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Sholpan [36]
3 years ago
5

luke jogged a total of 56.5 miles last week. if luke jogged the same distance,d, each day for 3 days and half that distance on e

ach of the reamaining days, which algebraic equation can he use to determine how many miles he jogged each day
Mathematics
1 answer:
Leokris [45]3 years ago
3 0
The algebraic equation used to answer this question is:
3d + 4(1/2d) = 56.5
This is because he ran d, the distance, three times that week. For the rest of the four days, he ran 1/2 the distance, or half d.
To find the answer, you first distribute:
3d + 2d = 56.5
Add like terms:
5d = 56.5
Divide both side by 5:
5d/5 = 56.5/5
d = 11.3
On the first three days, Luke jogged 11.3 miles. On the remaining days, where he jogged 1/2d, Luke jogged 5.65 miles.
Hope this helps!
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Need help please help me
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The solution is the point of intersection between the two equations.

Assuming you have a graphing calculator or a program to lets you graph equations (I use desmos) you simply put in the equetions and note down the coordinates of the point of intersection.

In the graph the first equation is in blue and the second in red.

The point of intersection = the solution = (-6 , -1)



If you dont have access to a graphing calculator you could draw the graphs by hand;

1) Draw a table of values for each equation; you do this by setting three or four values for x and calculating its image in y (you can use any values of x)

y = 0.5 x + 2 (Im writing 0.5 instead of 1/2 because I find its easier in this format)

x | y
-1 | 1.5 * y = 0.5 (-1) + 2 = 1.5
0 | 2 * y = 0.5 (0) + 2 = 2
1 | 2.5 * y = 0.5 (1) + 2 = 2.5
2 | 3 * y = 0.5 (2) + 2 = 3

y = x + 5

x | y
-1 | 4 * y = (-1) + 5 = 4
0 | 5 * y = (0) + 5 = 5
1 | 6 * y = (1) + 5 = 6
2 | 7 * y = (2) + 5 = 7

2) Plot these point on the graph
I suggest to use diffrent colored points or diffrent kinds of point markers (an x or a dot) to avoid confusion about which point belongs to which graph

3) Using a ruler draw a line connection all the dots of one graph and do the same for the other

4) The point of intersection is the solution

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