47/ 3 rad= 2820
have a great day
D is halfway between A and B
so the coordinates of D are (2,2)
E is halfway between A and C so the coordinates of E are (-1,1)
now you need to find the gradient/slope of DE and BC using the formula:

<h3>
<u>G</u><u>r</u><u>a</u><u>d</u><u>i</u><u>e</u><u>n</u><u>t</u><u> </u><u>o</u><u>f</u><u> </u><u>D</u><u>E</u><u>:</u><u> </u></h3>
SUB IN COORDINATES OF D AND E

therefore the gradient of DE is 1/3.
<h3>
<u>G</u><u>r</u><u>a</u><u>d</u><u>i</u><u>e</u><u>n</u><u>t</u><u> </u><u>o</u><u>f</u><u> </u><u>B</u><u>C</u><u>:</u></h3>
<em>S</em><em>U</em><em>B</em><em> </em><em>I</em><em>N</em><em> </em><em>C</em><em>O</em><em>O</em><em>R</em><em>D</em><em>I</em><em>N</em><em>A</em><em>T</em><em>E</em><em>S</em><em> </em><em>O</em><em>F</em><em> </em><em>B</em><em> </em><em>A</em><em>N</em><em>D</em><em> </em><em>C</em>
<em>
</em>
therefore the gradient of BC is -2/-6 which simplifies to 1/3.
<h3>
therefore, BC and DE are parallel as they both have a gradient/slope of 1/3 and parallel lines have the same gradient</h3>
Answer:
The equation of the line in point-slope form is
.
Step-by-step explanation:
According to the statement, let
and
. The equation of the line in point-slope form is defined by the following formula:
(1)
Where:
,
- Coordinates of the point A, dimensionless.
- Slope, dimensionless.
- Independent variable, dimensionless.
- Dependent variable, dimensionless.
In addition, the slope of the line is defined by:
(2)
If we know that
and
, then the equation of the line in point-slope form is:



From (2):


By (1):

The equation of the line in point-slope form is
.
Answer: negative 9 square root of 11
explanation:
5√11 - 12√11 - 2√11
collect like terms by calculating the sum or difference of their coefficients:
(5-12-2) √11
(5-12-2) = (5-14) = -9
so, -9√11