The given polynomial is
As this is a fourth degree polynomial, we can not directly use any formula, So here we will use hit and trial method
check whether x=-1 is a zero of the polynomial or not
Hence x=-1 is not a zero.
Now check whether x=-2 is a zero of the polynomial or not
Hence x=-2 is a zero of the polynomial.
Hence we can re-write the polynomial as
Hence the required binomial is (x+2)
Answer: The correct option is (D) P″(9, -12) and Q″(15, -3).
Step-by-step explanation: Given that triangle PQR is dilated by a scale factor of 1.5 to form triangle P′Q′R′. This triangle is then dilated by a scale factor of 2 to form triangle P″Q″R″.
The co-ordinates of vertices P and Q are (3, -4) and (5, -1) respectively.
We are to find the co-ordinates of the vertices P″ and Q″.
<u>Case I :</u> ΔPQR dilated to ΔP'Q'R'
The co-ordinates of P' and Q' are given by
<u>Case II :</u> ΔP'Q'R' dilated to ΔP''Q''R''
The co-ordinates of P'' and Q'' are given by
Thus, the co-ordinates of the vertices P'' and Q'' are (9, -12) and (15, -3).
Option (D) is CORRECT.
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Answer:
B
Step-by-step explanation:
Recommended for housing: 30%
X/100 = 612/1700
= 36%
Recommended for food: 10%
X/100 = 238/1700
= 14%
Recommended for transportation 15%
X/100 = 370/1700
= 21.8%
We are to show that if X ⊆ Y then (X ∪ Z) ⊆ (Y ∪ Z) for sets X, Y, Z.
Assume that a is a representative element of X, that is, a ∈ X. By the definition of union, a ∈ X ∪ Z. Now because X ⊆ Y and we assumed a ∈ X, then a ∈ Y by the definition of subset. And because a ∈ Y, then a ∈ Y ∪ Z by definition of union.
We chose our representative element, a, and showed that a ∈ X ∪ Y implies that a ∈ Y ∪ Z and this completes the proof.
Point D, as shown i the figure, is the intersection of the angle bisectors. This point is the Incircle, or the center of the inscribed circle.
All 3 angle bisectors meet at D, so drawing the angle bisector of C is useless. (Thus step 4 is not the one).
Since we have the center of the inscribed circle, we want to open the compass so that it touches all 3 sides at one point only, that is, we want the 3 sides to be tangent to this circle.
The segments joining the tangency points and D are 3 radii of the circle. We know that a radius is perpendicular to the tangent it touches.
Thus, we need to draw an altitude from D to any of the sides.
Answer: 1
