Sequence: 5/2, 5/4, 5/8, 5/16
a8=?
a1=5/2
a2=5/4
a3=5/8
a4=5/16
a2/a1=(5/4)/(5/2)=(5/4)*(2/5)=(5*2)/(4*5)=2/4=1/2
a3/a2=(5/8)/(5/4)=(5/8)*(4/5)=(5*4)/(8*5)=4/8=1/2
a4/a3=(5/16)/(5/8)=(5/16)*(8/5)=(5*8)/(16*5)=8/16=1/2
Ratio: r=a2/a1=a3/a2=a4/a3→r=1/2
an=a1*r^(n-1)
a1=5/2, r=1/2
an=(5/2)*(1/2)^(n-1)
an=(5/2)*[1^(n-1)/2^(n-1)]
an=(5/2)*[1/2^(n-1)]
an=(5*1)/[2*2^(n-1)]
an=5/2^(1+n-1)
an=5/2^n
n=8→a8=5/2^8
a8=5/256
Answers:
The formula for the general term or nth term for the sequence is an=5/2^n
a8=5/256
Answer : The value of x and y is 8 and 10 respectively.
Step-by-step explanation :
As we known that if two parallel lines are cut by a transversal line then consecutive interior angles are supplementary.
From the given figure we conclude that:
...........(1)
............(2)
...........(3)
...........(4)
Now we adding equation 2 and 4, we get the value of x.
Now we are putting the value of x in 4, we get the value of y.
Therefore, the value of x and y is 8 and 10 respectively.
Answer:
302
Step-by-step explanation:
IQR= 18 is the interquartile range of his scores.
255/56 = 4.55....
You can't have 0.55... buses so you round that up to 1.
Therefore 4+1 = 5 so 5 buses would be the minimum.
