Answer:
the 19th term of the arithmetic sequence is 73.
Step-by-step explanation:
Given:
The arithmetic sequence whose common difference is
And first term is

Find the 19th term of the arithmetic sequence

Now we know that,

Put all the value in above AP.




So,
term of the sequence is 73.
<span>13-((4/5)+(6/8))
Make your fractions have common denominators
</span>13-((32/40)+(30/40))
Add your fractions and simplify
13-(62/40)
or
13-(31/20)
or
13-(1 11/20)
Then turn 13 into a fraction with a common denominator! Im going to use the second fraction method (31/20)
13 written as a fraction is 13/1, its LCD with 31/20 is 20. I now multiply the top and bottom by 20
260/20
Now I rewrite the problem again
(260/20)-(31/20)
Which equals
229/20!
This is your unsimplified answer
Finally you simplify and get
11 9/20
From the instructions i gather it might be like this
The answer to the question is 1-5
Answer:
The students can group themselves in 360360 ways
Step-by-step explanation:
For this exercise we need to use the following equation:

This equation give us the number of assignation of n elements in k cell, where n1, n2, ..nk are the element that are in every cell
In this case we have 15 student that need to be assign in three vehicles with an specific capacity. This vehicles would be the equivalent to cells, so we can write the equation as:

Because the first vehicle have 7 seating, the second vehicle have 5 seating and the third vehicle have 3 seating.
Solving the equation we get 360360 ways to organized 15 students in three vehicles with capacity of 7, 5 and 3 seating.