All categories

3 years ago

Find the area and the circumference of a circle with diameter 6 m.

Mathematics

1 answer:

Strike441 [17]3 years ago

7 0

Zxzddsseddzssaaazzsasa

You might be interested in

The distance between 5,3 and 1,3

mel-nik [20]

Answer:

(4,0)

Step-by-step explanation:

(5,3) (1,3)

You need to subtract from the x axis on both coordinates and the y axis in both coordinates.

(5-1) (3-3)

(4,0)

I also checked on a graph as well, so it's correct.

4 0

2 years ago

Which expression is equivalent to 4 times the sum of 3 and x

vovikov84 [41]

Answer:

4(3+x)

Step-by-step explanation:

If any thing is in the brackets you need to do it first

so if you do 3 + x × 4 it will give yo 4x + 3 or 3 +4x

8 0

3 years ago

Let f(x)=2x2−8 .

MrRa [10]

Move down c units means minus 2 from whole function

f(x) moved down 2 units means f(x)-2
so

f(x)-2=2x²-8-2
f(x)-2=2x²-10

g(x)=2x²-10

3 0

2 years ago

Read 2 more answers

Surface integrals using an explicit description. Evaluate the surface integral \iint_{S}^{}f(x,y,z)dS using an explicit represen

Jobisdone [24]

Parameterize $S$ by the vector function

$\vec r(x,y)=x\,\vec\imath+y\,\vec\jmath+f(x,y)\,\vec k$

so that the normal vector to $S$ is given by

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=\left(\vec\imath+\dfrac{\partial f}{\partial x}\,\vec k\right)\times\left(\vec\jmath+\dfrac{\partial f}{\partial y}\,\vec k\right)=-\dfrac{\partial f}{\partial x}\vec\imath-\dfrac{\partial f}{\partial y}\vec\jmath+\vec k$

with magnitude

$\left\|\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}\right\|=\sqrt{\left(\dfrac{\partial f}{\partial x}\right)^2+\left(\dfrac{\partial f}{\partial y}\right)^2+1}$

In this case, the normal vector is

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=-\dfrac{\partial(8-x-2y)}{\partial x}\,\vec\imath-\dfrac{\partial(8-x-2y)}{\partial y}\,\vec\jmath+\vec k=\vec\imath+2\,\vec\jmath+\vec k$

with magnitude $\sqrt{1^2+2^2+1^2}=\sqrt6$ . The integral of $f(x,y,z)=e^z$ over $S$ is then

$\displaystyle\iint_Se^z\,\mathrm d\Sigma=\sqrt6\iint_Te^{8-x-2y}\,\mathrm dy\,\mathrm dx$

where $T$ is the region in the $x,y$ plane over which $S$ is defined. In this case, it's the triangle in the plane $z=0$ which we can capture with $0\le x\le8$ and $0\le y\le\frac{8-x}2$ , so that we have

$\displaystyle\sqrt6\iint_Te^{8-x-2y}\,\mathrm dx\,\mathrm dy=\sqrt6\int_0^8\int_0^{(8-x)/2}e^{8-x-2y}\,\mathrm dy\,\mathrm dx=\boxed{\sqrt{\frac32}(e^8-9)}$

5 0

2 years ago

28,062 divided by 3

Galina-37 [17]

28 062 ÷ 3
9 354
9 354 x 3
28062

3 0

3 years ago

Read 2 more answers