Multiply the first equation by -2 gives:_
-2y = -8 + 2x
2y = 8 - 2x
adding these 2 equations:-
0 = 0
This shows that the 2 equations are equal so there are infinite solutions
Answer:
Seth saved $90.
Step-by-step explanation:
1.Find out how much would be spent without buying the movie pass.
Multiply the amount of money that one would spend for each movie, by the number of movies one saw.
8 * 30 = 240
2.Find out the amount that was actually spent
Multiply the amount that was actually spent on each movie, by the number of movies that were seen. Then add the additional cost of the pass.
(4 * 30) + 30
= 120 + 30
= 150
3. Find the difference between the two values.
Subtract the amount that one was supposed to spend by the amount one did spend.
240 - 150
= 90
The correct answer is that there is more variability in the heights of the volleyball team members.
The mean absolute deviation shows us how spread out the data is, so the larger the mean absolute deviation the higher the variability.
Both teams have players that are 76 inches tall, so the last two statements cannot be true.
C!
One dollar = $1
And in a math equation is usually a decimal
56 cents is .56 because it’s not a whole new If you add them together you get $1.56 :) hope it helps
First of all, when I do all the math on this, I get the coordinates for the max point to be (1/3, 14/27). But anyway, we need to find the derivative to see where those values fall in a table of intervals where the function is increasing or decreasing. The first derivative of the function is

. Set the derivative equal to 0 and factor to find the critical numbers.

, so x = -3 and x = 1/3. We set up a table of intervals using those critical numbers, test a value within each interval, and the resulting sign, positive or negative, tells us where the function is increasing or decreasing. From there we will look at our points to determine which fall into the "decreasing" category. Our intervals will be -∞<x<-3, -3<x<1/3, 1/3<x<∞. In the first interval test -4. f'(-4)=-13; therefore, the function is decreasing on this interval. In the second interval test 0. f'(0)=3; therefore, the function is increasing on this interval. In the third interval test 1. f'(1)=-8; therefore, the function is decreasing on this interval. In order to determine where our points in question fall, look to the x value. The ones that fall into the "decreasing" category are (2, -18), (1, -2), and (-4, -12). The point (-3, -18) is already a min value.