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2 years ago

Second piece help ASAPP..

Mathematics

2 answers:

Evgesh-ka [11]2 years ago

7 0

Answer:

#2 80

Step-by-step explanation:

you start from the back and you have to put the place value

Ivan2 years ago

6 0

Answer:

I can help you but I'm not see it clear

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Philip has $2,000 and spends $23.75 on supplies. He divides the remaining amount equally among his 8 employees. How much does ea

Harrizon [31]

Answer:

Step-by-step explanation:

5 0

2 years ago

Solve -8x < 2.<br> A- <br> B<br> C<br> D

Anarel [89]

-8x <2

-8x÷2=4x

4× > 1

hope it helps

4 0

2 years ago

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

2 years ago

Given the functions f(x) = x2 + 6x − 1, g(x) = –x2 + 2, and h(x) = 2x2 − 4x + 3, rank them from least to greatest based on their

kotegsom [21]

You can find an axis of symmmetry of quadratic function by using this formula:

$x=-\dfrac{b}{2a} \\ \\ \hbox{Where function is} \ \ f(x)=ax^2+bx+c \ \ \hbox{assuming} \ \ a\neq 0$

An axis of f(x):

$x=-\dfrac{6}{2 \cdot 1}=-\dfrac{6}{2}=-3$

Of g(x):

$x=-\dfrac{0}{2 \cdot (-1)}=\dfrac{0}{2}=0$

Of h(x):

$x=-\dfrac{-4}{2 \cdot 2}=\dfrac{4}{4}=1$

At least to greatest you've got -3, 0, 1 so f(x), g(x), h(x). Answer a

6 0

2 years ago

Which quadrant does the coordinate (-8, 12) lie?

marysya [2.9K]

In quadrant 2 is where it lies.

3 0

2 years ago