Name the plane parallel to the plane AEF

30° + 180° (2n + 1), its radian measure is equivalent to?

Nuetrik [128]

Answer:

Example: Find the radian measure of the angles −70° and 120°.

Solution: To find the radian measure of −70° we multiply −70 by the conversion factor /180. We get

Similarly, for 120° we obtain

Note that when we write an angle as a fractional amount of , for example 2/3 times we write the result either as the numerator times divided by the denominator or as the fraction times . So the two values

are equivalent ways of writing the same number. You will see both methods used in the text and in the exercises.

Example: Find the degree measure of /12.

Solution: The conversion factor for going from radians to degrees is 180/. We get

and so the radian measure of /12 is 15°.

Step-by-step explanation:

7 0

3 years ago

Is the relation ship shown by the data linear? if so model the data with an equation pls help

svet-max [94.6K]

It is not linear. Check to see if the points fit in any of the equations. Use 3 for x and 1 for y. None of them work. You get false statements like 3=1.

3 0

4 years ago

A,B and C are the vertices of a triangle A has coordinates (4,6) B has coordinates (2,-2) C has coordinates (-2,-4) D is the mid

polet [3.4K]

Answer:

Segments DE and BC have equal slopes, showing that segments DE and BC are parallel

Step-by-step explanation:

Here we have the coordinates as follows

The coordinates of A is (4, 6)

The coordinates of B is (2, -2)

The coordinates of C is (-2, -4)

Therefore, the coordinates of D the midpoint AB is ((4 + 2)/2, (6 - 2)/2) which gives;

The coordinates of D is (3, 2)

Similarly, the coordinates of E the midpoint AC is ((4 - 2)/2, (6 - 4)/2) which gives;

The coordinates of E is (1, 1)

To prove that segment DE is parallel to segment BC, e show that the slopes of the two segments are equal as follows;

$Slope \, of \, a \, segment = \frac{Change \, in \, the\ y \, coordinates}{Change \, in \, the\, x \, coordinates}$

$Slope \, of \, segment \ DE =\frac{2 - 1}{3-1} = \frac{1}{2}$

$Slope \, of \, segment \ BC =\frac{-2 - (-4)}{2-(-2)} = \frac{2}{4} =\frac{1}{2}$

Therefore, the slopes of segments DE and BC are equal, which shows that segment DE is parallel to BC.

8 0

4 years ago

Show all work to identify the discontinuity and zero of the function f of x equals 3 x over quantity x squared minus 9.

Novosadov [1.4K]

Answer:

Step-by-step explanation: 4x+2¥=12 Wittenberg in slope-intercept form

8 0

4 years ago

Read 2 more answers

What is the center and radius?<br><br> X^2+y^2+6x-6y=-14

Zolol [24]

Answer:

Therefore,

$Center\ is\ C(-3,3)\\and\\Radius= 2\ units$

Step-by-step explanation:

Given:

A Equation of a Circle that is

$x^{2} +y^{2}+ 6x-6y=-14$

Which can be written as

$x^{2} +y^{2}+ 6x-6y+14=0$

To Find:

Center C( -g , -f ) = ?

radius = r =?

Solution:

General Equation of a Circle is given as

$x^{2} +y^{2}+ 2gx+2fy+c=0$

On Comparing the Given equation with the General equation we get the following values,

$2g=6\ and\ 2f=-6\ and\ c=14\\\therefore g=\frac{6}{2}=3\\ \\\therefore f=\frac{-6}{2}=-3\\$

Now, Center and Radius for above circle is given as

$C(-g,-f)=(-3,-(-3))=(-3,3)\\and\\radius = r=\sqrt{(g^{2}+f^{2}-c )} = \sqrt{(3^{2}+(-3)^{2}-14)}\\\\\therefore radius = r=\sqrt{(18-14)} \\\\\therefore radius = r=\sqrt{4}=2\ unit\\$

Therefore,

$Center\ is\ C(-3,3)\\and\\Radius= 2\ units$

6 0

4 years ago