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3 years ago

-12-6-(-2) combine like terms

Mathematics

2 answers:

dexar [7]3 years ago

8 0

Answer:

-12-6-(-2)

-12-6-+2

-18+2

-16

Hope This Helps!!!

denis-greek [22]3 years ago

6 0

Answer:

"Like terms" are terms whose variables (and their exponents such as the 2 in x2) are the same. In other words, terms that are "like" each other. Note: the coefficients (the numbers you multiply by, such as "5" in 5x) can be different.

-12-6-(-2)

-12-6-+2

-18+2

-16

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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

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

5 0

3 years ago

Help me please need it fast

Sonja [21]

Answer:I am pretty sure it is A

Step-by-step explanation:

6 0

3 years ago

Determine which numbers the equations need to be multiplied by to form opposite terms of the y-variable. 3x - 1 4 y = 15 2 3 x -

sesenic [268]

In the elimination method, one variable is eliminated from the system of equations to find the value of the other.

To create the opposite terms of the x-variable-

The first equation be multiplied by the number 4.

Second equation be multiplied by the number -6.

<h3>What is the elimination method?</h3>

To solve the system of equations and find the value of the variables, the elimination method is used.

In this method, one variable is eliminated from the system of equations to find the value of the other.

Given information-

The system of the equation given in the problem is,

$\rm 3x-\dfrac{1}{4}y=15$

The second equation given in the problem is;

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

To eliminate the x term, make the coefficient of x of both equations equal.

It can be done by multiplying the coefficient of x term of the first equation to the second equation and the coefficient of x term of the second equation to the first equation.

The first equation is multiplied by;

$\rm 4\times 3x-4\times \dfrac{1}{4}y=15\times 4\\\\12x-y=50$