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2 years ago

What are the zeros of the polynomial function

Mathematics

1 answer:

solniwko [45]2 years ago

3 0

Option B

The zeros of the polynomial function are x = 0 , x = 6, x = -1

Solution:

Given function is:

$f(x) = x^3 -5x^2 - 6x$

To find: zeros of the polynomial function

To find the zeros of polynomial function, set the function equal to zero and then solve for x.

$x^3 -5x^2 - 6x = 0$

Taking "x" as common term, we get

$x(x^2 - 5x - 6) = 0$

Equating to zero,

$x = 0 \text{ and } x^2 - 5x - 6 = 0$

So one of the zeros of polynomial is x = 0

Let us solve $x^2 - 5x - 6 = 0$ to find other zeros

Let us solve using quadratic formula,

$\text {For a quadratic equation } a x^{2}+b x+c=0, \text { where } a \neq 0\\\\x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Using the above formula,

$\text{ for } x^2 - 5x - 6 = 0 \text{ we have } a = 1 ; b = -5 ; c = -6$

$\text{ The discriminant } b^2 - 4ac > 0 , \text{ so, there are two real roots }$

Substituting the values of a, b, c in above formula,

$x=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(1)(-6)}}{2(1)}\\\\x=\frac{5 \pm \sqrt{25+24}}{2}=\frac{5 \pm \sqrt{49}}{2}\\\\x=\frac{5 \pm 7}{2}\\\\x=\frac{5+7}{2} \text{ or } x=\frac{5-7}{2}\\\\x=\frac{12}{2} \text{ or } x=\frac{-2}{2}\\\\x=6 \text{ or } x=-1$

Thus the zeros of the polynomial function are x = 0 , x = 6, x = -1

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A dose of medicine is made by diluting one part of concentrate to 5 parts of water. How much of the concentrate should there be

Anastasy [175]

Answer:

15 ml

Step-by-step explanation:

If we mix 1 part concentrate and 5 parts water, we have 6 parts in total.

To create 90ml medicine, how many "6 parts" are there?

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2 years ago

A box contains 24 transistors,4 of which are defective. If 4 are sold at random,find the following probabilities. i. Exactly 2 a

zavuch27 [327]

SOLUTION

This is a binomial probability. For i, we will apply the Binomial probability formula

i. Exactly 2 are defective

Using the formula, we have

$\begin{gathered} P_x=^nC_x\left(p^x\right?\left(q^{n-x}\right) \\ Where\text{ } \\ P_x=binomial\text{ probability} \\ x=number\text{ of times for a specific outcome with n trials =2} \\ p=\text{ probability of success = }\frac{4}{24}=\frac{1}{6} \\ q=probability\text{ of failure =1-}\frac{1}{6}=\frac{5}{6} \\ ^nC_x=\text{ number of combinations = }^4C_2 \\ n=\text{ number of trials = 4} \end{gathered}$

Note that I made the probability of being defective as the probability of success = p

and probability of none defective as probability of failure = q

Exactly 2 are defective becomes the binomial probability

$\begin{gathered} P_x=^4C_2\times\lparen\frac{1}{6})^2\times\lparen\frac{5}{6})^{4-2} \\ P_x=6\times\frac{1}{36}\times\frac{25}{36} \\ P_x=\frac{25}{216} \\ =0.1157 \end{gathered}$

Hence the answer is 0.1157

(ii) None is defective becomes

$\begin{gathered} \lparen\frac{5}{6})^4=\frac{625}{1296} \\ =0.4823 \end{gathered}$

hence the answer is 0.4823

(iii) All are defective

$\begin{gathered} \lparen\frac{1}{6})^4=\frac{1}{1296} \\ =0.00077 \end{gathered}$

(iv) At least one is defective

This is 1 - probability that none is defective

$\begin{gathered} 1-\lparen\frac{5}{6})^4 \\ =1-\frac{625}{1296} \\ =\frac{671}{1296} \\ =0.5177 \end{gathered}$

Hence the answer is 0.5177

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6 months ago

Problem 4: Let F = (2z + 2)k be the flow field. Answer the following to verify the divergence theorem: a) Use definition to find

Viktor [21]

Given that you mention the divergence theorem, and that part (b) is asking you to find the downward flux through the disk $x^2+y^2\le3$ , I think it's same to assume that the hemisphere referred to in part (a) is the upper half of the sphere $x^2+y^2+z^2=3$ .

a. Let $C$ denote the hemispherical cap $z=\sqrt{3-x^2-y^2}$ , parameterized by

$\vec r(u,v)=\sqrt3\cos u\sin v\,\vec\imath+\sqrt3\sin u\sin v\,\vec\jmath+\sqrt3\cos v\,\vec k$

with $0\le u\le2\pi$ and $0\le v\le\frac\pi2$ . Take the normal vector to $C$ to be

$\vec r_v\times\vec r_u=3\cos u\sin^2v\,\vec\imath+3\sin u\sin^2v\,\vec\jmath+3\sin v\cos v\,\vec k$

Then the upward flux of $\vec F=(2z+2)\,\vec k$ through $C$ is

$\displaystyle\iint_C\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^{\pi/2}((2\sqrt3\cos v+2)\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm dv\,\mathrm du$

$\displaystyle=3\int_0^{2\pi}\int_0^{\pi/2}\sin2v(\sqrt3\cos v+1)\,\mathrm dv\,\mathrm du$

$=\boxed{2(3+2\sqrt3)\pi}$

b. Let $D$ be the disk that closes off the hemisphere $C$ , parameterized by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath$

with $0\le u\le\sqrt3$ and $0\le v\le2\pi$ . Take the normal to $D$ to be

$\vec s_v\times\vec s_u=-u\,\vec k$

Then the downward flux of $\vec F$ through $D$ is

$\displaystyle\int_0^{2\pi}\int_0^{\sqrt3}(2\,\vec k)\cdot(\vec s_v\times\vec s_u)\,\mathrm du\,\mathrm dv=-2\int_0^{2\pi}\int_0^{\sqrt3}u\,\mathrm du\,\mathrm dv$

$=\boxed{-6\pi}$

c. The net flux is then $\boxed{4\sqrt3\pi}$ .

d. By the divergence theorem, the flux of $\vec F$ across the closed hemisphere $H$ with boundary $C\cup D$ is equal to the integral of $\mathrm{div}\vec F$ over its interior:

$\displaystyle\iint_{C\cup D}\vec F\cdot\mathrm d\vec S=\iiint_H\mathrm{div}\vec F\,\mathrm dV$

We have

$\mathrm{div}\vec F=\dfrac{\partial(2z+2)}{\partial z}=2$

so the volume integral is

$2\displaystyle\iiint_H\mathrm dV$

which is 2 times the volume of the hemisphere $H$ , so that the net flux is $\boxed{4\sqrt3\pi}$ . Just to confirm, we could compute the integral in spherical coordinates:

$\displaystyle2\int_0^{\pi/2}\int_0^{2\pi}\int_0^{\sqrt3}\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=4\sqrt3\pi$

4 0

3 years ago