5(6-i)+(6-i)(3-i). show work.
1 answer:
47 -14i
You can work this out in the straight-forward way, or you can recognize that (6-i) is a common factor. In the latter case, you have ...
... = (6-i)(5 + 3-i)
... = (6 -i)(8 -i)
This product of binomials is found in the usual way. Each term of one factor is multiplied by each term of the other factor and the results summed. Of course, i = √-1, so i² = -1.
... = 6·8 -6i -8i +i²
... = 48 -14i -1
... =
_____
A suitable graphing calculator will work these complex number problems easily.
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Let's begin noting that a triangle is isosceles if and only if two of its angles are congruent. We can thus find the angle <ABP, recalling that the sum of the interior angles of a triangle is equal to 180°.
Finally, let point K be the intersection between segments BC and PQ, and let's note that the triangle PQB is a right isosceles triangle, since all the angles in a square are equal to 90°, and the two triangles APB and BQC are congruent.
Therefore, the angle BKQ is equal to 180-50-45=85°.
Of course angle BKP=180-85=95°.
Hope this helps :)