Answer:
Vol. of the composite figure is 189 m³
Step-by-step explanation:
Find the volume of the larger figure by mult. together its 3 dimensions:
V = (9 m)(8 m)(3 m) = 216 m³.
Next, find the volume of the "notch," which is (3 m)³, or 27 m³.
Finally, subtract the "notch" volume from the 216 m³ volume found earlier:
216 m³ - 27 m³ = 189 m³
Answer:
y =-5/4x +11/2
Step-by-step explanation:
First we need to find the slope of the line
4x-5y = 10
Solve for y
Subtract 4x from each side
4x-4x-5y =-4x-10
-5y = -4x-10
Divide by -5
-5y/-5 = -4x/-5 -10/-5
y = 4/5x +2
The slope is 4/5
The slope of a line that is perpendicular is the negative reciprocal
m perpendicular = - (5/4)
We know have a slope = -5/4 and a point (2,3)
We can use the point slope form of a line
y-y1 = m(x-x1)
y-3 = -5/4 (x-2)
Distribute
y-3 =-5/4x +10/4
Add 3 to each side
y-3+3 = -5/4 x +5/2 +3
Get a common denominator
y = -5/4x +5/2 +6/2
y =-5/4x +11/2
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.
Answer:
a
Step-by-step explanation:
Your answer is
- Slope = -1.000/2.000 = -0.500
- x-intercept = 7/1 = 7.00000
- y-intercept = 7/2 = 3.50000
Step-by-step explanation:
