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>
The answer for your question is a
To convert into decimal, move the decimal to the left by 6 positions.
It will be 0.00000563
Answer is A.
Answer:
x=-6
Step-by-step explanation:
Let x be the number
3x = x^2 -54
Subtract 3x
3x-3x = x^2 -3x-54
0 = x^2 -3x-54
Factor
What 2 numbers multiply to -54 and add to -3
-9*6 = -54
-9+6 = -3
0=(x-9) (x+6)
Using the zero product property
x-9=0 x+6=0
x=9 x=-6
The questions asks for the negative solution
x=-6
≥The solution of an inequality is an interval, i.e. a range.
To prove that the interval found as solution, you must consider several cases.
1) In the case that the ineguailty is ≥ or ≤, first use the limits of the interval to prove they are valid solutions. This is, replace the limit values, one at a time, and verifiy the inequality.
2) If the sign is ≥ or > use a value to the right of the limit value to show that the values to the right are solution, and use a value to the left to show that they are not solution.
3) If the sign is ≤ or <, use a value to the left of the limit value to show that it is a solution and a value to the right of the limit value to show that it is not a solution.
