I would use elimination. I would multiply 4x+3y=2 by 3 so the it becomes 12x+9y=6, and I would multiply the bottom equation by 4 so that it becomes 12x+8y=4. Because I am using elimination I have to get rid of one variable, I chose to get rid of X hence that the coefficient of X in each equation is now 12. i simply subtract both equations and get y by itself. y=2. I plug in Y to find x in either of the original equations, where I get x=-1. To check my work substitute the values of x and y in the other equation. 4(-1)+3(2)=2 -------> 2=2
Answer:
C. an= 9(1/3)n-1
Step-by-step explanation:
The decreasing curve means the base of the exponential factor is less than 1, eliminating choices B and D. The value of the term is 9 at n=1, so eliminating choice A.
The general form of the n-th term is ...
an = a1·r^(n-1) . . . . first term a1, common ratio r
so we expect to see an exponent of (n-1) when a1 is the value for n=1. If you match the pattern above to the answer choices, you see the only decision required is whether r=3 or r=1/3. The decaying curve of the graph shows you that r < 1, so the choice is r=1/3 and the formula is ...
an = 9(1/3)^(n-1) . . . . . . . parentheses are needed on the exponent
<span>7m + 2/8 = 2m - 1/3
(7m X 24) + (2/8 X 24) = (2m X 24) - (1/3 X 24)
168m + 6 = 48m - 8
120m = -14
m = -0.1166</span>
The answer is c.
I took 130 and multiplied by 1.7 (.7 because of the mark up and 1 because I don't want to add the original back to the answer) and I got 221
You could also take 130 times .7 then add 130 to get the same answer.
I hope this helps.
So the problem wants to compute what could David should score in the fourth math test to ensure an average score of at least 90. So base on the score the possible total of the four score must be 360 to make their average of minimum of 90 so the answer would be letter B. X>93