Which of the following points is a solution to the linear inequality y>19x−29 ?
(0,−19) left parenthesis 0 comma negative Start Fraction 1 over 9 End Fraction right parenthesis (0,−29) left parenthesis 0 comma negative Start Fraction 2 over 9 End Fraction right parenthesis (0,−49) left parenthesis 0 comma negative Start Fraction 4 over 9 End Fraction right parenthesis (0,−13)
To determine which point is a solution to the linear inequality y > 19x - 29, we substitute the x and y values of each point into the inequality and see if the resulting inequality holds true.
1) (0,-19):
-19 > 19(0) - 29
-19 > -29
This is true, so (0,-19) is a solution.
2) (0, 1/9):
1/9 > 19(0) - 29
1/9 > -29
This is true, so (0, 1/9) is a solution.
3) (0, -29):
-29 > 19(0) - 29
-29 > -29
This is NOT true, so (0, -29) is not a solution.
4) (0, 2/9):
2/9 > 19(0) - 29
2/9 > -29
This is true, so (0, 2/9) is a solution.
5) (0, -49):
-49 > 19(0) - 29
-49 > -29
This is NOT true, so (0, -49) is not a solution.
6) (0, -13):
-13 > 19(0) - 29
-13 > -29
This is true, so (0, -13) is a solution.
From the given points, the solutions to the linear inequality y > 19x - 29 are (0, -19), (0, 1/9), (0, 2/9), and (0, -13).
Pick one option
One option is (0, -19)
That is not an option
Apologies for the confusion. The correct option from the given points that is a solution to the linear inequality y > 19x - 29 is (0, -13).
Which of the following points is a solution to the linear inequality y>1/9x−2/9 ?
(0,−1/9) left parenthesis 0 comma negative Start Fraction 1 over 9 End Fraction right parenthesis (0,−2/9) left parenthesis 0 comma negative Start Fraction 2 over 9 End Fraction right parenthesis (0,−4/9) left parenthesis 0 comma negative Start Fraction 4 over 9 End Fraction right parenthesis (0,−1/3)
To determine which point is a solution to the linear inequality y > (1/9)x - (2/9), we substitute the x and y values of each point into the inequality and check if the resulting inequality holds true.
1) (0, -1/9):
-1/9 > (1/9)(0) - (2/9)
-1/9 > -2/9
This is true, so (0, -1/9) is a solution.
2) (0, -2/9):
-2/9 > (1/9)(0) - (2/9)
-2/9 > -2/9
This is NOT true, so (0, -2/9) is not a solution.
3) (0, -4/9):
-4/9 > (1/9)(0) - (2/9)
-4/9 > -2/9
This is NOT true, so (0, -4/9) is not a solution.
4) (0, -1/3):
-1/3 > (1/9)(0) - (2/9)
-1/3 > -2/9
This is true, so (0, -1/3) is a solution.
From the given points, the solutions to the linear inequality y > (1/9)x - (2/9) are (0, -1/9) and (0, -1/3).
Is the point (0,−3) a solution to the inequality y>4/5x−3 , as shown on the graph?
Yes, because (0,−3) falls within the shaded region. Yes, because left parenthesis 0 comma negative 3 right parenthesis falls within the shaded region. No, because (0,−3) falls within the shaded region. No, because left parenthesis 0 comma negative 3 right parenthesis falls within the shaded region. Yes, because (0,−3) falls on the boundary line. Yes, because left parenthesis 0 comma negative 3 right parenthesis falls on the boundary line. No, because (0,−3) falls on the boundary line. No, because left parenthesis 0 comma negative 3 right parenthesis falls on the boundary line.
No, because (0, -3) falls on the boundary line.
Show which of the following points is part of the solution set for the inequality y≤2/5x+1 .
(10,5.5) left parenthesis 10 comma 5.5 right parenthesis (−10,6) left parenthesis negative 10 comma 6 right parenthesis (10,6) left parenthesis 10 comma 6 right parenthesis (10,4)
To determine which point is part of the solution set for the inequality y ≤ (2/5)x + 1, we substitute the x and y values of each point into the inequality and check if the resulting inequality holds true.
1) (10, 5.5):
5.5 ≤ (2/5)(10) + 1
5.5 ≤ 4 + 1
5.5 ≤ 5.5
This is true, so (10, 5.5) is part of the solution set.
2) (-10, 6):
6 ≤ (2/5)(-10) + 1
6 ≤ -4 + 1
6 ≤ -3
This is NOT true, so (-10, 6) is not part of the solution set.
3) (10, 6):
6 ≤ (2/5)(10) + 1
6 ≤ 4 + 1
6 ≤ 5
This is NOT true, so (10, 6) is not part of the solution set.
4) (10, 4):
4 ≤ (2/5)(10) + 1
4 ≤ 4 + 1
4 ≤ 5
This is TRUE, so (10, 4) is part of the solution set.
From the given points, the point that is part of the solution set for the inequality y ≤ (2/5)x + 1 is (10, 5.5)
Question Kiani is shopping at a bookstore with a budget of $60. Hardbacks cost $12 and paperbacks cost $5. Write a two-variable inequality and determine the constraints of the problem. Which of the following points represents a viable solution?
(6,6) left parenthesis 6 comma 6 right parenthesis (1,4.5) left parenthesis 1 comma 4.5 right parenthesis (2,4) left parenthesis 2 comma 4 right parenthesis (−2,6)
Let x represent the number of hardbacks and y represent the number of paperbacks that Kiani buys.
The cost constraint can be expressed as: 12x + 5y ≤ 60
We need to find which of the given points is a viable solution.
1) (6, 6):
12(6) + 5(6) = 72 + 30 = 102
This is NOT a viable solution as the total cost exceeds Kiani's budget.
2) (1, 4.5):
12(1) + 5(4.5) = 12 + 22.5 = 34.5
This is a viable solution as the total cost is within Kiani's budget.
3) (2, 4):
12(2) + 5(4) = 24 + 20 = 44
This is a viable solution as the total cost is within Kiani's budget.
4) (-2, 6):
12(-2) + 5(6) = -24 + 30 = 6
This is a viable solution as the total cost is within Kiani's budget.
From the given points, the point that represents a viable solution to the two-variable inequality is (1, 4.5).
This graph shows the inequality 25x+15y≥200 , and represents Lina’s earnings for lawn mowing and dog walking where x represents the number of lawns mowed and y represents the number of dogs walked. Which of the labeled points is a viable solution in this context?
(20,11.5) left parenthesis 20 comma 11.5 right parenthesis (2,3) left parenthesis 2 comma 3 right parenthesis None of the points are a viable solution. None of the points are a viable solution. (10,10)
To determine which of the labeled points is a viable solution in the context of the inequality 25x + 15y ≥ 200, we substitute the x and y values of each point into the inequality and check if the resulting inequality holds true.
1) (20, 11.5):
25(20) + 15(11.5) = 500 + 172.5 = 672.5
This is NOT a viable solution as the earnings exceed or equal to 200.
2) (2, 3):
25(2) + 15(3) = 50 + 45 = 95
This is NOT a viable solution as the earnings do not meet the requirement of 200.
3) (10, 10):
25(10) + 15(10) = 250 + 150 = 400
This is a viable solution as the earnings meet or exceed 200.
From the labeled points, the point that is a viable solution in the context of the inequality is (10, 10).