2/3[8(5-2)^2 +3*2] How do I distribute the 2/3? And then solve the rest!

Mathematics

1 answer:

jolli1 [7]3 years ago

8 0

To distribute the 2/3, this is what you do.

2/3[8(5-2)^2 + 3*2] = (2/3)(8)(5-2)^2 + (2/3)(3*2) =

Now you can continue. You need to follow the correct order of operations.

= (2/3)(8)(3^2) + (2/3)(6)

= (2/3)(8)(9) + 4

= (2/3)(9)(8) + 4

= (6)(8) + 4

= 48 + 4

= 52

This is how I would have done it.

2/3[8(5-2)^2 + 3*2] =

= 2/3[8(3)^2 + 6]

= 2/3[8 * 9 + 6]

= 2/3[72 + 6]

= 2/3[78]

= 2 * 1/3(78)

= 2 * 26

= 52

You might be interested in

200 is 10 times as much as blank and 1 tenth of

anzhelika [568]

Answer:

200 is 10 times as much as 20 and 1 tenth of 2,000

Step-by-step explanation:

Divide 200 by 10 and you'll get 20 and multiply it by 10 to get 2,000.

8 0

3 years ago

Read 2 more answers

Find the curl of ~V ~V = sin(x) cos(y) tan(z) i + x^2y^2z^2 j + x^4y^4z^4 k

ch4aika [34]

Given

$\vec v = f(x,y,z)\,\vec\imath+g(x,y,z)\,\vec\jmath+h(x,y,z)\,\vec k \\\\ \vec v = \sin(x)\cos(y)\tan(z)\,\vec\imath + x^2y^2z^2\,\vec\jmath+x^4y^4z^4\,\vec k$

the curl of $\vec v$ is

$\displaystyle \nabla\times\vec v = \left(\frac{\partial h}{\partial y}-\frac{\partial g}{\partial z}\right)\,\vec\imath - \left(\frac{\partial h}{\partial x}-\frac{\partial f}{\partial z}\right)\,\vec\jmath + \left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)\,\vec k$

$\nabla\times\vec v = \left(4x^4y^3z^4-2x^2y^2z\right)\,\vec\imath \\\\ - \left(4x^3y^4z^4-\sin(x)\cos(y)\sec^2(z)\right)\,\vec\jmath \\\\ + \left(2xy^2z^2+\sin(x)\sin(y)\tan(z)\right)\,\vec k$

$\nabla\times\vec v = \left(4x^4y^3z^4-2x^2y^2z\right)\,\vec\imath \\\\ + \left(\sin(x)\cos(y)\sec^2(z)-4x^3y^4z^4\right)\,\vec\jmath \\\\ + \left(2xy^2z^2+\sin(x)\sin(y)\tan(z)\right)\,\vec k$