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2 years ago

Which polynomial is a quintic binomial?

Mathematics

1 answer:

kirza4 [7]2 years ago

5 0

"quintic" = degree is 5

"binomial" = there are 2 terms

The only polynomial that fits this description is 3x⁵ + 2. The degree is clearly 5, and there are 2 terms (3x⁵ and 2).

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Given f(x)=x^2+5 and g(x)=2x^3-1 find (f/g)(-3)

AlekseyPX

For this case we have the following functions:

$f (x) = x ^ 2 + 5\\g (x) = 2x ^ 3-1$

By definition of composition of functions we have:

$(f / g) (x) = \frac {f (x)} {g (x)}$

Then substituting:

$f (-3) = (- 3) ^ 2 + 5 = 9 + 5 = 14\\g (-3) = 2 (-3) ^ 3-1 = 2 (-27) -1 = -54-1 = -55$

So:

$(f / g) (- 3) = \frac {14} {- 55} = - 0.25$

Answer:

$(f / g) (- 3) = \frac {14} {- 55} = - 0.25$

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3 years ago

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Step-by-step explanation:

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2 years ago

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3 years ago

Read 2 more answers

Factor completely: 3x^3+33x^2+54x=0

kap26 [50]

Answer:

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2 years ago

General solutions of sin(x-90)+cos(x+270)=-1<br> {both 90 and 270 are in degrees}

mixer [17]

Answer:

$\left[\begin{array}{l}x=2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{2}+2\pi k,\ k\in Z\end{array}\right.$

Step-by-step explanation:

Given:

$\sin (x-90^{\circ})+\cos(x+270^{\circ})=-1$

First, note that

$\sin (x-90^{\circ})=-\cos x\\ \\\cos(x+270^{\circ})=\sin x$

So, the equation is

$-\cos x+\sin x= -1$

Multiply this equation by $\frac{\sqrt{2}}{2}:$

$-\dfrac{\sqrt{2}}{2}\cos x+\dfrac{\sqrt{2}}{2}\sin x= -\dfrac{\sqrt{2}}{2}\\ \\\dfrac{\sqrt{2}}{2}\cos x-\dfrac{\sqrt{2}}{2}\sin x=\dfrac{\sqrt{2}}{2}\\ \\\cos 45^{\circ}\cos x-\sin 45^{\circ}\sin x=\dfrac{\sqrt{2}}{2}\\ \\\cos (x+45^{\circ})=\dfrac{\sqrt{2}}{2}$

The general solution is

$x+45^{\circ}=\pm \arccos \left(\dfrac{\sqrt{2}}{2}\right)+2\pi k,\ \ k\in Z\\ \\x+\dfrac{\pi }{4}=\pm \dfrac{\pi }{4}+2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{4}\pm \dfrac{\pi }{4}+2\pi k,\ \ k\in Z\\ \\\left[\begin{array}{l}x=2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{2}+2\pi k,\ k\in Z\end{array}\right.$

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3 years ago