d = 3 , a₁₂ = 40 and S
= 7775
In an arithmetic sequence the nth term and sum to n terms are
<h3>• a
= a₁ + (n-1)d</h3><h3>• S
=
[2a + (n-1)d]</h3><h3>
where d is the common difference</h3><h3>a₆ = a₁ + 5d = 22 ⇒ 7 + 5d = 22 ⇒ 5d = 15 ⇔ d = 3</h3><h3>a₁₂ = 7 + 11d = 7 +( 11× 3) = 7 + 33 = 40</h3><h3>S₁₀₀ =
[(2×7) +(99×3)</h3><h3> = 25(14 + 297) = 25(311)= 7775</h3>
A: 2,5
B: 3,1
C: -2,4
Explanation:
When you’re moving right and up you would add however many numbers you moved up to the original points because going right on a graph makes the X a larger number, and going up makes it larger.
Answer:
Step-by-step explanation:
Given that among 500 freshmen pursuing a business degree at a university, 315 are enrolled in an economics course, 213 are enrolled in a mathematics course, and 123 are enrolled in both an economics and a mathematics course.
From the above we find that
a) either economics of Math course is
Out of 500 students 405 have taken either Math or Economics
Hence
c) student who have taken neither = 
Exactly one course is either math or economics - both
= 
<span>A has earned twice as much as B so: A=2B
B has averaged 50% more than C so: B=C+0.5C
Combined is 220K, So A+B+C=220,000
The last equation can be re-written as:
2B+1.5C+C=220K
AND since A=2B, it can be rewritten again as:
3C+1.5C+C=220K
Now you can get C
finally, since B=1.5C you can get B, then to get the average, just divide that by 3 (years) and you should be done.
So B = 60000 for 3 years</span>
