The graphs that display the same data are a. I and II
<h3>What is Data Representation?</h3>
This refers to the different ways in which a set of data are displayed in a certain place such as a bar graph, frequency table, line graph, etc.
Hence, we can see that from the evaluation of the data, there is the use of the same data in I and II as the bar and line graphs contain the same country voter turnout.
Read more about bar graphs here:
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It's basically 5,000*2=10,000
and 10,000/2=5,000. your answer would be either they multiplied 5,000 by 2 and 10,000 divided by 2, they increased the sales by 5,000
Answer:
0
Step-by-step explanation:
Write out the 2 equations:
2x + y = -3
-2y = 6+4x
-You then pick a variable that you want to solve for. I chose to solve for y because it would be a bit easier.
-2y = 6 + 4x
-Divide both sides by -2 to isolate the variable
(-2y) / -2 = (6 + 4x) / -2
-it comes out to y = -3 -2x
-Now that you have the y you can plug in the value into the second equation.
2x + y = -3 ---> 2x + (-3 -2x) = -3
This can be simplified to:
2x - 3 - 2x = -3
The 2x's cancel out because there is a +2x and a -2x. You are left with a
-3 = -3. If you add 3 to both sides of the equation you end up with 0 = 0, which can be simplified to just 0.
Answer:
Austin will take 9 hours to paint a room and dale with take 3 hours to paint a room
Step-by-step explanation:
Answer:
- 12 ft parallel to the river
- 6 ft perpendicular to the river
Step-by-step explanation:
The least fence is used when half the total fence is parallel to the river. That is, the shape of the rectangle is twice as long as it is wide.
72 = W(2W)
36 = W²
6 = W . . . . . . the width perpendicular to the river
12 = 2W . . . . the length parallel to the river
_____
<em>Development of this relation</em>
Let T represent the total length of the fence for some area A. Then if x is the length along the river, the width is y=(T-x)/2, and the area is ...
A = xy = x(T -x)/2
Note that the equation for area is that of a parabola with zeros at x=0 and at x=T. That is, for some fence length T, the area will be a maximum at the vertex of this parabola. That vertex is located halfway between the zeros, at ...
x = (0 +T)/2 = T/2
The corresponding area width (y) is ...
y = (T -T/2)/2 = T/4
Equivalently, the fence length T will be a minimum for some area A when x=T/2 and y=T/4. This is the result we used above.
