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2 years ago

Factor the following: 2x2+9x+9

Mathematics

2 answers:

ryzh [129]2 years ago

8 0

Answer: (2x+3)(x+3)

Step-by-step explanation:

Looking at this, you know that it must look something like

(? +3)(?+3) , because they must multiply to 9. The ?s must multiply to 2x^2, the most plausible values being 2x and x, ending us up with (2x+3)(x+3)

Semenov [28]2 years ago

6 0

Answer:

(x + 3) (2x + 3)

Step-by-step explanation:

2x^2+9x+9

2x^2 + 3x + 6x + 9

x(2x +3) + 3 (2x + 3)

(x + 3) (2x + 3)

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If x -1 = 2y/3,then y=

earnstyle [38]

Answer:

The answer is

Step-by-step explanation:

To solve for y cross multiply

That's

3( x - 1) = 2y

2y = 3x - 3

Divide both sides by 2 to make y stand alone

That's

We have the final answer as

Hope this helps you

6 0

2 years ago

Read 2 more answers

Determine the values of the constants r and s such that i(x, y) = x rys is an integrating factor for the given differential equa

garri49 [273]

$\underbrace{y(7xy^2+6)}_{M(x,y)}\,\mathrm dx+\underbrace{x(xy^2-1)}_{N(x,y)}\,\mathrm dy=0$

For the ODE to be exact, we require that $M_y=N_x$ , which we'll verify is not the case here.

$M_y=21xy^2+6$
$N_x=2xy^2-1$

So we distribute an integrating factor $i(x,y)$ across both sides of the ODE to get

$iM\,\mathrm dx+iN\,\mathrm dy=0$

Now for the ODE to be exact, we require $(iM)_y=(iN)_x$ , which in turn means

$i_yM+iM_y=i_xN+iN_x\implies i(M_y-N_x)=i_xN-i_yM$

Suppose $i(x,y)=x^ry^s$ . Then substituting everything into the PDE above, we have

$x^ry^s(19xy^2+7)=rx^{r-1}y^s(x^2y^2-x)-sx^ry^{s-1}(7xy^3+6y)$
$19x^{r+1}y^{s+2}+7x^ry^s=rx^{r+1}y^{s+2}-rx^ry^s-7sx^{r+1}y^{s+2}-6sx^ry^s$
$19x^{r+1}y^{s+2}+7x^ry^s=(r-7s)x^{r+1}y^{s+2}-(r+6s)x^ry^s$
$\implies\begin{cases}r-7s=19\\r+6s=-7\end{cases}\implies r=5,s=-2$

so that our integrating factor is $i(x,y)=x^5y^{-2}$ . Our ODE is now

$(7x^6y+6x^5y^{-1})\,\mathrm dx+(x^7-x^6y^{-2})\,\mathrm dy=0$

Renaming $M(x,y)$ and $N(x,y)$ to our current coefficients, we end up with partial derivatives

$M_y=7x^6-6x^5y^{-2}$
$N_x=7x^6-6x^5y^{-2}$

as desired, so our new ODE is indeed exact.

Next, we're looking for a solution of the form $\Psi(x,y)=C$ . By the chain rule, we have

$\Psi_x=7x^6y+6x^5y^{-1}\implies\Psi=x^7y+x^6y^{-1}+f(y)$

Differentiating with respect to $y$ yields

$\Psi_y=x^7-x^6y^{-2}=x^7-x^6y^{-2}+\dfrac{\mathrm df}{\mathrm dy}$
$\implies\dfrac{\mathrm df}{\mathrm dy}=0\implies f(y)=C$

Thus the solution to the ODE is

$\Psi(x,y)=x^7y+x^6y^{-1}=C$

4 0

3 years ago

The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e

Korolek [52]

Answer:

A sample of 499 is needed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is given by:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

In this question, we have that:

$\pi = 0.21$

90% confidence level

So $\alpha = 0.1$ , z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$ , so $Z = 1.645$ .

How large a sample would be required in order to estimate the fraction of tenth graders reading at or below the eighth grade level at the 90% confidence level with an error of at most 0.03

We need a sample of n, which is found when $M = 0.03$ . So