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:
See below.
Step-by-step explanation:
The graph of the quadratic equation is a parabola or u shaped graph. It has a vertex at (-1,-3) since h=-1 and k=-3 in the vertex form. It is also facing down since the leading coefficient is negative.
13 divided by 11.3 is 1.150442477876106. So 15 by 1.150442477876106 and you get 13.03846162772485. Just round it to the nearest tenth and you get 13.0.
the angles on the opposite sides of a quadrilateral touching the circumference adds up to 180 degrees
so,
(21x-2) + (38x +5) = 180
21x+38x −2+5=180
59x+3=180
59x=177
x=3
Answer:
can you click a full photo of this question.