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Mamont248 [21]
3 years ago
7

Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and the zeros −3 and 5

+i.
Mathematics
1 answer:
love history [14]3 years ago
7 0

Answer:

<em>f(x) = x² (2+i)x-15-3i</em>

Step-by-step explanation:

Since the zeros of the equation are -3 and 5+i, hence the factors of the polynomial in x is (x+3) and (x-(5+i))

Multiplying both factors

f(x) =  (x+3)(x-(5+i))

f(x) = (x²-(5+i)x+3x -3(5+i))

f(x) = x² - (5+i- 3)x -15-3i

f(x) = x² (2+i)x-15-3i

<em>hence the required polynomial function in x is f(x) = x² (2+i)x-15-3i</em>

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15 is at least the quotient of a number and 6 increased by 3
crimeas [40]

Answer:

The number is x\geq 72

Step-by-step explanation:

Given:

Quotient \geq 15

Let the number be x

The quotient is 15 of a number and 6 which is increaase by 3

Hence forming the inequality expression from the question given we get;

\frac{x}{6}+3\geq 15

Now Solving the above equation we get,

\frac{x}{6}+3\geq 15\\\\\frac{x}{6}\geq 15-3\\\\\frac{x}{6}\geq 12\\\\x\geq 12\times6\\x\geq 72

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Step-by-step explanation:

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alexandr1967 [171]
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Find the vertex form for the following quadratic function. f(x)=x^2+14x+41
Luda [366]

f(x) = (x + 7)² - 8

the equation of a quadratic in vertex form is

y = a(x - h)² + k

where (h, k ) are the coordinates of the vertex and a is a multiplier

given the quadratic in standard form : y = ax² + bx + c ( a ≠ 0 )

then the x-coordinate of the vertex is

x_{vertex} = - \frac{b}{2a}

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with a = 1, b = 14 and c = 41

x_{vertex} = - \frac{14}{2} = - 7

substitute this value into the equation for y- coordinate

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