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3 years ago

X^3-2x^2+5x-10=0 how do you solve or factor this polynomial?

1 answer:

7 0

$\bf x^3-2x^2+5x-10=0\implies \stackrel{\textit{doing some grouping}}{(x^3-2x^2)+(5x-10)}=0 \\\\\\ \stackrel{\textit{common factoring}}{x^2\underline{(x-2)}+5\underline{(x-2)}}=0\implies \stackrel{\textit{more common factoring}}{\underline{(x-2)}(x^2+5)}=0 \\\\[-0.35em] ~\dotfill\\\\ x-2=0\implies x=2~~~~\textit{\large\checkmark} \\\\[-0.35em] ~\dotfill\\\\ x^2+5=0\implies x^2=-5\implies x=\sqrt{-5}\implies x=i\sqrt{5}\quad \bigotimes$

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What number cannot be used as the denominator of a fraction? Explain why.

koban [17]

Zero cannot be a denominator.

7 0

3 years ago

Read 2 more answers

An article reported that for a sample of 52 kitchens with gas cooking appliances monitored during a one-week period, the sample

kolbaska11 [484]

Answer:

a) $654.16-2.01\frac{164.55}{\sqrt{52}}=608.29$

$654.16+2.01\frac{164.55}{\sqrt{52}}=700.03$

And we can conclude that we are 95% confident that the true mean of Co2 level is between 608.29 and 700.03 ppm

b) $n=(\frac{1.960(175)}{25})^2 =188.23 \approx 189$

Step-by-step explanation:

Part a

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=52-1=51$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,51)".And we see that $t_{\alpha/2}=2.01$

Replacing we got:

$654.16-2.01\frac{164.55}{\sqrt{52}}=608.29$

$654.16+2.01\frac{164.55}{\sqrt{52}}=700.03$

And we can conclude that we are 95% confident that the true mean of Co2 level is between 608.29 and 700.03 ppm

Part b

The margin of error is given by :

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

The desired margin of error is ME =50/2=25 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} s}{ME})^2$ (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got $z_{\alpha/2}=1.960$ , and we use an estimator of the population variance the value of 175 replacing into formula (b) we got:

$n=(\frac{1.960(175)}{25})^2 =188.23 \approx 189$

6 0

3 years ago

You buy a cylindrical container of salt with a diameter of 3.25 inches and a height of 5 inches. Your salt shaker is a cylinder

Sphinxa [80]

First calculate volume of the cylindrical container: Pi * R2 * H; where Pi is 22/7 or 3.147, R is the radius and H is the height of the container.

Vol of the cylindrical container: 3.147 * (3.25/2) * (3.25/2)* 5 = 41.55 inch 3

Volume of the shaker using the above principle:
3.147 * (1/2) * (1/2) * 1.5 = 1.18 inch 3

Number of times it can fill the salt shaker will be the ratio of the volumes:

= Volume of the cylindrical container divided by the volume of the shaker = 41.55 divided by 1.18

Answer is 35 times if rounded in absolute number.

3 0

3 years ago

Use stoke's theorem to evaluate∬m(∇×f)⋅ds where m is the hemisphere x^2+y^2+z^2=9, x≥0, with the normal in the direction of the

ludmilkaskok [199]

By Stokes' theorem,

$\displaystyle\int_{\partial\mathcal M}\mathbf f\cdot\mathrm d\mathbf r=\iint_{\mathcal M}\nabla\times\mathbf f\cdot\mathrm d\mathbf S$

where $\mathcal C$ is the circular boundary of the hemisphere $\mathcal M$ in the $y$ - $z$ plane. We can parameterize the boundary via the "standard" choice of polar coordinates, setting

$\mathbf r(t)=\langle 0,3\cos t,3\sin t\rangle$

where $0\le t\le2\pi$ . Then the line integral is

$\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=\int_{t=0}^{t=2\pi}\mathbf f(x(t),y(t),z(t))\cdot\dfrac{\mathrm d}{\mathrm dt}\langle x(t),y(t),z(t)\rangle\,\mathrm dt$
$=\displaystyle\int_0^{2\pi}\langle0,0,3\cos t\rangle\cdot\langle0,-3\sin t,3\cos t\rangle\,\mathrm dt=9\int_0^{2\pi}\cos^2t\,\mathrm dt=9\pi$

We can check this result by evaluating the equivalent surface integral. We have

$\nabla\times\mathbf f=\langle1,0,0\rangle$

and we can parameterize $\mathcal M$ by

$\mathbf s(u,v)=\langle3\cos v,3\cos u\sin v,3\sin u\sin v\rangle$

so that

$\mathrm d\mathbf S=(\mathbf s_v\times\mathbf s_u)\,\mathrm du\,\mathrm dv=\langle9\cos v\sin v,9\cos u\sin^2v,9\sin u\sin^2v\rangle\,\mathrm du\,\mathrm dv$

where $0\le v\le\dfrac\pi2$ and $0\le u\le2\pi$ . Then,

$\displaystyle\iint_{\mathcal M}\nabla\times\mathbf f\cdot\mathrm d\mathbf S=\int_{v=0}^{v=\pi/2}\int_{u=0}^{u=2\pi}9\cos v\sin v\,\mathrm du\,\mathrm dv=9\pi$

as expected.

7 0

3 years ago

Colin's shopping basket has items costing $7.90,

jeyben [28]

Answer:

$20 i think

Step-by-step explanation:

7 0

3 years ago