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3 years ago

What are the possible rational zeros for f(x)=3x^4+x^3-13x^2-2x+9?

Mathematics

1 answer:

zaharov [31]3 years ago

4 0

Answer:

Option 4 (±1, ±1/3, ±3, ±9) is the correct option.

Step-by-step explanation:

The given expression is $f(x)=3x^{4}+x^{3}-13x^{2}-2x+9$

We have to find the possible rational zeros for the function.

So by the rational zero theorem factors will be

=±(Factors of constant term 9)/±factors of coefficient of $x^{4}$

=±(Factors of 9)/±(Factors of 3)

=±(1, 3, 9)/±(1, 3)

=±(1, 3, 9, 1/3)

So option 4 is the correct answer.

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What are the solutions to the equation x² + 16x + 14 = 0?

poizon [28]

Answer:

-8+5√2 and -8-5√2

Step-by-step explanation:

Given the expression x² + 16x + 14 = 0

USing the general formulas

x = -16±√16²-4(14)/2

x = -16±√256-56/2

x = -16±√200/2

x = -16±10√2/2

x = -8±5√2

Hence the required solutions are -8+5√2 and -8-5√2

3 0

3 years ago

I have been stuck on number 10 (this problem) for a while now. Can anyone help me with it and walk me through every step in orde

kotykmax [81]

When you multiply a same number but with different powers, you can simply add the powers together. So, in your question, add the powers -1 and -7 together.
7^(-1) x 7^(-7) = 7^(-8)

When you divide a same number but with different powers, you subtract the power at the top with the power from the denominator. So, -8 - (-7) = -1.
7^(-8) / 7^(-7) = 7^(-1)

So your answer would be 7^(-1).
Hopefully my explanation was clear?

5 0

3 years ago

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This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive an

sleet_krkn [62]

Answer:

-6re−r [sin(6θ) - cos(6θ)]

Step-by-step explanation:

the Jacobian is ∂(x, y) /∂(r, θ) = δx/δθ × δy/δr - δx/δr × δy/δθ

x = e−r sin(6θ), y = er cos(6θ)

δx/δθ = -6rcos(6θ)e−r sin(6θ), δx/δr = -sin(6θ)e−r sin(6θ)

δy/δθ = -6rsin(6θ)er cos(6θ), δy/δr = cos(6θ)er cos(6θ)

∂(x, y) /∂(r, θ) = δx/δθ × δy/δr - δx/δr × δy/δθ

= -6rcos(6θ)e−r sin(6θ) × cos(6θ)er cos(6θ) - [-sin(6θ)e−r sin(6θ) × -6rsin(6θ)er cos(6θ)]

= -6rcos²(6θ)e−r (sin(6θ) - cos(6θ)) - 6rsin²(6θ)e−r (sin(6θ) - cos(6θ))

= -6re−r (sin(6θ) - cos(6θ)) [cos²(6θ) + sin²(6θ)]

= -6re−r [sin(6θ) - cos(6θ)] since [cos²(6θ) + sin²(6θ)] = 1

6 0

4 years ago

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Kisachek [45]

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3 years ago

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shtirl [24]

Answer:

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Step-by-step explanation:

3 0

3 years ago

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