@jain4444 2.4?
@jain4444 said:aur sir jii hum apse sikhte aa rahe hai Sohan and Rohan cut a regular hexagon and regular octagon out of circular papers of radius 20 each.find difference between perimeters of hexagon and octagon1.62.422.6
2.4??not sure
@jain4444 said:aur sir jii hum apse sikhte aa rahe hai Sohan and Rohan cut a regular hexagon and regular octagon out of circular papers of radius 20 each.find difference between perimeters of hexagon and octagon1.62.422.6
2.4
@jain4444 said:aur sir jii hum apse sikhte aa rahe hai Sohan and Rohan cut a regular hexagon and regular octagon out of circular papers of radius 20 each.find difference between perimeters of hexagon and octagon1.62.422.6
2.4
1. Given a set A={1,2,3,4,5} B= {a,b,c} f:A->B
Find the :
a. Total Number of functions
b. Number of Surjective functions
c. Number of into functions
d. Number of bijective functions
e. Number of injective functions
Any idea about this ??
Petya's mother sends him to the market with a red and a green bag to buy 10 carrots and 6 radishes.On his way home, Petya distributes the vegetables into 2 bags in such a way that no bag is empty. In how many ways can he do this?
@grkkrg said:Petya's mother sends him to the market with a red and green bag to buy 10 carrots and 6 radishes.On the way home, Petya distributes the vegetables into 2 bags in such a way that no bag is empty .In how ways can he can do this?
30 ??
@grkkrg said:Petya's mother sends him to the market with a red and a green bag to buy 10 carrots and 6 radishes.On his way home, Petya distributes the vegetables into 2 bags in such a way that no bag is empty. In how many ways can he do this?
64 ?
@Brooklyn said:
Bhai I have used cosine rule
For Hexagon :
cos (360/6) = (20^2+20^2-m^2)/(2*20*20)
m=20=>perimeter=20*6=120
For octagon:
cos (360/8)=(20^2+20^2-n^2)/(2*20*20)
n=sqrt(234.3)=>perimeter=8*sqrt(234.3)=8*15.3=122.4
(sqrt(234) is approx 15.3)
Difference in perimeter=122.4-120=2.4
@grkkrg said:Petya's mother sends him to the market with a red and a green bag to buy 10 carrots and 6 radishes.On his way home, Petya distributes the vegetables into 2 bags in such a way that no bag is empty. In how many ways can he do this?
a+b=10
=>11c1=11
0,6
1,5
2,4
3,3
4,2
5,1
6,0
so 11*7=77
77-2=75
=>11c1=11
0,6
1,5
2,4
3,3
4,2
5,1
6,0
so 11*7=77
77-2=75
@Brooklyn said:didnt get u
distribute 10 carrots in 2 bags => (10 + 2 - 1)C(2 -1) = 11
distribute 6 radishes in 2 bags => (6 + 2 - 1)C(2 - 1) = 7
total = 11 * 7 = 77
But the red bag is empty in one case and the green bag in another.
So 77 - 2 = 75
distribute 6 radishes in 2 bags => (6 + 2 - 1)C(2 - 1) = 7
total = 11 * 7 = 77
But the red bag is empty in one case and the green bag in another.
So 77 - 2 = 75
@Brooklyn said:didnt get u
first distribute 10 carrots , so 11 ways, including 0,10 and 10,0
now for each 11 case, there are 7 ways of distributing radishes , so total = 11*7=77
but these includes 2 cases which are not valid
that is when red bag had 0,10 and 0,6 and same for green bag, so 77-2=75
now for each 11 case, there are 7 ways of distributing radishes , so total = 11*7=77
but these includes 2 cases which are not valid
that is when red bag had 0,10 and 0,6 and same for green bag, so 77-2=75
@jain4444 said:aur sir jii hum apse sikhte aa rahe hai Sohan and Rohan cut a regular hexagon and regular octagon out of circular papers of radius 20 each.find difference between perimeters of hexagon and octagon1.62.422.6
cos 60=(2r^2-x^2)/(2r^2)
=>x=r
cos 45=(2r^2-y^2)/(2r^2)
=>y=r*_/(2-_/2)=0.7653*r
8*0.7653*20 - 6*20
=122.448 - 120
=2.448
2.4
=>x=r
cos 45=(2r^2-y^2)/(2r^2)
=>y=r*_/(2-_/2)=0.7653*r
8*0.7653*20 - 6*20
=122.448 - 120
=2.448
2.4