@sowmyanarayanan said:S is the region described by the following:1. |x|2. |y|3. x > yFind the Perimeter of the region formed by the intersection of S and the unit circle with center at (1,1)1. 82. 43. between 1 and 2.754. between 2.75 and 43rd -> 1+ (1.4/2) +(2*pi/8)
@sowmyanarayanan
@vijay_chandola said:If x is the side of the largest equilateral triangle that can be drawn inside a square of side 1, what can be said about x?1) 0.7 2) x = 13) 1 4) x > 1.1
3)option....side is 2sqrt2/(1+sqrt3).......
a single hand wind screen wiper of a car is of length 25 cm and it rotates through an angle of 'x' units which can be adjusted. if the desired wiping area is419.071 square cm , then the angle 'x' is -
45;60;90;120
@vijay_chandola said:5A(n+1) + 1 = 5*An + n. Given A5 = 55, find A55.(n+1) is subscript.Why this thread is so dead today..?
A(n+1) - A(n) = (n-1)/5
A(55)-A(54)=53/2
A(54)-A(53)=52/5
.
.
.
A(6)-A(5)=4/5
adding all we get
A(55)-A(5)=(53+52+....+4)/5
A(55)=285+55=340
Let A1 be a square whose side is "a" metres. Circle C1 circumscribes the square A such that all its vertices are on C1. Another square A2 circumscribes C1. Circle C2 circumscribes A2, and A3circumscribes C2, and so on. If DN is the area between the square AN and the circle CN, where N is a natural number, then the ratio of the sum of all DN to D1 is:
A. 1
B. π/2 -1
C. Infinity
D. None of the above
B. π/2 -1
C. Infinity
D. None of the above
@deedeedudu B covers 60 mts in 12 second speed--> 5m/s
the A would have covered 1000m in 188s--> 250m in 47s (B)
@pankaj1988 said:3)option....side is 2sqrt2/(1+sqrt3).......
@gautam22 said:3?
Correct!
Tell your approach.
@Ashmukh said:Let A1 be a square whose side is "a" metres. Circle C1 circumscribes the square A such that all its vertices are on C1. Another square A2 circumscribes C1. Circle C2 circumscribes A2, and A3circumscribes C2, and so on. If DN is the area between the square AN and the circle CN, where N is a natural number, then the ratio of the sum of all DN to D1 is:A. 1
B. π/2 -1
C. Infinity
D. None of the above
getting gp
for n =1, 1
n=2, 1+2
n=3, 1+2+4
so we are getting Gp sum depends on N
it should be (2^n-1)
if Dn tends to infinity sum tends to infinity
@invincible2910 said:a single hand wind screen wiper of a car is of length 25 cm and it rotates through an angle of 'x' units which can be adjusted. if the desired wiping area is419.071 cubic cm , then the angle 'x' is -
45;60;90;120
area in cubic cm ???
5 |x €“ y| How many integer solutions exist for the given set of inequalities?
a) 20
b) 10
c) 24
d) 40
here the circle
"circumscribes" the square. and again the square "circumscribes" the circle .hence the annular area increases and wont converge... In this case it is impossible to find the sum of area unless the number n is given..
what is d OA?
@krum @gautam22 @techsurge
@sowmyanarayanan said:S is the region described by the following:1. |x| 2. |y| 3. x > yFind the Perimeter of the region formed by the intersection of S and the unit circle with center at (1,1)1. 82. 43. between 1 and 2.754. between 2.75 and 4
OA Option 4
@chandrakant.k The numbers x and y are three-digit positive integers, and x + y is a four-digit integer. The tens digit of x equals 7 and the tens digit of y equals 5. If x
3rd one is definitly true
and either 1 among 1 and 2 has to b tru :)
@vijay_chandola said:5 a) 20b) 10c) 24d) 40
20?
4,3; -4,-3; - 4 sol
2,3; -2,-3; - 4 sol
1,3; -1,-3; - 4 sol
2,4; -2,-4; - 4 sol
2,2; -2,-2; - 2 sol
3,3; -3,-3; - 2 sol
@techsurge
4,3; -4,-3; - 4 sol
2,3; -2,-3; - 4 sol
1,3; -1,-3; - 4 sol
2,4; -2,-4; - 4 sol
2,2; -2,-2; - 2 sol
3,3; -3,-3; - 2 sol
@techsurge