x = 4
8 multiplied by itself is 64.
And so, 2 times itself is 4.
The sum of any arithmetic sequence is the average of the first and last terms times the number of terms.
Any term in an arithmetic sequence is:
a(n)=a+d(n-1), where a=initial term, d=common difference, n=term number
So the first term is a, and the last term is a+d(n-1) so the sum can be expressed as:
s(n)(a+a+d(n-1))(n/2)
s(n)=(2a+dn-d)(n/2)
s(n)=(2an+dn^2-dn)/2
However we need to know how many terms are in the sequence.
a(n)=a+d(n-1), and we know a=3 and d=2 and a(n)=21 so
21=3+2(n-1)
18=2(n-1)
9=n-1
10=n so there are 10 terms in the sequence.
s(n)=(2an+dn^2-dn)/2, becomes, a=3, d=2, n=10
s(10)=(2*3*10+2*10^2-2*10)/2
s(10)=(60+200-20)/2
s(10)=240/2
s(10)=120
Answer: $ 17.72
Step-by-step explanation:
Cost of soda = $2.15 , Cost of chicken = $8.77 , Cost of slice of cake = $4.10.
Total bill amount = (Cost of soda ) + (Cost of chicken) +(Cost of slice of cake)
= $ (2.15 + 8.77 +4.10)
= $ 15.02
Tip amount = 18% of Total bill amount
= 0.18 (15.02)
= $ 2.7036 ≈ $2.70
Entire bill = Total bill amount + Tip amount
= $ ( 15.02+2.70)
= $ 17.72
Hence, entire bill = $ 17.72
Answer: 40320 sequence to test.
Step-by-step explanation:
we have given the following data:
No of wires which need to be attached to a circuit board = 8
What We need to find is the number of possible sequences of assembly that must be tested.
Using our multiplication rule of counting we arrive at,
Number of possible sequences of assembly is given by
8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
= 40320
i.e. there are 8 choices for the first wire, 7 choices for the second wire, 6 choices for the third wire, 5 choice for the fourth wire same till the 8th wire.
Hence, there are 40320 possible sequences of assembly.
Answer:
x<_ 4
Step-by-step explanation:
first you distribute 5 to equal 5-35x+5 then you would carry the non variables to one side to be -35x>_-140 and the you would flip the sides because te variable is negative so it would be -35x<_ -140 the divid to get x<_4
