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3 years ago

Ughhhh...People wanted it again Lol if ur last jokes on u

Mathematics

2 answers:

Alik [6]3 years ago

7 0

Huh..? Wanted what…??

nevsk [136]3 years ago

5 0

Answer:

what do you want tell the question

Step-by-step explanation:

You might be interested in

Which of the values shown are potential roots of f(x) = 3x3 â€"" 13x2 â€"" 3x 45? Select all that apply.

Verdich [7]

The potential roots of the function are, $\pm1, \ \pm3, \ \pm5, \ \pm9,\ \pm15, \ \pm45,\ \pm \dfrac{1}{3},\ \pm \dfrac{5}{3}$

And the accurate root is 3 it can be determined by using rules of the rational root equation.

<h2>Given that,</h2>

Function; $\rm f(x) = 3x^3 - 13x^2 -3x + 45$

<h3>We have to determine,</h3>

Which of the values shown are potential roots of the given equation?

<h3>According to the question,</h3>

Potential roots of the polynomial are all possible roots of f(x).

$\rm f(x) = 3x^3 - 13x^2 -3x + 45$

Using rational root theorem test. We will find all the possible or potential roots of the polynomial.

$\rm p=\dfrac{All\ the \ positive}{Negative\ factors \ of\ 45}$

$\rm q=\dfrac{All\ the \ positive}{Negative\ factors \ of\ 3}$

The factor of the term 45 are,

$\pm1, \ \pm3, \ \pm5, \ \pm9,\ \pm15, \ \pm45$

And The factor of 3 are,

$\pm1, \ \pm3$

All the possible roots are,

$\dfrac{p}{q} = \pm1, \ \pm3, \ \pm5, \ \pm9,\ \pm15, \ \pm45,\ \pm \dfrac{1}{3},\ \pm \dfrac{5}{3}$

Now check for all the rational roots which are possible for the given function,

$\rm f(x) = 3x^3 - 13x^2 -3x + 45\\\\ f(1) = 3(1)^3 - 13(1)^2 -3(1) + 45 = 3-13-3+45 = 32\neq 0\\\\ f(-1) = 3(-1)^3 - 13(-1)^2 -3(-1) + 45 =- 3-13+3+45 = 32\neq 0\\\\ f(3) = 3(3)^3 - 13(3)^2 -3(3) + 45 = 81-117-9+45 =0\\\\ f(-3) = 3(-3)^3 - 13(-3)^2 -3(-3) + 45 = -81+117+9+45 =-144\neq 0$

Therefore, x = 3 is the potential root of the given function.

Hence, The potential roots of the function are, $\pm1, \ \pm3, \ \pm5, \ \pm9,\ \pm15, \ \pm45,\ \pm \dfrac{1}{3},\ \pm \dfrac{5}{3}$ .

For more details about Potential roots refer to the link given below.

brainly.com/question/25873992

8 0

2 years ago

What is the value of x?<br><br> Enter your answer in the box.

Keith_Richards [23]

The answer would be 5, because you divide 40 by 8 like they did to get the sides

7 0

3 years ago

What is the fraction for 57%

kaheart [24]

Answer:

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Write the equation of the line

meriva

Answer:

gradient is the same as slope

so m=9

Step-by-step explanation:

y=9x+C. insert point(1,4)

4=9*1+C

4-9=C

C=-5

so the equation of the line is y=9x-5

7 0

3 years ago

Read 2 more answers

Help calculus module 8 DBQ<br><br> please show work

igor_vitrenko [27]

1. The four subintervals are [0, 2], [2, 3], [3, 7], and [7, 8]. We construct trapezoids with "heights" equal to the lengths of each subinterval - 2, 1, 4, and 1, respectively - and the average of the corresponding "bases" equal to the average of the values of $R(t)$ at the endpoints of each subinterval. The sum is then

$\dfrac{R(0)+R(2)}2(2-0)+\dfrac{R(2)+R(3)}2(3-2)+\dfrac{R(3)+R(7)}2(7-3)+\dfrac{R(7)+R(8)}2(7-8)=\boxed{24.83}$

which is measured in units of gallons, hence representing the amount of water that flows into the tank.

2. Since $R$ is differentiable, the mean value theorem holds on any subinterval of its domain. Then for any interval $[a,b]$ , it guarantees the existence of some $c\in(a,b)$ such that

$\dfrac{R(b)-R(a)}{b-a)=R'(c)$

Computing the difference quotient over each subinterval above gives values of 0.275, 0.3, 0.3, and 0.26. But just because these values are non-zero doesn't guarantee that there is definitely no such $c$ for which $R'(c)=0$ . I would chalk this up to not having enough information.

3. $R(t)$ gives the rate of water flow, and $R(t)\approx W(t)$ , so that the average rate of water flow over [0, 8] is the average value of $W(t)$ , given by the integral

$R_{\rm avg}=\displaystyle\frac1{8-0}\int_0^8\ln(t^2+7)\,\mathrm dt$

If doing this by hand, you can integrate by parts, setting

$u=\ln(t^2+7)\implies\mathrm du=\dfrac{2t}{t^2+7}\,\mathrm dt$

$\mathrm dv=\mathrm dt\implies v=t$

$R_{\rm avg}=\displaystyle\frac18\left(t\ln(t^2+7)\bigg|_{t=0}^{t=8}-\int_0^8\frac{2t^2}{t^2+7}\,\mathrm dt\right)$

For the remaining integral, consider the trigonometric substitution $t=\sqrt 7\tan s$ , so that $\mathrm dt=\sqrt 7\sec^2s\,\mathrm ds$ . Then

$R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\int_0^{\tan^{-1}(8/\sqrt7)}\frac{7\tan^2s}{7\tan^2s+7}\sec^2s\,\mathrm ds$

$R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\int_0^{\tan^{-1}(8/\sqrt7)}\tan^2s\,\mathrm ds$

$R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\int_0^{\tan^{-1}(8/\sqrt7)}(\sec^2s-1)\,\mathrm ds$

$R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\left(\tan s-s\right)\bigg|_{s=0}^{s=\tan^{-1}(8/\sqrt7)}$

$R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\left(\tan\left(\tan^{-1}\frac8{\sqrt7}\right)-\tan^{-1}\frac8{\sqrt7}\right)$

$\boxed{R_{\rm avg}=\displaystyle\ln71-2+\frac{\sqrt7}4\tan^{-1}\frac8{\sqrt7}}$

or approximately 3.0904, measured in gallons per hour (because this is the average value of $R$ ).

4. By the fundamental theorem of calculus,

$g'(x)=f(x)$

and $g(x)$ is increasing whenever $g'(x)=f(x)>0$ . This happens over the interval (-2, 3), since $f(x)=3$ on [-2, 0), and $-x+3>0$ on [0, 3).

5. First, by additivity of the definite integral,

$\displaystyle\int_{-2}^xf(t)\,\mathrm dt=\int_{-2}^0f(t)\,\mathrm dt+\int_0^xf(t)\,\mathrm dt$

Over the interval [-2, 0), we have $f(x)=3$ , and over the interval [0, 6], $f(x)=-x+3$ . So the integral above is

$\displaystyle\int_{-2}^03\,\mathrm dt+\int_0^x(-t+3)\,\mathrm dt=3t\bigg|_{t=-2}^{t=0}+\left(-\dfrac{t^2}2+3t\right)\bigg|_{t=0}^{t=x}=\boxed{6+3x-\dfrac{x^2}2}$

6 0

3 years ago