = 1 + 4\cdot \sin t$ .

b) $x = 4\cdot \cos t$ , $y = 1 + 4\cdot \sin t$ .

c) $x = 4\cdot \cos t''$ , $y = 1 + 4\cdot \sin t''$

Where:

$4\cdot \cos t' = 0$

$1 + 4\cdot \sin t' = 5$

The solution is $t' = \frac{\pi}{2}$

The parametric equations are:

$x = 4\cdot \cos \left(t+\frac{\pi}{2} \right)$

$y = 1 + 4\cdot \sin \left(t + \frac{\pi}{2} \right)$

7 0

3 years ago

Find an equation for the perpendicular bisector of the line segment whose endpoints

TEA [102]

Answer:

y= -2x -8

Step-by-step explanation:

I will be writing the equation of the perpendicular bisector in the slope-intercept form which is y=mx +c, where m is the gradient and c is the y-intercept.

A perpendicular bisector is a line that cuts through the other line perpendicularly (at 90°) and into 2 equal parts (and thus passes through the midpoint of the line).

Let's find the gradient of the given line.

$\boxed{gradient = \frac{y1 -y 2}{x1 - x2} }$

Gradient of given line

$= \frac{1 - ( - 5)}{3 - ( - 9)}$

$= \frac{1 + 5}{3 + 9}$

$= \frac{6}{12}$

$= \frac{1}{2}$

The product of the gradients of 2 perpendicular lines is -1.

(½)(gradient of perpendicular bisector)= -1

Gradient of perpendicular bisector

= -1 ÷(½)

= -1(2)

= -2

Substitute m= -2 into the equation:

y= -2x +c

To find the value of c, we need to substitute a pair of coordinates that the line passes through into the equation. Since the perpendicular bisector passes through the midpoint of the given line, let's find the coordinates of the midpoint.

$\boxed{midpoint = ( \frac{x1 + x2}{2} , \frac{y1 + y2}{2}) }$