The answer is 5 cm for the circumference of the smaller square
$725.78
First timer. Do I need to write detailed solution?
Answer:
y = (3/4)x + 2
Step-by-step explanation:
Slope-intercept form is y=mx+b where (x, y) is a point on the linear graph, m is the slope (rise/run), and b is the y-intercept (the y-value at which the graph passes through the y-axis).
Looking at the graph, we can see that the point at which the line crosses the y-axis is (0, 2) which makes it the y-intercept. Thus, the b in the slope-intercept form is 2.
Next, we are looking for the slope of the line. To do this, we can calculate the rise/run of the line by choosing to points on it. Since we already have the point (0, 2), we just need one more.
For example, the point (-4, -1) can be used. The slope can be found by ((y-y)/(x-x)) in which the first y and x values correspond with the first point and that of the second correspond with the second set. So in this case, m = (2-(-1))/(0-(-4)) = 3/4
Plugging in the calculated m and b value in the slope intercept equation, we get y = (3/4)x + 2
Answer: y=−12x+3.
Step-by-step explanation:
1) The average increase in the level of CO2 emissions per year from years 2 to 4 is:
Average=[f(4)-f(2)]/(4-2)=(29,172.15-26,460)/2=2,712.15/2=1,356.075 metric tons. The first is false.
2) The average increase in the level of CO2 emissions per year from years 6 to 8 is:
Average=[f(8)-f(6)]/(8-6)=(35,458.93-32,162.29)/2=3,296.64/2=1,648.32 metric tons. The second is false.
3) The average increase in the level of CO2 emissions per year from years 4 to 6 is:
Average=[f(6)-f(4)]/(6-4)=(32,162.29-29,172.15)/2=2,990.14/2=1,495.07 metric tons. The third is false.
4) The average increase in the level of CO2 emissions per year from years 8 to 10 is:
Average=[f(10)-f(8)]/(10-8)=(39,093.47-35,458.93)/2=3,634.54/2=1,817.27 metric tons. The fourth is true.
Answer: Fourth option: The average increase in the level of CO2 emissions per year from years 8 to 10 is 1,817.27 metric tons.
