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2 years ago

Hello, can anyone help, please?

Mathematics

1 answer:

vazorg [7]2 years ago

6 0

Answer: dude you can do this yourself.

Step-by-step explanation:

toss coins ;-;

You might be interested in

40 oz > 3 lb true or false?

OleMash [197]

1lbs is 16oz, so do the math: 16x3 is 48. 3lbs is 48oz, meaning this is false.

7 0

3 years ago

Find AB if A(0,3) and B(2,7)<br><br> AB= ?

Snezhnost [94]

To get you answer multiply .3 and 2.7

7 0

3 years ago

Solve the following equation for y.

labwork [276]

Answer:

Step-by-step explanation:

2y+2=36

Let's solve this equation for 2y first. This means get 2y by itself.

So to get 2y by itself I need to undo that addition of 2 next to it.

The inverse operation of addition is subtraction.

I'm going to subtract 2 on both sides:

2y =36-2

2y =34

Now we want to get y by itself so we divide both sides by 2 since division and multiplication are inverse operations:

y= 34/2

y= 17

3 0

3 years ago

Read 2 more answers

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

There are 53 regular pencils in the box and some colored pencils in the box. There are a total of 84 pencils. How many colored p

vodka [1.7K]

Let be "x" the number of colored pencils in the box.

According to the information given in the exercise, there are 84 pencils in the box and 53 of them are regular pencils.

Knowing this information, you can set up the following equation:

$x+53=84$

Finally, you have to solve for the variable "x" in order to find its value. In order to do this, you can apply the Subtraction property of equality by subtracting 53 from both sides of the equation:

$\begin{gathered} x+53-(53)=84-(53) \\ x=31 \end{gathered}$

The answer is: There are 31 colored pencils in the box.

7 0

1 year ago