A rectangle is 8cm long and b cm broad. Find the range of values of b if the perimeter of the rectangle is not greater than 50 cm and not less than 18 cm.
2L + 2b = 18 cm, min.
16 + 2b = 18,
b = 1 cm, min.
2L + 2b = 50 cm, max.
16 + 2b = 50,
b = 17 cm, max.
18 <= 2(8+b) <= 50
9 <= 8+b <= 25
1 <= b <= 17
hello, please I don't understand the solution, new here
18 <=2(8+b)<50
9<=8+b<=25
Therefore the lowest possible number is 1<=b
Highest possible number is b<17
😶
A rectangle is 10 cm long and x cm broad find the range of x cm if the perimeter of the rectangle is not greater than 60 cm and not less than 20 cm
The perimeter of a rectangle is given by the formula P = 2(L + b), where P is the perimeter, L is the length, and b is the breadth.
Given that the length L = 10 cm, we can write the inequality for the perimeter as:
20 ≤ 2(10 + x) ≤ 60
Dividing both sides by 2, we get:
10 ≤ 10 + x ≤ 30
Subtracting 10 from all sides, we get:
0 ≤ x ≤ 20
Therefore, the range of values for x is 0 cm ≤ x ≤ 20 cm.
A woman is 14 years older than her son. In 5 years time the will not be more than 36 years old. What is the likely range of the present age of the son?
Let the present age of the son be x years. Then, the present age of the woman is x + 14 years.
In 5 years time, the age of the son will be x+5 years and the age of the woman will be x+14+5 = x+19 years.
According to the problem, the sum of their ages in 5 years time will not exceed 36 years. Therefore,
(x + 5) + (x + 19) ≤ 36
2x + 24 ≤ 36
2x ≤ 12
x ≤ 6
So, the present age of the son is less than or equal to 6.
Since the woman is 14 years older than the son, her age cannot be less than 20 years. Therefore, the present age of the son is between 0 and 6 years.
Hence, the likely range of the present age of the son is 0 ≤ x ≤ 6.
If I add 5 to a number and then multiply the result by 3 the result is not less than 30. Find the range of the unknown numbr
Let the unknown number be x. According to the given condition,
3(x + 5) ≥ 30
Dividing both sides by 3, we get:
x + 5 ≥ 10
Subtracting 5 from both sides, we get:
x ≥ 5
Therefore, the range of the unknown number is x ≥ 5. This means that the unknown number can be any number greater than or equal to 5.
The cost of kerosine rises #1.50 a litre last week a driver bought 30 litres at the old price. This week he bought 10 litres at the new price. All together the kerosine cost not less than #300 what was the range of the old price for a litre of kerosine
Let the old price of kerosene per liter be x.
Last week, the driver bought 30 liters of kerosene at the old price, so he spent 30x naira.
This week, the driver bought 10 liters of kerosene at the new price, which is x+1.50 naira per liter. Therefore, he spent 10(x+1.50) naira.
The total amount of money he spent on kerosene is not less than 300 naira. So we can write:
30x + 10(x+1.50) ≥ 300
Simplifying and solving for x, we get:
40x + 15 ≥ 300
40x ≥ 285
x ≥ 7.125
Therefore, the old price of kerosene per liter was not less than #7.125.
Note that we rounded off to three decimal places, but the actual value could be slightly higher depending on how many decimal places we use.
Solve these inequalities. Sketch a graph of each solution
12≥3x
To solve the inequality 12 ≥ 3x, we need to isolate x on one side of the inequality. We can do this by dividing both sides by 3:
12/3 ≥ 3x/3
4 ≥ x
So the solution to the inequality is x ≤ 4.
To sketch a graph of this solution, start by drawing a number line and marking a closed circle at 4 (since x can be equal to 4) and shading to the left of the circle, because all values of x less than or equal to 4 make the inequality true. The graph should look like this:
```
----------------●================>
4
```
The ● represents a closed circle at 4, and the shaded region to the left of the circle represents the solution to the inequality.
Solve these inequalities.sketch a graph of each solution
1. -10<5x
2. 4x≥-9
3. 8≤3x
4. 3x+1<13
5. 5x-2≥8
6. -5>4x+15
7. 3≤17+2x
8. 4x-2>19
9. 3≤3+5
1. -10 < 5x
To solve the inequality -10 < 5x, we need to isolate x on one side of the inequality. We can do this by dividing both sides by 5:
-10/5 < 5x/5
-2 < x
So the solution to the inequality is x > -2.
To sketch a graph of this solution, start by drawing a number line and marking an open circle at -2 (since x cannot be equal to -2) and shading to the right of the circle, because all values of x greater than -2 make the inequality true. The graph should look like this:
```
<================●----------------
-2
```
The ● represents an open circle at -2, and the shaded region to the right of the circle represents the solution to the inequality.
2. 4x ≥ -9
To solve the inequality 4x ≥ -9, we need to isolate x on one side of the inequality. We can do this by dividing both sides by 4:
4x/4 ≥ -9/4
x ≥ -9/4
So the solution to the inequality is x ≥ -2.25.
To sketch a graph of this solution, start by drawing a number line and marking a closed circle at -2.25 and shading to the right of the circle, because all values of x greater than or equal to -2.25 make the inequality true. The graph should look like this:
```
<================●----------------
-2.25
```
The ● represents a closed circle at -2.25, and the shaded region to the right of the circle represents the solution to the inequality.
3. 8 ≤ 3x
To solve the inequality 8 ≤ 3x, we need to isolate x on one side of the inequality. We can do this by dividing both sides by 3:
8/3 ≤ 3x/3
8/3 ≤ x
So the solution to the inequality is x ≥ 8/3.
To sketch a graph of this solution, start by drawing a number line and marking a closed circle at 8/3 and shading to the right of the circle, because all values of x greater than or equal to 8/3 make the inequality true. The graph should look like this:
```
<================●----------------
8/3
```
The ● represents a closed circle at 8/3, and the shaded region to the right of the circle represents the solution to the inequality.
4. 3x + 1 < 13
To solve the inequality 3x + 1 < 13, we need to isolate x on one side of the inequality. We can do this by subtracting 1 from both sides:
3x < 12
x < 4
So the solution to the inequality is x < 4.
To sketch a graph of this solution, start by drawing a number line and marking an open circle at 4 (since x cannot be equal to 4) and shading to the left of the circle, because all values of x less than 4 make the inequality true. The graph should look like this:
```
----------------●================>
4
```
The ● represents an open circle at 4, and the shaded region to the left of the circle represents the solution to the inequality.
5. 5x - 2 ≥ 8
To solve the inequality 5x - 2 ≥ 8, we need to isolate x on one side of the inequality. We can do this by adding 2 to both sides and then dividing by 5:
5x - 2 + 2 ≥ 8 + 2
5x/5 ≥ 10/5
x ≥ 2
So the solution to the inequality is x ≥ 2.
To sketch a graph of this solution, start by drawing a number line and marking a closed circle at 2 and shading to the right of the circle, because all values of x greater than or equal to 2 make the inequality true. The graph should look like this:
```
<================●----------------
2
```
The ● represents a closed circle at 2, and the shaded region to the right of the circle represents the solution to the inequality.
6. -5 > 4x + 15
To solve the inequality -5 > 4x + 15, we need to isolate x on one side of the inequality. We can do this by subtracting 15 from both sides and then dividing by 4 (and reversing the direction of the inequality):
-5 - 15 < 4x
-20/4 < x
-5 < x
So the solution to the inequality is x > -5.
To sketch a graph of this solution, start by drawing a number line and marking an open circle at -5 (since x cannot be equal to -5) and shading to the right of the circle, because all values of x greater than -5 make the inequality true. The graph should look like this:
```
<================●----------------
-5
```
The ● represents an open circle at -5, and the shaded region to the right of the circle represents the solution to the inequality.
7. 3 ≤ 17 + 2x
To solve the inequality 3 ≤ 17 + 2x, we need to isolate x on one side of the inequality. We can do this by subtracting 17 from both sides and then dividing by 2:
3 - 17 ≤ 2x
-14/2 ≤ x
-7 ≤ x
So the solution to the inequality is x ≥ -7.
To sketch a graph of this solution, start by drawing a number line and marking a closed circle at -7 and shading to the right of the circle, because all values of x greater than or equal to -7 make the inequality true. The graph should look like this:
```
<================●----------------
-7
```
The ● represents a closed circle at -7, and the shaded region to the right of the circle represents the solution to the inequality.
8. 4x - 2 > 19
To solve the inequality 4x - 2 > 19, we need to isolate x on one side of the inequality. We can do this by adding 2 to both sides and then dividing by 4:
4x - 2 + 2 > 19 + 2
4x/4 > 21/4
x > 5.25
So the solution to the inequality is x > 5.25.
To sketch a graph of this solution, start by drawing a number line and marking an open circle at 5.25 and shading to the right of the circle, because all values of x greater than 5.25 make the inequality true. The graph should look like this:
```
<================●----------------
5.25
```
The ● represents an open circle at 5.25, and the shaded region to the right of the circle represents the solution to the inequality.
9. 3 ≤ 3 + 5
To solve the inequality 3 ≤ 3 + 5, we don't need to do anything! Both sides of the inequality are equal, so any value of x will make the inequality true.
To sketch a graph of this solution, start by drawing a number line and mark an example value of x (such as x=0) with a closed circle. There is no shading needed, because the inequality is true for all values of x. The graph should look like this:
```
----------------●================>
0
```
The ● represents a closed circle at x=0, and there is no shaded region because the inequality is true for all values of x.
Solve this inequality.sketch a graph
3≤3x+5
To solve the inequality 3 ≤ 3x + 5, we need to isolate x on one side of the inequality. We can do this by subtracting 5 from both sides:
3 - 5 ≤ 3x
-2 ≤ 3x
Dividing both sides by 3, we get:
-2/3 ≤ x
So the solution to the inequality is x ≥ -2/3.
To sketch a graph of this solution, start by drawing a number line and marking a closed circle at -2/3 and shading to the right of the circle, because all values of x greater than or equal to -2/3 make the inequality true. The graph should look like this:
```
<================●----------------
-2/3
```
The ● represents a closed circle at -2/3, and the shaded region to the right of the circle represents the solution to the inequality.
If I add 5 to a number and then multiply the result by 3, the result is not less than 30. Find the range of the unknown number
Let the unknown number be x. According to the given condition:
3(x + 5) ≥ 30
Dividing both sides by 3, we get:
x + 5 ≥ 10
Subtracting 5 from both sides, we get:
x ≥ 5
Therefore, the range of the unknown number is x ≥ 5. This means that the unknown number can be any number greater than or equal to 5.
The cost of kerosine rises by #1.50 a litre last week a driver bought 30 litres at the old price. This week he bought 10 litres at the new price. All together the kerosine cost not less than #300, what was the range of the old price for a litre of kerosine
Let the old price of kerosene per liter be x.
Last week, the driver bought 30 liters of kerosene at the old price, so he spent 30x naira.
This week, the driver bought 10 liters of kerosene at the new price, which is x+1.50 naira per liter. Therefore, he spent 10(x+1.50) naira.
The total amount of money he spent on kerosene is not less than 300 naira. So we can write:
30x + 10(x+1.50) ≥ 300
Simplifying and solving for x, we get:
40x + 15 ≥ 300
40x ≥ 285
x ≥ 7.125
Therefore, the old price of kerosene per liter was not less than #7.125.
Note that we rounded off to three decimal places, but the actual value could be slightly higher depending on how many decimal places we use.
A rectngle is 10 cm long and x cm broad find the range of x cm if the perimeter of the rectangle is not greater than 60 cm and not less than 20 cm
The perimeter of a rectangle is given by the formula P = 2(L + b), where P is the perimeter, L is the length, and b is the breadth.
Given that the length L = 10 cm, we can write the inequality for the perimeter as:
20 ≤ 2(10 + x) ≤ 60
Dividing both sides by 2, we get:
10 ≤ 10 + x ≤ 30
Subtracting 10 from all sides, we get:
0 ≤ x ≤ 20
Therefore, the range of values for x is 0 cm ≤ x ≤ 20 cm.
A woman is 14 years older than her son.In 5 years time the son will not be more than 36 years old. What is the likely range of the present age of the son?
Let the present age of the son be x years. Then, the present age of the woman is x + 14 years.
In 5 years time, the age of the son will be x+5 years and the age of the woman will be x+14+5 = x+19 years.
According to the problem, the son's age in 5 years time will not exceed 36 years. Therefore,
x + 5 ≤ 36
Simplifying and solving for x, we get:
x ≤ 31
So, the present age of the son is less than or equal to 31.
Since the woman is 14 years older than the son, her age cannot be less than 14 years more than the son's age. Therefore, the present age of the son is between 0 and 31 years.
Hence, the likely range of the present age of the son is 0 ≤ x ≤ 31.