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3 years ago

After birth, a baby's weight increased at a steady rate for several weeks.

1 answer:

Volgvan3 years ago

5 0

Y=mx+b first we will find m or the slope.

m=(9.2-7.2)/(6-1)=2/5=0.4 now we have:

y=0.4x+b, using either point we can now solve for b, the "y-intercept", which in this case is the weight at birth :) I'll use point (1, 7.2) so our equation is:

7.2=0.4+b, b=6.8

y=0.4x+6.8

So the weight at birth was 6.8 lb

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3 years ago

What is less 2/10 or 10/6?

Ann [662]

This is obvious, 2/10 is less than one
10/6 is more than 1 so
2/10<1<10/6
therfor 2/10 is less

the legit way to do it is
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2/10 times 3/3=6/10
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6/10<50/30

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3 years ago

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DerKrebs [107]

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2 years ago

Read 2 more answers

2.1 repeating as a mixed number?

Schach [20]

Answer:

2 1/10

Step-by-step explanation:

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3 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

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

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3 years ago