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3 years ago

I need help this is my homework I will brainless you

Mathematics

1 answer:

hammer [34]3 years ago

5 0

Answer:

4-10.5

5-no because the nearest thousandth is not 5 or more

6- Tu is correct because the tenth is 6 which is more than 5 so it can be rounded up to $79

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koban [17]

Answer:

D) $2.81

Step-by-step explanation:

3.8 x .74 = 2.81

8 0

2 years ago

40×+4y=-24 what is the equation for y

aksik [14]

Answer:

y=-6-10 x

Step-by-step explanation:

make y the subject

8 0

3 years ago

What is your current debt ratio as a percentage to the nearest hundredth?

olasank [31]

Answer:

Step-by-step explanation: see attachment below

4 0

4 years ago

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

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

5 0

4 years ago

9 and 5/12 - 3 and 2/3

lana [24]

Recall to always convert the mixed fractions to "improper" fractions first,

$\bf \stackrel{mixed}{9\frac{5}{12}}\implies \cfrac{9\cdot 12+5}{12}\implies \stackrel{improper}{\cfrac{113}{12}}
\\\\\\
\stackrel{mixed}{3\frac{2}{3}}\implies \cfrac{3\cdot 3+2}{3}\implies \stackrel{improper}{\cfrac{11}{3}}\\\\
-------------------------------\\\\
\cfrac{113}{12}-\cfrac{11}{3}\impliedby \textit{clearly our LCD is 12}\implies \cfrac{113-(4)11}{12}
\\\\\\
\cfrac{113-44}{12}\implies \cfrac{69}{12}\implies 5\frac{9}{12}$

5 0

3 years ago