Let’s define x as the amount invested by Sam in the first year.
Here are the corresponding expressions derived from the provided descriptions for Sam's investments.
For Sam:
2nd year: investment = 5x/2 - 2000
3rd year: investment = x/5 + 1000
The total Sam invested is:
x + (5x/2 - 2000) + (x/5 + 1000)
Next, we can form the expressions for Sally’s investments.
For Sally
1st year: investment = 3x/2 - 1000
2nd year: investment = 2x - 1500
3rd year: investment = x/4 + 1400
Thus, Sally's total investment is,
total = (3x/2 - 1000) + (2x - 1500) + (x/4 + 1400)
Setting both totals equal gives us:
(x) + (5x/2 - 2000) + (x/5 + 1000) = (3x/2 - 1000) + (2x - 1500) + (x/4 + 1400)
Solving for x,
x = 2000
For Sally's investment for the third year:
investment = x/4 + 1400 = (2000/4 + 1400) = 1900
RESULTS:
Sam's first year = $2000
Sally's third year = $1900
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
La respuesta es 4,13 al problema.
