This problem tackles the place values of numbers. From the rightmost end of the number to the leftmost side, these place values are ones, tens, hundreds, thousands, ten thousands, hundred thousands, millions, ten millions, one hundred millions, and so on and so forth. My idea for the solution of this problem is to add up all like multiples. In this problem, there are 5 multiples expressed in ones, thousands, hundred thousands, tens and hundreds. Hence, you will add up 5 like terms. The solution is as follows

30(1) + 82(1,000) + 4(100,000) + 60(10) + 100(100)

The total answer is 492,630. Therefore, the number's identity is 492,630.

7 0

2 years ago

I need help setting up my problem

Genrish500 [490]

The measure of angle A is 144.3 degrees and the angle to cut the molding is 54.3 degrees

<h3>How to solve for angle A?</h3>

Start by solving the acute part of angle A using the following sine function

sin(Ax) = (30 - 4)/32

Evaluate the quotient

sin(Ax) = 0.8125

Take the arc sin of both sides

Ax = 54.3

The measure of angle A is then calculated as:

A = 90 + Ax

This gives

A = 90 + 54.3

Evaluate

A = 144.3

Hence, the measure of angle A is 144.3 degrees

<h3>The angle to cut the molding</h3>

In (a), we have:

Ax = 54.3

This represents the angle where the molding would be cut

Hence, the angle to cut the molding is 54.3 degrees

Read more about angles at:

brainly.com/question/1592456

#SPJ1

4 0

1 year 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

2 years ago

Use long division to find the quotient below.<br> (x2 + 3x - 18) + (x-3)

Vanyuwa [196]

Answer:

6x-21

Step-by-step explanation:

6 0

2 years ago