The ordered pair which makes both inequalities true is: D. (3, 0).
<h3>How to determine ordered pair?</h3>
In Mathematics, an inequality can be used to show the relationship between two (2) or more integers and variables in an equation.
In order to determine ordered pair which makes both inequalities true, we would substitute the points into the inequalities as follows:
At (0, 0), we have:
y > -2x + 3
0 > -2(0) + 3
0 > 3 (false).
y < x – 2
0 < 0 - 2
0 < -2 (false)
At (0, -1), we have:
y > -2x + 3
-1 > -2(0) + 3
-1 > 3 (false).
y < x – 2
-1 < 0 - 2
-1 < -2 (false)
At (1, 1), we have:
y > -2x + 3
1 > -2(1) + 3
1 > -1 (true).
y < x – 2
1 < 1 - 2
1 < -1 (false)
At (3, 0), we have:
y > -2x + 3
0 > -2(3) + 3
0 > -3 (true).
y < x – 2
0 < 3 - 2
0 < 1 (true).
Read more on inequalities here: brainly.com/question/24372553
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Answer:
Step-by-step explanation:
ok. you said 2000 and 2500 so I'll just solve for both.
this is very simple, mis/mia wants to earn 2000/2500 so let's solve
this can be done two ways. an equation way, or a dividing way. I personally like the dividing way.
2000 divided by 55= 36.36-- so she will have to tune up 37 bicycles
2500 divided by 55= 45.45-- so she will have to tune 46 bicycles. this is because 46 bicycles, times the amount of money == just a little over 2500 so she has reached her goal
Answer:
Option B.
Step-by-step explanation:
We need to find the range of numbers is most appropriate for the y-axis scale and interval on a graph for given table.
From the given table it is clear that the minimum value of y is 10 and the maximum value of y is 34.
It means 10 and 34 must be included in the Range.
In option C and D 34 in not included in the range, so these options are incorrect.
In option A, 10 is the minimum value of range and interval of 10 is not possible for the range 10-35, because it contains only multiple of 10 on the y-axis. So, this option is incorrect.
In option B, both 10 and 34 are included and interval of 5 is possible for range 0-35.
Therefore, the correct option is B.
Sorry not sure about this