<span>Solving the equation for Y = 1.
5 * (3 * 1 - 2) = 15 * 1 - 10
5 * (3 - 2) = 15 -10
5 * 1 = 5
5 = 5
Solving the equation for Y = 2.
5 * (3 * 2 - 2) = 15 * 2 - 10
5 * (6 - 2) = 30 -10
5 * 4 = 20
20 = 20
Solving the equation for Y = 4.
5 * (3 * 4 - 2) = 15 * 4 - 10
5 * (12 - 2) = 60 -10
5 * 10 = 50
50 = 50
Solving the equation for Y = 8.
5 * (3 * 8 - 2) = 15 * 8 - 10
5 * (24 - 2) = 120 -10
5 * 22 = 110
110 = 110
Solving the equation for Y = 9.
5 * (3 * 9 - 2) = 15 * 9 - 10
5 * (27 - 2) = 135 -10
5 * 25 = 125
125 = 125
This proves that the equation holds good for at least 5 values of 'y', which are 1, 2, 4, 8 and 9.
However, it can be proved that the equation holds good for any value of y.
Expression 5(3y-2) can be simplified to 15y -10 which is the same expression on the right had side of the equation provided.
So, equation 5(3y-2)=15y-10 is actually 15y-10=15y-10 and since this is true for all values of y, it has been proved that it is true for at least 5 values of y.</span>
Answer:
Step-by-step explanation:
Given that alpha and beta be conjugate complex numbers
such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}.
Let
since they are conjugates
Imaginary part of the above =0
i.e. 
So the value of alpha =
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Answer:
4
Step-by-step explanation:
the Fundamental Theorem of Algebra states that for any polynomial of degree n, there are n roots, some of which may be complex
The polynomial shown is of degree 4 ( highest exponent of x )
Hence the polynomial has 4 roots/ zeros
Answer:
it is the bottom one
Step-by-step explanation:
i think i could be wrong
