3[(11-6) 2 squared divided by5]

Which equation represents the sentence? the sum of 10 and a number is 3 times the number.

uranmaximum [27]

Answer:

If a neutral object loses 1.5x106 electrons, then what will be its charge? ... Felect = (9x109 N/m2/C2)•(4.65 x 10-6 C)•(7.28 x 10-6 C)/(0.658 m)2 = 0.704 N.

Step-by-step explanation:

5 0

2 years ago

A cube has a volume of 729 cubic centimeters. What can be concluded about the cube? Check all that apply.

Alona [7]

Answer:

This is a perfect cube.

The side length is 9 centimeters.

Taking the cube root of the volume will determine the side length

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Can someone plz help me with this one problem plz!!!! (I Will Mark Brainliest)!!!!!

Vinil7 [7]

Answer:

it would be in the middle? like starting at 0

6 0

3 years ago

Use Green's Theorem to evaluate the line integral along the given positively oriented curve. âˆ«C (3y +5eâˆšx)dx + (10x + 3 cos

elena-s [515]

By Green's theorem, the line integral

$\displaystyle \int_C f(x,y)\,\mathrm dx + g(x,y)\,\mathrm dy$

is equivalent to the double integral

$\displaystyle \iint_D \frac{\partial g}{\partial x} - \frac{\partial f}{\partial y} \,\mathrm dx\,\mathrm dy$

where D is the region bounded by the curve C, provided that this integrand has no singularities anywhere within D or on its boundary.

It's a bit difficult to make out what your integral should say, but I'd hazard a guess of

$\displaystyle \int_C \left(3y+5e^{-x}\right)\,\mathrm dx + \left(10x+3\cos\left(y^2\right)\right)\,\mathrm dy$

Then the region D is

D = {(x, y) : 0 ≤ x ≤ 1 and x ² ≤ y ≤ √x}

so the line integral is equal to

$\displaystyle \int_0^1\int_{x^2}^{\sqrt x} \frac{\partial\left(10x+3\cos\left(y^2\right)\right)}{\partial x} - \frac{\partial\left(3y+5e^{-x}\right)}{\partial y}\,\mathrm dy\,\mathrm dx \\\\ = \int_0^1 \int_{x^2}^{\sqrt x} (10-3)\,\mathrm dy\,\mathrm dx \\\\ = 7\int_0^1 \int_{x^2}^{\sqrt x} \mathrm dy\,\mathrm dx$

which in this case is 7 times the area of D.

The remaining integral is trivial:

$\displaystyle 7\int_0^1\int_{x^2}^{\sqrt x}\mathrm dy\,\mathrm dx = 7\int_0^1y\bigg|_{y=x^2}^{y=\sqrt x}\,\mathrm dx \\\\ = 7 \int_0^1\left(\sqrt x-x^2\right)\,\mathrm dx \\\\ = 7 \left(\frac23x^{3/2}-\frac13x^3\right)\bigg|_{x=0}^{x=1} = 7\left(\frac23-\frac13\right) = \boxed{\frac73}$

6 0

3 years ago

If i = −1 and a and b are non-zero real numbers, what is 1a+bi ?

MariettaO [177]

The i you normally think of (the one = to sqrt(-1) is not the same i they are using here. They are using this i the same way you would use x = - 1 in any normal equation or relation.

1a + b(-1)
a - b is your answer <<<< answer

6 0

2 years ago