The only ones that would work are if the factor is composite and if 1 is added to it it can be divided into 48. So 15 would work. The factor 15 is composite, and if 1 is added it can be divided into 48.Therefore the son is 15.
Soccer game theory is that one of the first people to play the game and play it they will
S = d/t
st = d
t = d/s
The time going is t1.
The time returning is t2.
The total time is 4 hours, so we have t1 + t2 = 4
The speed of the current is c.
The speed going is 9 + c.
The speed returning is 9 - c.
t1 = 16/(9 + c)
t2 = 16/(9 - c)
t1 + t2 = 16/(9 + c) + 16/(9 - c)
4 = 16/(9 - c) + 16/(9 + c)
1 = 4/(9 - c) + 4/(9 + c)
(9 + c)(9 - c) = 4(9 - c) + 4(9 + c)
81 - c^2 = 36 - 4c + 36 + 4c
81 - c^2 = 72
c^2 = 9
c^2 - 9 = 0
(c + 3)(c - 3) = 0
c + 3 = 0 or c - 3 = 0
c = -3 or c = 3
We discard the negative answer, and we get c = 3.
The speed of the current is 3 mph.
ANSWER
x = ±1 and y = -4.
Either x = +1 or x = -1 will work
EXPLANATION
If -3 + ix²y and x² + y + 4i are complex conjugates, then one of them can be written in the form a + bi and the other in the form a - bi. In other words, between conjugates, the imaginary parts are same in absolute value but different in sign (b and -b). The real parts are the same
For -3 + ix²y
⇒ real part: -3
⇒ imaginary part: x²y
For x² + y + 4i
⇒ real part: x² + y (since x, y are real numbers)
⇒ imaginary part: 4
Therefore, for the two expressions to be conjugates, we must satisfy the two conditions.
Condition 1: Imaginary parts are same in absolute value but different in sign. We can set the imaginary part of -3 + ix²y to be the negative imaginary part of x² + y + 4i so that the
x²y = -4 ... (I)
Condition 2: Real parts are the same
x² + y = -3 ... (II)
We have a system of equations since both conditions must be satisfied
x²y = -4 ... (I)
x² + y = -3 ... (II)
We can rearrange equation (II) so that we have
y = -3 - x² ... (II)
Substituting into equation (I)
x²y = -4 ... (I)
x²(-3 - x²) = -4
-3x² - x⁴ = -4
x⁴ + 3x² - 4 = 0
(x² + 4)(x² - 1) = 0
(x² + 4)(x-1)(x+1) = 0
Therefore, x = ±1.
Leave alone (x² + 4) as it gives no real solutions.
Solve for y:
y = -3 - x² ... (II)
y = -3 - (±1)²
y = -3 - 1
y = -4
So x = ±1 and y = -4. We can confirm this results in conjugates by substituting into the expressions:
-3 + ix²y
= -3 + i(±1)²(-4)
= -3 - 4i
x² + y + 4i
= (±1)² - 4 + 4i
= 1 - 4 + 4i
= -3 + 4i
They result in conjugates
