Yes, if points P and Q are contained in a plane, then the line segment connecting P and Q, denoted as PQ, is also entirely contained in that plane. This is a fundamental property of planes in Euclidean geometry, where any line segment formed by two points within the same plane must lie entirely within that plane. Therefore, the assertion is correct.
Yes, the perpendicular bisector of the line segment connecting points ( p ) and ( q ) is indeed the set of all points that are equidistant from both ( p ) and ( q ). This line is perpendicular to the segment ( pq ) at its midpoint, ensuring that any point on the bisector maintains equal distance to both ( p ) and ( q ). Thus, it effectively divides the segment into two equal halves while being perpendicular to it.
False. In order for the line PQ to lie in plane B, then both P and Q must lie in plane B.
Since rs is the perpendicular bisector of pq, it follows that point s is the midpoint of segment pq, meaning that ps is equal to qs. Additionally, because rs is perpendicular to pq, the angles formed at the intersection (∠prs and ∠qrs) are both right angles (90 degrees). Consequently, any point on line rs is equidistant from points p and q.
2 + pq
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true
apex it’s true on god
True
PQ
False. In order for the line PQ to lie in plane B, then both P and Q must lie in plane B.
Province de Quebec
PQ Monthly was created in 2012.
2 + pq
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If M P and Q are collinear and MP plus PQ equals MQ then P is between M and Q.
The Method To Add an element in Circular Queue # define MAXQUEUE 100 struct queue{ int items[MAXQUEUE]; int front, rear; } struct queue q; q.front=q.rear=MAXQUEUE -1; void ENQ(struct queue *pq, int x) { /* make room for new element*/ if(pq ->rear = MAXQUEUE - 1) pq-> rear = 0; else (pq->rear)++; /* check for overflow */ if(pq ->rear==pq->front) { printf("queue overflow); exit(1); } pq->items[pq->rear]=x; return; }/* end of ENQ*/ A Method to Delete an element from Circular Queue int DQ(struct queue *pq) { if(pq-> rear == pq-> front) { printf("queue underflow"); exit(1); }/*end if*/ if(pq->front = = MAXQUEUE-1) pq->front=0; else (pq->front)++; return(pq->items[pq->front]);
Because b is the mid point of pq, pb = qb. pb is half as long as pq Eq#1....pb = 1/2 pq Eq#2....pq = pb +8 Substitute Eq#1 into Eq #2 pq = 1/2 pq + 8 subtracting1/2 pq from both sides 1/2 pq = 8 pq = 16 problem here: you can't subtract 1/2 ... you would have to divide.