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stira [4]
2 years ago
8

Jacob finished his test 10 minutes before Mark. Which equation represents the time, j, Jacob spent on the test compared to the t

ime, m, Mark spent on the test?
Mathematics
1 answer:
Marina CMI [18]2 years ago
6 0

Answer:

The time Jacob spent on the test represented by J = (m + 10) minutes

where m is the number of minutes spent by Mark on the test

Step-by-step explanation:

Here, we want to give an equation which represents the time taken for Jacob to finish his test compared to the time m which Mark spent on the test.

Since Jacob finished 10 minutes before Mark, it means Mark actually took longer on the test.

Now, we are told that Mark spent m minutes on the test, the time spent by Jacob on the test is thus (m + 10) minutes

Let the amount of time spent by Jacob be J minutes. This in relation to the amount of time spent by Mark will be ;

J = (m + 10) minutes

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Money spent by Kate = K

Money spent by Lauren = L

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K = L + 1

Together they spent $5...

L + K = 5

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Answer:

Patricia will run 132 kilometers with 11 liters of gas.

Step-by-step explanation:

The given table is representing the distance driven by Patricia for every liter of gas she puts in her car.

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soldier1979 [14.2K]

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1. Graphs

I used Excel to calculate the pH values and draw the graphs (see the Figure).

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The points at which pH = 0 and pH = 1 are indicated by the large red dots.

2. x = 0.5

When x = 0.5, pH ≈0.30. The point is indicated by the red diamond.

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This function has no y-intercept, because log(0) is undefined.

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Point T, the midpoint of segment RS, can be found using the formulas x = (6 – 2) + 2 and y = (4 – 6) + 6. What are the coordinat
Fiesta28 [93]

Question:

Point T, the midpoint of segment RS, can be found using the formulas x = (1/2) (6 – 2) + 2 and y = (1/2) (4 – 6) + 6. What are the coordinates of point T?

Answer:

T(4,5)

Step-by-step explanation:

Given

x = \frac{1}{2}(6 - 2) + 2

y = \frac{1}{2}(4 - 6) + 6

Required

Determine the coordinates of T

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Hence, the coordinates of T(x,y) is:

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Move the decimal back to its point of origin

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