Q: What is the total distance around a figure? -General Knowledge

Q: What is the total distance around a figure? -General Knowledge
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Answer :

Perimeter of a surface
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The figure shows the front view of a convex lens, which originally had only one edge. Five holes of different shapes, namely triangle, square, pentagon, hexagon and circle, were drilled through it at points P, Q, R, S and T in such a way that the holes were parallel to each other, perpendicular to the edge and all the holes were still within the lens edge. What is the total number of edges in the lens after the holes were drilled?

Description : The figure shows the front view of a convex lens, which originally had only one edge. Five holes of different shapes, namely triangle, square, pentagon, hexagon and circle, were drilled through it at points P ... . What is the total number of edges in the lens after the holes were drilled? 

Last Answer : 57

Q: What is a figure with 6, 7, 8, 9, 10 sides called? -General Knowledge

Description : Q: What is a figure with 6, 7, 8, 9, 10 sides called? -General Knowledge

Last Answer : Hexagon, Heptagon, Octagon, Nonagon, Decagon respectively

l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

Description : l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

Last Answer : Though E, draw a line parallel to p intersecting L at G and n at H respectively. Since l | | m ⇒ AG | | BE and AB | | GE [by construction] ∴ Opposite sides of quadrilateral AGEB are ... ∠DGE = ∠FHE [alternate interior angles] By ASA congruence axiom, we have △DEG ≅ △FEH Hence, DE = EF

P is the mid - point of side AB of a parallelogram ABCD. A line through B parallel to PD meets DC at Q and AD produced at R (see figure). -Maths 9th

Description : P is the mid - point of side AB of a parallelogram ABCD. A line through B parallel to PD meets DC at Q and AD produced at R (see figure). -Maths 9th

Last Answer : (i) In △ARB,P is the mid point of AB and PD || BR. ∴ D is a mid - point of AR [converse of mid - point theorem] ∴ AR = 2AD But BC = AD [opp sides of ||gm ABCD] Thus, AR = 2BC (ii) ∴ ABCD is a ... a mid - point of AR and DQ || AB ∴ Q is a mid point of BR [converse of mid - point theorem] ⇒ BR = 2BQ

The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Description : The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Last Answer : Join AC and QP, also it is given that AQ || CP ∴ △ACQ and △APQ are on the same base AQ and lie between the same parallels AQ || CP. ∴ ar(△ACQ) = ar(△APQ) or ar(△ABC) + ar(△ABQ) = ar(△BPQ) + ar(△ABQ) or ar(△ABC) = ar( △BPQ) or 1/2 ar(||gm ABCD) = 1/2 ar(||gm PBQR) or ar(||gm ABCD) = ar(||gm PBQR)

In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP. -Maths 9th

Description : In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP. -Maths 9th

Last Answer : In △PAD, ∠A = 90° and DA = PA = PB ⇒ ∠ADP = ∠APD = 90° / 2 = 45° Similarly, in △QBC, ∠B = 90° and BQ = BC = AB ⇒∠BCQ = ∠BQC = 90° / 2 = 45° In △PAD and △QBC , we have PA = QB [given] ∠A = ... [each = 90° + 45° = 135°] ⇒ △PDC = △QCD [by SAS congruence rule] ⇒ PC = QD or DQ = CP

Write the coordinates of each of the points P, Q, R, S, T and 0 from the figure . -Maths 9th

Description : Write the coordinates of each of the points P, Q, R, S, T and 0 from the figure . -Maths 9th

Last Answer : Here, points P and S lie in I quadrant so their both coordinates will be positive. Now, perpendicular distance of P from both axes is 1, so coordinates of P are (1, 1). Also, perpendicular distance of S ... 0 is the intersection of both axes, so it is the origin and its coordinates are O (0,0).

In the figure, PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR= RS and PA || QB || RC. -Maths 9th

Description : In the figure, PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR= RS and PA || QB || RC. -Maths 9th

Last Answer : Given In a parallelogram PSDA, points 0 and R are on PS such that PQ = QR = RS and PA || QB || RC. To prove ar (PQE) = ar (CFD) Proof In parallelogram PABQ, and PA||QB [given] So, ... = ΔDCF [by ASA congruence rule] ∴ ar (ΔPQE) = ar (ΔCFD) [since,congruent figures have equal area] Hence proved.

In figure, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. -Maths 9th

Description : In figure, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. -Maths 9th

Last Answer : Given ABCDE is a pentagon. BP || AC and EQ|| AD. To prove ar (ABCDE) = ar (APQ) Proof We know that, triangles on the same base and between the same parallels are equal in area. Here, ΔADQ and ΔADE lie on the ... ar (ΔACD) = ar (ΔADE) + ar (ΔACB) + ar (ΔACD) ⇒ ar (ΔAPQ) = ar (ABCDE) Hence proved.

l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

Description : l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

Last Answer : Though E, draw a line parallel to p intersecting L at G and n at H respectively. Since l | | m ⇒ AG | | BE and AB | | GE [by construction] ∴ Opposite sides of quadrilateral AGEB are ... ∠DGE = ∠FHE [alternate interior angles] By ASA congruence axiom, we have △DEG ≅ △FEH Hence, DE = EF

P is the mid - point of side AB of a parallelogram ABCD. A line through B parallel to PD meets DC at Q and AD produced at R (see figure). -Maths 9th

Description : P is the mid - point of side AB of a parallelogram ABCD. A line through B parallel to PD meets DC at Q and AD produced at R (see figure). -Maths 9th

Last Answer : (i) In △ARB,P is the mid point of AB and PD || BR. ∴ D is a mid - point of AR [converse of mid - point theorem] ∴ AR = 2AD But BC = AD [opp sides of ||gm ABCD] Thus, AR = 2BC (ii) ∴ ABCD is a ... a mid - point of AR and DQ || AB ∴ Q is a mid point of BR [converse of mid - point theorem] ⇒ BR = 2BQ

The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Description : The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Last Answer : Join AC and QP, also it is given that AQ || CP ∴ △ACQ and △APQ are on the same base AQ and lie between the same parallels AQ || CP. ∴ ar(△ACQ) = ar(△APQ) or ar(△ABC) + ar(△ABQ) = ar(△BPQ) + ar(△ABQ) or ar(△ABC) = ar( △BPQ) or 1/2 ar(||gm ABCD) = 1/2 ar(||gm PBQR) or ar(||gm ABCD) = ar(||gm PBQR)

In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP. -Maths 9th

Description : In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP. -Maths 9th

Last Answer : In △PAD, ∠A = 90° and DA = PA = PB ⇒ ∠ADP = ∠APD = 90° / 2 = 45° Similarly, in △QBC, ∠B = 90° and BQ = BC = AB ⇒∠BCQ = ∠BQC = 90° / 2 = 45° In △PAD and △QBC , we have PA = QB [given] ∠A = ... [each = 90° + 45° = 135°] ⇒ △PDC = △QCD [by SAS congruence rule] ⇒ PC = QD or DQ = CP

Write the coordinates of each of the points P, Q, R, S, T and 0 from the figure . -Maths 9th

Description : Write the coordinates of each of the points P, Q, R, S, T and 0 from the figure . -Maths 9th

Last Answer : Here, points P and S lie in I quadrant so their both coordinates will be positive. Now, perpendicular distance of P from both axes is 1, so coordinates of P are (1, 1). Also, perpendicular distance of S ... 0 is the intersection of both axes, so it is the origin and its coordinates are O (0,0).

In the figure, PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR= RS and PA || QB || RC. -Maths 9th

Description : In the figure, PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR= RS and PA || QB || RC. -Maths 9th

Last Answer : Given In a parallelogram PSDA, points 0 and R are on PS such that PQ = QR = RS and PA || QB || RC. To prove ar (PQE) = ar (CFD) Proof In parallelogram PABQ, and PA||QB [given] So, ... = ΔDCF [by ASA congruence rule] ∴ ar (ΔPQE) = ar (ΔCFD) [since,congruent figures have equal area] Hence proved.

In figure, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. -Maths 9th

Description : In figure, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. -Maths 9th

Last Answer : Given ABCDE is a pentagon. BP || AC and EQ|| AD. To prove ar (ABCDE) = ar (APQ) Proof We know that, triangles on the same base and between the same parallels are equal in area. Here, ΔADQ and ΔADE lie on the ... ar (ΔACD) = ar (ΔADE) + ar (ΔACB) + ar (ΔACD) ⇒ ar (ΔAPQ) = ar (ABCDE) Hence proved.

In the adjoining figure, P and Q have co-ordinates (4, 6)and (0, 3) respectively. Find (i) the co-ordinates of R (ii) Area of quadrilateral OAPQ. -Maths 9th

Description : In the adjoining figure, P and Q have co-ordinates (4, 6)and (0, 3) respectively. Find (i) the co-ordinates of R (ii) Area of quadrilateral OAPQ. -Maths 9th

Last Answer : Let the line 2x + 3y - 30 = 0 divide the join of A(3, 4) and B(7, 8) at point C(p, q) in the ratio k : 1. Then,p = \(rac{7k+3}{k+1}\), q = \(rac{8k+4}{k+1}\)As the point C lies on the line 2x + 3y - 30 ... {3}{2}+1},rac{8 imesrac{3}{2}+4}{rac{3}{2}+1}\bigg)\) = \(\big(rac{27}{5},rac{32}{5}\big)\).

Five point charges each of charge + q are placed on five vertices of a regular of side h as shown in the figure. Then

Description : Five point charges each of charge + q are placed on five vertices of a regular of side h as shown in the figure. Then

Last Answer : Five point charges each of charge + q are placed on five vertices of a regular of side h as shown in the ... (4piepsilon_(0))q^(2)/h^(2)` along OC.

In the above figure, AB is parallel to CD. P and R are the points on AB and CD respectively. Q is in between AB and CD. Find the value of x in degrees

Description : In the above figure, AB is parallel to CD. P and R are the points on AB and CD respectively. Q is in between AB and CD. Find the value of x in degrees

Last Answer : In the above figure, AB is parallel to CD. P and R are the points on AB and CD respectively. Q is in between AB and CD. Find the value of x in degrees

A long solenoid carrying a current I is placed with its axis vertical as shown in the figure. A particle of mass m and charge q is released from the t

Description : A long solenoid carrying a current I is placed with its axis vertical as shown in the figure. A particle of mass m and charge q is released from the t

Last Answer : A long solenoid carrying a current I is placed with its axis vertical as shown in the figure. A particle of ... than g C. equal to g D. None of these

The equivalent resistance between P and Q in the figure is approximately

Description : The equivalent resistance between P and Q in the figure is approximately

Last Answer : The equivalent resistance between P and Q in the figure is approximately A. 6`Omega` B. 5.333`Omega` C. `7.5Omega` D. 20`Omega`

In the circuit shown in figure q varies with time t as `q=(t^(2)=16)`. Here q is in coulomb and t in second. Find `V_(ab) "at" t=5s`

Description : In the circuit shown in figure q varies with time t as `q=(t^(2)=16)`. Here q is in coulomb and t in second. Find `V_(ab) "at" t=5s`

Last Answer : In the circuit shown in figure q varies with time t as `q=(t^(2)=16)`. Here q is in coulomb and t in second. ... 50 V B. 35.5 V C. 46.5 V D. 40.2 V

In the circuit shown in figure q varies with time t as `q=(t^(2)=16)`. Here q is in coulomb and t in second. Find `V_(ab)=(V_(a)-V_(b)) at t=3s`

Description : In the circuit shown in figure q varies with time t as `q=(t^(2)=16)`. Here q is in coulomb and t in second. Find `V_(ab)=(V_(a)-V_(b)) at t=3s`

Last Answer : In the circuit shown in figure q varies with time t as `q=(t^(2)=16)`. Here q is in coulomb and t in second. ... `18.5 V` C. `-25.5 V` D. `22.5 V`

Figure P shows how a point on a circle traces a path when it is rolled on the ground. The point in which of the polygons shown in the options creates the path in Figure Q?

Description : Figure P shows how a point on a circle traces a path when it is rolled on the ground. The point in which of the polygons shown in the options creates the path in Figure Q?  

Last Answer : D 

Figure P shows an isometric view of an object made with a single sheet of cardboard. Figure Q shows the front view of the same object after colouring. Which option shows the object after it is cut open?

Description : Figure P shows an isometric view of an object made with a single sheet of cardboard. Figure Q shows the front view of the same object after colouring. Which option shows the object after it is cut open?

Last Answer : C

Top and side views of a photographer’s setup for a product shoot are shown in the figure. P is the primary light source, S is the secondary light source, R is the reflector and Q is the camera position. If the Primary Light Source is at 100% intensity, and the secondary light is at 25% intensity, which one of the given options is the photograph taken using this setup?

Description : Top and side views of a photographer's setup for a product shoot are shown in the figure. P is the primary light source, S is the secondary light source, R is the reflector and Q is the camera ... 25% intensity, which one of the given options is the photograph taken using this setup?

Last Answer : D

Triangle P and pentagon Q have markings on them as shown in the figure. If they are placed over each other, which of the following arrangements is/are possible?

Description : Triangle P and pentagon Q have markings on them as shown in the figure. If they are placed over each other, which of the following arrangements is/are possible? 

Last Answer : A, B, D

For the circuit shown in figure, the Boolean expression for the output y in terms of inputs P, Q, R and S is

Description : For the circuit shown in figure, the Boolean expression for the output y in terms of inputs P, Q, R and S is

Last Answer : For the circuit shown in figure, the Boolean expression for the output y in terms of inputs P, Q, R and S is P+Q+R+S

Q: What is the total number of dots on a dice? -General Knowledge

Description : Q: What is the total number of dots on a dice? -General Knowledge

Last Answer : 21 dots.....

Find the distance between points P(8 2) and Q(-1 5) to the nearest tenth.?

Description : Find the distance between points P(8 2) and Q(-1 5) to the nearest tenth.?

Last Answer : Using the distance formula it is the square root of 61 which is about 7.8 to the nearest tenth

A boat takes 4hours for traveling downstream from point P to point Q and coming back to point P upstream. If the velocity of the stream is 2km ph and the speed of the boat in still water is 4kmph, what is the distance between P and Q? A.9 km B.7 km C.5 km D.6km

Description : A boat takes 4hours for traveling downstream from point P to point Q and coming back to point P upstream. If the velocity of the stream is 2km ph and the speed of the boat in still water is 4kmph, what is the distance between P and Q? A.9 km B.7 km C.5 km D.6km

Last Answer : Answer- D Basic Formula: Speed of stream = ½ (a-b) km/hr Speed of still water = ½ (a+b) km/hr Explanation: Time taken by boat to travel upstream and downstream = 4 hours Velocity of the stream, ½ (a-b) = 2km/hr a ... + x / 6 = 4 3x + x / 6 = 4 4x = 24 so,x = 6 distance between P and Q = 6km

The thickness of a laminar boundary layer over a flat plate at two different sections P and Q are 0.8 cm and 2.4 cm respectively. If the section Q is 3.6 m downstream of P, the distance of section P from the leading edge of the plate is (a) 0.32 m (b) 0.22 m (c) 0.40 m (d) 0.53 m

Description : The thickness of a laminar boundary layer over a flat plate at two different sections P and Q are 0.8 cm and 2.4 cm respectively. If the section Q is 3.6 m downstream of P, the distance of section P from the leading edge of the plate is (a) 0.32 m (b) 0.22 m (c) 0.40 m (d) 0.53 m

Last Answer : Ans: (c) Correct Answer is: 0.45 m

Maximum distance covered in a day by a Patrolman should not normally exceed q. 2 km. r. 5 km. s. 10 km. t. 20 km .*

Description : Maximum distance covered in a day by a Patrolman should not normally exceed q. 2 km. r. 5 km. s. 10 km. t. 20 km .*

Last Answer : t. 20 km

In covering a definite distance, the speeds of P & Q are in the ratio of 5:6. P takes 45 mins more than Q to reach the target. The time taken by P to reach the target is. a) 4 hrs 20 mins b) 4 hrs 45 mins c) 2hrs 30 mins d) 4hrs 30 mins

Description : In covering a definite distance, the speeds of P & Q are in the ratio of 5:6. P takes 45 mins more than Q to reach the target. The time taken by P to reach the target is. a) 4 hrs 20 mins b) 4 hrs 45 mins c) 2hrs 30 mins d) 4hrs 30 mins

Last Answer : D Ratio of speeds = 5:6 Ratio of time =6:5 Suppose P takes 6x hrs and Q takes 5x hrs to reach the target. Then, 6x – 5x = 45/60 X=3/4 Time taken by P = 6x hrs =(6*3/4)hrs =4.5 hrs i.e., the time taken by P to reach the target is 4hrs 30mins

Distance between two stations P & Q is 1556km. A passenger train covers the journey from P to Q at 168km per hr and return back to P with a uniform speed of 112km/hr. Find the Average speed of the train during the whole journey? A) 124.4 km/hr B) 130.4km/hr C) 134.4 km/hr D) 130.0 km/hr

Description : Distance between two stations P & Q is 1556km. A passenger train covers the journey from P to Q at 168km per hr and return back to P with a uniform speed of 112km/hr. Find the Average speed of the train during the whole journey? A) 124.4 km/hr B) 130.4km/hr C) 134.4 km/hr D) 130.0 km/hr

Last Answer : C) Given x= 168 , y = 112 Required average speed = (2xy / x + y) km/hr = 2 * 168 *112 / 168 + 112 = 37632 / 280 =134.4 km/hr

A boat takes 38 hrs for travelling downstream from point p to point q and coming back to a point r midway between p and q. If the velocity of the stream is 8 km/hr and the speed of the boat in still water is 28 km/hr. What is the distance between p and q? A) 720 B) 640 C) 510 D) 450

Description : A boat takes 38 hrs for travelling downstream from point p to point q and coming back to a point r midway between p and q. If the velocity of the stream is 8 km/hr and the speed of the boat in still water is 28 km/hr. What is the distance between p and q? A) 720 B) 640 C) 510 D) 450

Last Answer : ANSWER: A Explanation: Speed downstream = (28 + 8)km/hr = 36 km/hr Speed upstream = (28 – 8)km/hr = 20 km/hr Let the distance between p and q be ‘x’km Then, x/36 + (x/2)/20 = 38 x/36 +x/40 = 38 19x = 13680 X = 720 km Therefore the distance between p and q is 720km

How to figure distance vs. TV screen size?

Description : How to figure distance vs. TV screen size?

Last Answer : Google is your friend – check these sites out: http://reviews.cnet.com/8301-33199_7-57468333-221/how-big-a-tv-should-i-buy/ http://www.crutchfield.com/S-QvMQm0VlU0i/learn/learningcenter/home/TV_placement.html http://howto.wired.com/wiki/Calculate_the_Optimum_TV_Screen_Size_for_Your_Room

In figure LM is a line parallel to the Y-axis at a distance of 3 units. -Maths 9th

Description : In figure LM is a line parallel to the Y-axis at a distance of 3 units. -Maths 9th

Last Answer : Given, LM is a line parallel to the Y-axis and its perpendicular distance from Y-axis is 3 units. (i) Coordinate of point P = (3, 2) [since, its perpendicular distance from X-axis is 2] Coordinate of ... 3, abscissa of point M = 3 Difference between the abscissa of the points L and M = 3 -3 = 0

In figure LM is a line parallel to the Y-axis at a distance of 3 units. -Maths 9th

Description : In figure LM is a line parallel to the Y-axis at a distance of 3 units. -Maths 9th

Last Answer : Given, LM is a line parallel to the Y-axis and its perpendicular distance from Y-axis is 3 units. (i) Coordinate of point P = (3, 2) [since, its perpendicular distance from X-axis is 2] Coordinate of ... 3, abscissa of point M = 3 Difference between the abscissa of the points L and M = 3 -3 = 0

In the figure, what is the distance from point P to line a?

Description : In the figure, what is the distance from point P to line a?

Last Answer : 3v/ 2 ~ 4.2

what- In the figure, what is the distance from point P to line a?

Description : what- In the figure, what is the distance from point P to line a?

Last Answer : \sqrt(9.8)~ 3.13

A straight rod of mass m and lenth `L` is suspended from the identical spring as shown in the figure The spring stretched by a distance of `x_(0)` due

Description : A straight rod of mass m and lenth `L` is suspended from the identical spring as shown in the figure The spring stretched by a distance of `x_(0)` due

Last Answer : A straight rod of mass m and lenth `L` is suspended from the identical spring as shown in the figure The spring ... `(mgR)/(2LE)` D. `(mgR)/(E)`

Assertion Upper wire shown in figure is fixed. At a certain distance x,lower wire can remain in equilibrium. Reason The above equilibrium of lower wir

Description : Assertion Upper wire shown in figure is fixed. At a certain distance x,lower wire can remain in equilibrium. Reason The above equilibrium of lower wir

Last Answer : Assertion Upper wire shown in figure is fixed. At a certain distance x,lower wire can ... above equilibrium of lower wire is stable equilibrium.

Which of the following figure shown the magnetic flux denstiy b at a distance r from a long straight rod carrying a steady current I ?

Description : Which of the following figure shown the magnetic flux denstiy b at a distance r from a long straight rod carrying a steady current I ?

Last Answer : Which of the following figure shown the magnetic flux denstiy b at a distance r from a long straight rod carrying a steady current I ? A. B. C. D.

A vertically immersed surface is shown in the below figure. The distance of its centre of pressure from the water surface is (A) (bd²/12) + (B) (d²/12 ) + (C) b²/12 + (D) d²/12 +

Description : A vertically immersed surface is shown in the below figure. The distance of its centre of pressure from the water surface is (A) (bd²/12) + (B) (d²/12 ) + (C) b²/12 + (D) d²/12 +

Last Answer : Answer: Option B

Figure out the components of non-verbal communication in a classroom from the following : (A) Facial expression, cultural space and seating arrangement (B) Speed of utterance, feel good factor and acoustics (C) High sound, physical ambience and teacher-learner distance (D) Facial expression, kinesics and personal space

Description : Figure out the components of non-verbal communication in a classroom from the following : (A) Facial expression, cultural space and seating arrangement (B) Speed of utterance, feel good ... sound, physical ambience and teacher-learner distance (D) Facial expression, kinesics and personal space

Last Answer : (D) Facial expression, kinesics and personal space 

The following question is based on the diagram given below. If the two small circles represent formal class-room education and distance education and the big circle stands for university system of education, which figure represents the university systems. (A) (B) (C) (D)

Description : The following question is based on the diagram given below. If the two small circles represent formal class-room education and distance education and the big circle stands for university system of education, which figure represents the university systems. (A) (B) (C) (D) 

Last Answer : Answer: B In Figure 2 big circle stands for University system of Education and two seprate small circle inside the Big circle represent formal class-room education and distance education.

I'm a seed that's underground. Many like to spread me around. But my family it's said is quite insane. And if you've a low I.Q. you've this kind of brain. Eat me raw and I'll make you sick. Eat too much and your tongue will stick.What am I? -Riddles

Description : I'm a seed that's underground. Many like to spread me around. But my family it's said is quite insane. And if you've a low I.Q. you've this kind of brain. Eat me raw and I'll make you sick. Eat too much and your tongue will stick.What am I? -Riddles

Last Answer : A peanut.

The electron with change `(q=1.6xx10^(-19)C)` moves in an orbit of radius `5xx10^(-11)`m with a speed of `2.2xx10^(6)ms^(-1)`, around an atom. The equ

Description : The electron with change `(q=1.6xx10^(-19)C)` moves in an orbit of radius `5xx10^(-11)`m with a speed of `2.2xx10^(6)ms^(-1)`, around an atom. The equ

Last Answer : The electron with change `(q=1.6xx10^(-19)C)` moves in an orbit of radius `5xx10^(-11)`m with a speed of `2. ... -3)A` C. `1.12xx10^(-9)A` D. `1.12A`

P walks around a circular field at the rate of two rounds per hour while Q runs around it at the rate of eight rounds per hour. They start in the opposite direction from the same point at 6.45 a.m. They shall first cross each other at? a) 6.45 a.m b) 6.50 a.m c) 7.51 a.m d) 6.51 a.m e) 7.45 am

Description : P walks around a circular field at the rate of two rounds per hour while Q runs around it at the rate of eight rounds per hour. They start in the opposite direction from the same point at 6.45 a.m. They shall first cross each other at? a) 6.45 a.m b) 6.50 a.m c) 7.51 a.m d) 6.51 a.m e) 7.45 am

Last Answer : d Since P and Q move in the opposite direction along the circular field, so they will first meet each other when there is a difference of one round between the two. Relative speed of P and Q = 8+2 = 10 rounds per hour. Time ... 1/10 hr = 6 min. 6 Hrs 45 min + 6 min = 6 hrs 51 min i.e, 6.51 a.m