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
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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.