Answer:
39
Step-by-step explanation:
Let the three consecutive odd numbers be:
x+2, x+4, x+6.
The sum of the consecutive numbers is 123 i.e
x+2 + (x+4) + (x+6) = 123
x + 2 + x + 4 + x + 6 = 123
3x + 12 = 123
Collect like terms
3x = 123 — 12
3x = 111
Divide both side by the coefficient of x i.e 3
x = 111/3
x = 37
Now let us find the value of the three odd numbers. This is illustrated:
1st : x + 2 = 37 + 2 = 39
2nd : x + 4 = 37 + 4 = 41
3rd : x + 6 = 37 + 6 = 43
The smallest of the three consecutive odd numbers is 39
Answer:
Answer below
Step-by-step explanation:
4(x−1)
2
=10−2
4(x−1)
2
=8
(x−1)
2
=
4
8
(x−1)
2
=2
x −1=±√
2
x−1=√
2
x-1=-\sqrt{2}x−1=−√
2
x=√
2
+1
x=−√
2
+1
x=√
2
+1,−√
2
+1
Suppose we wish to determine whether or not two given polynomials with complex coefficients have a common root. Given two first-degree polynomials a0 + a1x and b0 + b1x, we seek a single value of x such that
Solving each of these equations for x we get x = -a0/a1 and x = -b0/b1 respectively, so in order for both equations to be satisfied simultaneously we must have a0/a1 = b0/b1, which can also be written as a0b1 - a1b0 = 0. Formally we can regard this system as two linear equations in the two quantities x0 and x1, and write them in matrix form as
Hence a non-trivial solution requires the vanishing of the determinant of the coefficient matrix, which again gives a0b1 - a1b0 = 0.
Now consider two polynomials of degree 2. In this case we seek a single value of x such that
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