Solve the inequality. Graph the solution set and write it in interval notation.
1/4 (x + 4) < 1/5 (3x + 2)
To solve the inequality, we will first simplify both sides:
1/4 (x + 4) < 1/5 (3x + 2)
Distribute the 1/4 and 1/5:
(x + 4)/4 < (3x + 2)/5
Multiply both sides by 4:
x + 4 < 4/5 (3x + 2)
Multiply both sides by 5:
5(x + 4) < 4 (3x + 2)
Distribute the 5 and 4:
5x + 20 < 12x + 8
Subtract 12x from both sides:
-7x + 20 < 8
Subtract 20 from both sides:
-7x < -12
Divide both sides by -7 (reversing the inequality):
x > 12/7
The solution set is all real numbers greater than 12/7.
Graphing this on a number line, we put an open circle at 12/7 and shade to the right to represent all the values greater than 12/7:
<=======o=========
12/7
To solve the inequality 1/4 (x + 4) < 1/5 (3x + 2), we need to simplify both sides and isolate the variable.
Let's start by simplifying the inequality:
1/4 (x + 4) < 1/5 (3x + 2)
Multiply both sides of the inequality by 20 (the least common multiple of 4 and 5) to eliminate the fractions:
20 * 1/4 (x + 4) < 20 * 1/5 (3x + 2)
5(x + 4) < 4(3x + 2)
5x + 20 < 12x + 8
Now, let's isolate the variable:
5x - 12x < 8 - 20
-7x < -12
Remember to divide by -7, but since we are dividing by a negative number, the inequality flips:
x > -12 / -7
x > 12/7
To graph the solution set, we represent the numbers on a number line. Since x is greater than 12/7, we indicate that the solution lies to the right of 12/7 on the number line.
Using interval notation, we can express the solution set as (12/7, +∞). This means that x is greater than 12/7 but not equal to it, going towards positive infinity.
Therefore, the solution set graphed on a number line and written in interval notation is (12/7, +∞).
To solve the inequality, we need to isolate the variable x.
Given: 1/4(x + 4) < 1/5(3x + 2)
First, we'll distribute the fractions on both sides of the inequality:
1/4 * x + 1/4 * 4 < 1/5 * 3x + 1/5 * 2
Simplifying further, we get:
1/4 * x + 1 < 3/5 * x + 2/5
Now, let's gather the x terms on one side of the inequality and the constant terms on the other side:
1/4 * x - 3/5 * x < 2/5 - 1
Combining the x terms and constant terms, we get:
(1/4 - 3/5) * x < 2/5 - 1
We can simplify further:
(-7/20) * x < -3/5
Now, we divide both sides by the coefficient of x, which is -7/20. Remember that when dividing by a negative number, the inequality sign flips:
x > (-3/5) / (-7/20)
Simplifying the right side:
x > (60/100) / (35/100)
x > 60/35
Simplifying the right side further:
x > 12/7
Now, we have the solution, x > 12/7. Let's represent this solution on a number line graph:
------o---------o------
12/7
The open circle indicates that 12/7 is not part of the solution since the inequality is strict (1/4(x + 4) is strictly less than 1/5(3x + 2)). The arrow to the right of the number line indicates values greater than 12/7.
In interval notation, we can write the solution as x ∈ (12/7, ∞). The interval (12/7, ∞) indicates all values of x greater than 12/7.
Write the answer in interval notation.
The answer in interval notation is (12/7, ∞).
Solve the inequality. Graph the solution set and write it in interval notation.
-6 < 2 (x - 3) <= 11
To solve the inequality, we will first simplify both sides:
-6 < 2(x - 3) <= 11
Distribute the 2:
-6 < 2x - 6 <= 11
Add 6 to all parts of the inequality:
0 < 2x <= 17
Divide all parts by 2 (reversing the inequality for negative numbers):
0/2 < 2x/2 <= 17/2
Simplify:
0 < x <= 17/2
The solution set is all real numbers greater than 0 and less than or equal to 17/2.
Graphing this on a number line, we put an open circle at 0 and a closed circle at 17/2, and shade between them to represent all the values greater than 0 and less than or equal to 17/2:
-------o=========o-----
0 17/2
In interval notation, we can express the solution set as (0, 17/2]. This means that x is greater than 0 and less than or equal to 17/2.
that was the incorrect answer
I apologize for the mistake. Let's redo the solution.
To solve the inequality, we will first simplify both sides:
-6 < 2(x - 3) ≤ 11
Distribute the 2:
-6 < 2x - 6 ≤ 11
Add 6 to all parts of the inequality:
0 < 2x ≤ 17
Divide all parts by 2:
0/2 < 2x/2 ≤ 17/2
Simplify:
0 < x ≤ 17/2
The solution set is all real numbers greater than 0 and less than or equal to 17/2.
Graphing this on a number line, we put an open circle at 0 and a closed circle at 17/2, and shade between them to represent all the values greater than 0 and less than or equal to 17/2:
-------o=========]
0 17/2
In interval notation, we can express the solution set as (0, 17/2]. This means that x is greater than 0 and less than or equal to 17/2.
The perimeter of a rectangle is to be no greater than 80 centimeters and the length must be 25 centimeters. Find the maximum width of the rectangle.
Let's denote the width of the rectangle as w.
The perimeter of a rectangle is given by the formula: P = 2(length + width)
Given that the length is 25 centimeters and the perimeter must be no greater than 80 centimeters, we can set up the following inequality:
2(25 + w) ≤ 80
Simplify this inequality:
50 + 2w ≤ 80
Subtract 50 from both sides:
2w ≤ 30
Divide both sides by 2:
w ≤ 15
Therefore, the maximum width of the rectangle should be 15 centimeters to ensure that the perimeter is no greater than 80 centimeters.
Steve and Michelle Neely are celebrating their 30th anniversary by having a reception at a local reception hall. They have budgeted $2,500 for their reception. If the reception hall charges a $80 cleanup fee plus $39 per person, find the greatest number of people that they may invite and still stay within their budget.
Let's denote the number of people they invite as "x".
The total cost of the reception hall is composed of two parts:
1. Cleanup fee: $80.
2. Cost per person: $39 * x (since each person costs $39).
We need to find the value of x that satisfies the budget constraint, where the total cost is no greater than $2,500:
80 + 39x ≤ 2500
Subtract 80 from both sides:
39x ≤ 2420
Divide both sides by 39:
x ≤ 62.051
Since the number of people must be a whole number, the greatest number of people they may invite is 62.
Therefore, they may invite a maximum of 62 people and still stay within their budget.