Solve the equation... 10x - 12 = 48 - 2x

1 answer:

hjlf3 years ago

4 0

If you would like to solve the equation 10 * x - 12 = 48 - 2 * x, you can calculate this using the following steps:

10 * x - 12 = 48 - 2 * x
10 * x + 2 * x = 48 + 12
12 * x = 60 /12
x = 60 / 12
x = 5

The correct result is 5.

You might be interested in

Which of the following situations represents a linear relationship?

dmitriy555 [2]

Answer:

Susie is losing 5 pounds every month on her diet.

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

A. 140° B. 300° C. 220° D. 180° E. 320°

kvasek [131]

Answer:

140

Step-by-step explanation:

3 0

2 years ago

Use cylindrical coordinates. find the volume of the solid that lies within both the cylinder x2 y2 = 9 and the sphere x2 y2 z2 =

marissa [1.9K]

In Cartesian coordinates, the region is given by $-3\le x\le3$ , $-\sqrt{9-x^2}\le y\le\sqrt{9-x^2}$ , and $-\sqrt{16-x^2-y^2}\le z\le\sqrt{16-x^2-y^2}$ . Converting to cylindrical coordinates, using

$\begin{cases}\mathbf x(r,\theta,\zeta)=r\cos\theta\\\mathbf y(r,\theta,\zeta)=r\sin\theta\\\mathbf z(r,\theta,\zeta)=\zeta\end{cases}$

we get a Jacobian determinant of $r$ , and the region is given in cylindrical coordinates by $0\le\theta\le2\pi$ , $0\le r\le3$ , and $-\sqrt{16-r^2}\le z\le\sqrt{16-r^2}$ .

The volume is then

$\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=3}\int_{z=-\sqrt{16-r^2}}^{z=\sqrt{16-r^2}}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=\dfrac{4(64-7\sqrt7)\pi}3$