well if it asks you to send to 4 people and there are 8 generations which includes yours that mean
1 sent to 4 - 4 recived
4 send to 4 each = 16 recived + the 4 before = 20 (generation 2)
16 send to 4 = 64 + 20 = 84 (generation 3)
64 send to 4 = 256 + 84= 320 (g4)
256 send to 4 = 1024 + 320 = 1344 (g5)
1344 s t 4 = 5376 + 1344= 6730 (g6)
so on and so forth till generation 8
Answer:
80+(15x)
Step-by-step explanation:
10 times 8=80
1.5 times 10=15
so she gets 80 dollars for the first 8 hours then for every extra hour she gets 15 dollars
So the given series is "16, 06, 68, 88, __"
Count all the cyclical opening in each of these numbers. For example in 16, there is a one cyclical loop present in it(the one in 6), similarly in 06 it is two(one in zero and one in 6), going ahead, in 68 it is 3(one in 6 and two in 8).
From here on things become simple: hence, the cyclical figures in these equations written down becomes 1,2,3,4,_,3.
Let's now try solving the above sequence, going by the logical reasoning the only number that can fill in the gap should be 4.
Answer:
D. 
Step-by-step explanation:
We are asked to find the GCF of 
.
Since we know that GCF of two numbers is the greatest number that is a factor of both of them.
First of all we will GCF of 44 and 121.
Factors of 44 are: 1, 2, 4, 11, 22, 44.
Factors of 121 are: 1, 11, 11, 121.
We can see that greatest common factor of 44 and 121 is 11.
Now let us find GCF of 
.
Factors of 
are: 
Factors of 
are: 
We can see that greatest common factor of is 
.
Now let us find GCF of 
.
Factors of 
are: 
Factors of 
are:
We can see that greatest common factor of is 
.
Upon combining our all GCFs we will get,
Therefore, GCF of is and option D is the correct choice.
