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SSSSS [86.1K]
4 years ago
9

Determine which of the following statements is true for the function g(x) = 4x^5 - 3x^6 + 2x^3 - 1.

Mathematics
1 answer:
jonny [76]4 years ago
8 0

Answer:

Option D. As x increases, y decreases. As x decreases, y decreases

Step-by-step explanation:

we have the function

g(x)=4x^{5}-3x^{6}+2x^{3}-1

we know that

g(x) is a polynomial of sixth degree

The value of the function will approach the same negative infinity for large positive or negative x. That infinity is determined by the sign of the coefficient of the 6th-degree term, which is negative.

therefore

As x increases -----> The value of y decreases

As x decreases ----> The value of y decreases

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7 choose 3 is going to be 35
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3 years ago
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14% of all cell phone users use their phone for internet access. In a random sample of 10 cell
EastWind [94]

Answer:

if number of useres are take to be 100(example)

Step-by-step explanation:

total users=100

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3 years ago
The answer of the problem
k0ka [10]

Answer:

x=16 and x=-3

Step-by-step explanation:

A <em>second-degree equation</em> or <em>quadratic equation of a variable</em> is an equation that it has the general expression:

ax^{2} +bx+c=0 with a\neq 0

Where x is the variable, and a, b and c constants; a is the quadratic coefficient (other than 0), b the linear coefficient and c is the independent term. This polynomial can be interpreted by means of the graph of a quadratic function, that is, by a parabola. This graphical representation is useful, because the abscissas of the intersections or point of tangency of this graph, in the case of existing, with the X axis are the real roots of the equation. If the parabola does not cut the X axis the roots are complex numbers, they correspond to a negative discriminant.

Second degree equation solutions

For a quadratic equation with real or complex coefficients there are always two solutions, not necessarily different, called roots, which can be real or complex (if the coefficients are real and there are two non-real solutions, then they must be complex conjugates). General formula for obtaining roots:

x=\frac{-b±\sqrt{b^{2} -4ac} }{2a}

The discriminant serves to analyze the nature of the roots that can be real or complex.

Δ=b^{2} -4ac

Solving the problem of the answer.

x^{2} -13x-48=0 with a = 1, b = -13, and c = -48

Substituting the values in the general formula for a quadratic equation.

x=\frac{-(-13)±\sqrt{(-13)^{2} -4(1)(-48)} }{2(1)}

x=\frac{13±\sqrt{169+192} }{2}

x=\frac{13±\sqrt{361} }{2}

Then, we obtain the roots:

x=\frac{13+\sqrt{361} }{2} and x=\frac{13-\sqrt{361} }{2}

Solving the roots:

x=\frac{13+\sqrt{361} }{2}\\x=\frac{13+19}{2}\\x=\frac{32}{2}\\x=16

x=\frac{13-\sqrt{361} }{2}\\x=\frac{13-19}{2}\\x=\frac{-6}{2}\\x=-3

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3 years ago
si la suma de 3 números impares consecutivos da como resultado 21 entonces ¿el numeromayor impar es?​
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Solve 5 over x minus 5 equals the quantity of x over x minus 5, minus five fourths for x and determine if the solution is extran
Vikentia [17]
Assuming the equation is:
5/(x-5) = x/(x-5) - 5x/4
We first multiply by the LCD: 4(x-5)
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Substituting x = 5 gives denominators of 0, which is extraneous.
Substituting x = 4/5 gives a valid equation, so this is the only correct solution.
3 0
3 years ago
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