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G Gautam harsolia
Given number is   Now, on rationalisation we will get                                                                                                           Therefore, answer is  
(-2,4) lies in second quadrant (3,-1) lies in fourth quadrant  (-1,0) lies in second quadrant (1,2) lies in first quadrant  (-3,-5) lies in third quadrant

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P Pankaj Sanodiya
The major divisions in kingdom Plantae are Thallophyta, Bryophyta, Pteridophyta, Gymnosperms, and Angiosperms. The following points constitute the basis of these divisions: (i) Presence or absence of distinct organdIes. (ii) Presence or absence of distinct and differentiated tissues, which can carry food and water. (iii) Presence or absence of seeds. (iv) Whether the seeds are enclosed...

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H Harsh Kankaria
Side of the square field =  Perimeter of the square = According to question, He completes 1 round in .  Speed of the farmer =  Distance covered in  = Now, Number of round trips completed travelling  = We know, in 3 round trips the displacement will be zero. In round, the farmer will reach diametrically opposite to his initial position.  Displacement =

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A Aadil Khan
Find the remainder when x power 3 plus 3x power 2 plus 3x plus 1 is divided by 5 plus 2x

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D Divya Prakash Singh
The steps of construction to be followed: Step 1: Draw a ray DY. Step 2: Draw an arc ACD with O as a center. Step 3: Now, with A as a centre, draw two arcs B and C on the arc ACD. Step 4: Taking B and C as centres, arcs are made to intersect at E and the angle formed is . Step 5: Take F and D as centres, draw arcs to intersect at point X or the bisector of angle EOD is made. Step 6: Join OX...

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R Ritika Kankaria
The steps of construction to be followed: Step 1: Draw a ray OY. Step 2: Then, taking O as a centre, draw an arc ABC. Step 3:  Now, with A as a centre, draw two arcs B and C which are made on the arc ABC. Step 4: Taking B and C as centres simulataneously, arcs are made to intersect at E and  is constructed. Step 5: With B and C as centres, arcs are made to intersect at X. Step 6: Join the OX...

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D Divya Prakash Singh
Steps to construction to follow: Step 1: Draw a ray OY. Step 2: Now, taking O as the centre draw an arc ABC. Step 3: On taking A as a centre, draw two arcs B and C on the arc ABC. Step 4: Now, taking B and C as centres, arcs are made to intersect at point E and the  is constructed. Step 5: Taking A and C as centres, arcs are made to intersect at D. Step 6: Now, join OD and hence,  is...

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D Devendra Khairwa
In the first part we have proved that   . Thus by  c.p.c.t. , we can conclude :                                                                      

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S seema garhwal
In  ABC,         A+ABC+ACB= A  = BDC =    (Angles in same segment)            

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D Devendra Khairwa
                                                                In the above figure, point B is the mean position about which the bob rotates. When the bob is released from point C, it attains some velocity while moving down (upto B) and deaccelerated and stops at point A. Thus point A and point C are the maximum height points. And the velocity of bob at point B will be maximum. The total...

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D Devendra Khairwa
From the table, the relation between time and the distance can be seen.                                                                         Thus the velocity of the particle is increasing with time.                                                                             and                                                                                    Hence acceleration increases...
The given data is not continous we therefore modify the limits of the class intervals as well to make the class intervals continous. To make the frequency polygon we first modify the table as follows To make the frequency polygon we mark the number of balls on the x-axis and the runs scored on the y-axis. The representation of the given information in the form of frequency polygon is as follows.

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D Divya Prakash Singh
First, we make  and then its bisector. The steps of constructions are: 1. Draw a ray OA. 2. Taking O as the center and any radius of your own choice, draw an arc cutting OA at B. 3. Now, taking B as center and with the same radius as before, draw an arc intersecting the previously drawn arc at point C. 4. Draw the ray OD passing through the C. Thus,  Now, we draw bisector of  5. Taking C and...

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M manish painkra
We have,  PQR is produced to a point S and  bisectors of PQR and PRS meet at point T, By exterior angle sum property, PRS = P + PQR Now,  ................(i) Since QT and QR are the bisectors of  PQR and PRS respectively. Now, in QRT, ..............(ii) From eq (i) and eq (ii),  we get Hence proved

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M manish painkra
We have,  PQ PS, PQ || SR, SQR = 28° and QRT = 65° Now, In  QRS, the side SR produced to T and PQ || RS therefore, QRT =  =   So,  Also, QRT = RSQ + SQR (By exterior angle property of a triangle) Therefore, RSQ = QRT - SQR                     Now, in  PQS, P + PQS + PSQ =   

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M manish painkra
We have, lines PQ and RS intersect at point T, such that PRT = 40°, RPT = 95° and TSQ = 75° In PRT, by using angle sum property PRT + PTR + TPR =  So, PTR  =     Since lines, PQ and RS intersect at point T therefore, PTR = QTS (Vertically opposite angles)                 QTS =  Now, in QTS, By using angle sum property TSQ + STQ + SQT =  So, SQT = 
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