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3 years ago

14

The line plot shows the weight gain, in pounds, of several dogs seen on Monday by a veterinarian at the animal clinic. A number

line with one x above .5, one x above 1.5, one x above 2, one x above 3, two xs above 3.5, two xs above 4, one x above 4.5, and one x above 8.5. How many dogs gained fewer than 3 lb? Enter your answer in the box.

2 answers:

denis-greek [22]3 years ago

6 0

Answer:

3 x's

Step-by-step explanation:

Lerok [7]3 years ago

3 0

There were 3 dogs that gained fewer than 3 pounds.

Counting the number of x's marked above numbers less than 3, we can see that there are 3 x's.

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Solution for dy/dx+xsin 2 y=x^3 cos^2y

vichka [17]

Rearrange the ODE as

$\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y=x^3\cos^2y$
$\sec^2y\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y\sec^2y=x^3$

Take $u=\tan y$ , so that $\dfrac{\mathrm du}{\mathrm dx}=\sec^2y\dfrac{\mathrm dy}{\mathrm dx}$ .

Supposing that $|y|$ , we have $\tan^{-1}u=y$ , from which it follows that

$\sin2y=2\sin y\cos y=2\dfrac u{\sqrt{u^2+1}}\dfrac1{\sqrt{u^2+1}}=\dfrac{2u}{u^2+1}$
$\sec^2y=1+\tan^2y=1+u^2$

So we can write the ODE as

$\dfrac{\mathrm du}{\mathrm dx}+2xu=x^3$

which is linear in $u$ . Multiplying both sides by $e^{x^2}$ , we have

$e^{x^2}\dfrac{\mathrm du}{\mathrm dx}+2xe^{x^2}u=x^3e^{x^2}$
$\dfrac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]=x^3e^{x^2}$

Integrate both sides with respect to $x$ :

$\displaystyle\int\frac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]\,\mathrm dx=\int x^3e^{x^2}\,\mathrm dx$
$e^{x^2}u=\displaystyle\int x^3e^{x^2}\,\mathrm dx$

Substitute $t=x^2$ , so that $\mathrm dt=2x\,\mathrm dx$ . Then

$\displaystyle\int x^3e^{x^2}\,\mathrm dx=\frac12\int 2xx^2e^{x^2}\,\mathrm dx=\frac12\int te^t\,\mathrm dt$

Integrate the right hand side by parts using

$f=t\implies\mathrm df=\mathrm dt$
$\mathrm dg=e^t\,\mathrm dt\implies g=e^t$
$\displaystyle\frac12\int te^t\,\mathrm dt=\frac12\left(te^t-\int e^t\,\mathrm dt\right)$

You should end up with

$e^{x^2}u=\dfrac12e^{x^2}(x^2-1)+C$
$u=\dfrac{x^2-1}2+Ce^{-x^2}$
$\tan y=\dfrac{x^2-1}2+Ce^{-x^2}$

and provided that we restrict $|y|$ , we can write

$y=\tan^{-1}\left(\dfrac{x^2-1}2+Ce^{-x^2}\right)$

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