Area of a rectangle = length x width
For this postcard:
length = 4 in
width = (3+b) in
area = 24 in^2
Substitute into the area formula:
24 = 4 x (3+b)
24 = 12 + 4b
24 - 12 = 4b
12 = 4b
b = 3 in
Therefore:
the length of the postcard = 4 inch
the width of the postcard = b+3 = 3 + 3 = 6 inch
Answer:
Step-by-step explanation:
Considering the equation
Sin(5x) = ½
5x = arcSin(½)
5x = 30°
Then,
The general formula for sin is
5θ = n180 + (-1)ⁿθ
Dividing throughout by 5
θ = n•36 + (-1)ⁿ30/5
θ = 36n + (-1)ⁿ6
The solution range is
0<θ<2π which means 0<θ<360
First solution
When n = 0
θ = 36n + (-1)ⁿθ
θ = 0×36 + (-1)^0×6
θ = 6°
When n = 1
θ = 36n + (-1)ⁿ6
θ = 36-6
θ = 30°
When n = 2
θ = 36n + (-1)ⁿ6
θ = 36×2 + 6
θ = 78°
When n =3
θ = 36n + (-1)ⁿ6
θ = 36×3 - 6
θ = 102°
When n=4
θ = 36n + (-1)ⁿ6
θ = 36×4 + 6
θ = 150
When n=5
θ = 36n + (-1)ⁿ6
θ = 36×5 - 6
θ = 174°
When n = 6
θ = 36n+ (-1)ⁿ6
θ = 36×6 + 6
θ = 222°
When n = 7
θ = 36n + (-1)ⁿ6
θ = 36×7 - 6
θ = 246°
When n =8
θ = 36n + (-1)ⁿ6
θ = 36×8 + 6
θ = 294°
When n =9
θ = 36n + (-1)ⁿ6
θ = 36×9 - 6
θ = 318°
When n =10
θ = 36n + (-1)ⁿ6
θ = 36×10 + 6
θ = 366°
When n = 10 surpasses the θ range
Thus, the solutions range from n =0 to n=9
Therefore, there are 10 solutions within the interval 0<θ<2π
Let Sanku be x. If Madhu's age is three times his, Madhu would be 3x. The age difference is 14, leading to the equation 3x - x = 14, which simplifies to 2x = 14, thus x = 7, indicating Sanku is 7 years old, while Madhu is 21.
