Step-by-step explanation:
1. Let x be the lesser even integer.
The greater even integer is x+2.
The inequality is: 3x ≤ 5(x+2)
Solving for x, we get x ≤ 5.
Possible values of x are 2, 4, and 6.
2. Let x be the first even integer.
The second even integer is x+2.
The inequality is: x + (x+2) < 25
Solving for x, we get x < 11.
Possible values of x are 2, 4, 6, 8, and 10.
3. Let B be Ben's age and V be Ven's age.
The inequality is: B + V > 24 and B+2 = 3(V+2)
Substituting B+2 = 3(V+2) to B, we get B = 3V+4.
Substituting B = 3V+4 to the first inequality, we get 4V > 16.
Solving for V, we get V > 4.
Possible values of V are 5, 6, 7, ..., and so on.
Then, B = 3V+4.
4. Let x be the lowest grade in the next test.
The inequality is: (75+82+78+80+x)/5 < 80
Solving for x, we get x < 83.
Ronnel's lowest grade in the next test must be less than 83.
5. Let P be the cost of the pants and B be the cost of the blouse.
The inequality is: P + B ≤ 1400 and P = 2B-100
Substituting P = 2B-100 to the first inequality, we get 3B ≤ 1500.
Solving for B, we get B ≤ 500.
The greatest amount that Regine can spend for the blouse is Php 500.
pa mark as brainlist! thanks
Copyright © 2024 EHUB.TIPS team's - All rights reserved.
Answers & Comments
Step-by-step explanation:
1. Let x be the lesser even integer.
The greater even integer is x+2.
The inequality is: 3x ≤ 5(x+2)
Solving for x, we get x ≤ 5.
Possible values of x are 2, 4, and 6.
2. Let x be the first even integer.
The second even integer is x+2.
The inequality is: x + (x+2) < 25
Solving for x, we get x < 11.
Possible values of x are 2, 4, 6, 8, and 10.
3. Let B be Ben's age and V be Ven's age.
The inequality is: B + V > 24 and B+2 = 3(V+2)
Substituting B+2 = 3(V+2) to B, we get B = 3V+4.
Substituting B = 3V+4 to the first inequality, we get 4V > 16.
Solving for V, we get V > 4.
Possible values of V are 5, 6, 7, ..., and so on.
Then, B = 3V+4.
4. Let x be the lowest grade in the next test.
The inequality is: (75+82+78+80+x)/5 < 80
Solving for x, we get x < 83.
Ronnel's lowest grade in the next test must be less than 83.
5. Let P be the cost of the pants and B be the cost of the blouse.
The inequality is: P + B ≤ 1400 and P = 2B-100
Substituting P = 2B-100 to the first inequality, we get 3B ≤ 1500.
Solving for B, we get B ≤ 500.
The greatest amount that Regine can spend for the blouse is Php 500.
pa mark as brainlist! thanks