Notice that
13 - 9 = 4
17 - 13 = 4
so it's likely that each pair of consecutive terms in the sum differ by 4. This means the last term, 149, is equal to 9 plus some multiple of 4 :
149 = 9 + 4k
140 = 4k
k = 140/4
k = 35
This tells you there are 35 + 1 = 36 terms in the sum (since the first term is 9 plus 0 times 4, and the last term is 9 plus 35 times 4). Among the given options, only the first choice contains the same amount of terms.
Put another way, we have
but if we make the sum start at k = 1, we need to replace every instance of k with k - 1, and accordingly adjust the upper limit in the sum.
Answer:
The answer is B.
Step-by-step explanation:
Answer:
Step-by-step explanation: inputs because am pro
Answer:
(x, y) = (4, 9)
Step-by-step explanation:
You can use the second equation to write an expression for y.
y = 10 - 1/4x . . . . subtract 1/4x from the given 2nd equation
Now, substitute that for y in the first equation:
1/2x + 1/3(10 -1/4x) = 5
1/2x + 10/3 -1/12x = 5 . . . . eliminate parentheses
You can work this a couple of ways from here. One is to multiply by 12 to eliminate fractions. Another is to work with the numbers as they are. We'll do that, because we don't get enough practice doing arithmetic with fractions.
5/12x = 5/3 . . . . . subtract 10/3
x = (5/3)/(5/12) = 12/3 = 4
From above, we can use our expression for y:
y = 10 - 1/4·4 = 9
The solution is (x, y) = (4, 9).
