D and E are the mid-points of the sides AB and AC of ΔABC and O is any point on side BC. O is joined to A. If P and Q are the mid-points of OB and OC
respectively, then prove that DEQP is a parallelogram.

Check out the diagram below.

Since D and E are midpoints of AB and AC this means midsegment DE is parallel to side BC. Furthermore, it means DE is parallel to PQ because PQ is entirely on BC.

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Note how I drew a line from A to O. This forms two triangles ABO and ACO

For triangle ABO, segment DP is a midsegment because D and P are midpoints of AB and OB respectively. This means side DP is parallel to side AO. Keep this in mind as we'll use it later.

For triangle ACO, segment EQ is parallel to AO because EQ is a midsegment for similar reasons as already mentioned (E and Q are midpoints of AC and OC respectively). So this must mean that AO is parallel to EQ

Now chain the arguments together: if DP || AO and AO || EQ, then DP || EQ by the transitive property.

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In short, we've proven two things

This is enough to conclude that DEQP is a parallelogram. By definition, a parallelogram has two pairs of opposite parallel sides.

answer: domain and range mean all possible values of x and y that could be substituted to the inequality. a solution means all possible values that make the inequality statement true. so while the range is still all real numbers, the solution is the shaded area, for which y will always be greater than the line of the function.

(- 1, 0 )

step-by-step explanation:

an x- intercept is where the function crosses the x- axis.

to find the x- intercept/s look for the value of x when f(x) = 0

from the given table there are 3 values of x where f(x) = 0

when x = - 1, x = 2, x = 3

the one that is listed is (- 1, 0 )