Please help I’m BEGGING

a circle has an area of 225 square inches. if the radius of the circle is multiples by 6, then what is the new area of the circl

Lesechka [4]

Answer:

D: 8,100π in²

Step-by-step explanation:

The formula for area of a circle is

A = πr²

The radius is squared, so if the radius or a new circle is 6 times larger (multiplied by 6), the difference in area will be 36 times greater. So the area is

225π(36) = 8,100π

3 0

2 years ago

From the measurements of the legs of a right triangle measuring 3cm, 4cm and the hypotenuse of the same 5cm, calculate the measu

WARRIOR [948]

Answer:

3cm and 4cm addition

7cm and 5cm subtract

2cm answer

3 0

3 years ago

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

Which ordered pair is a solution to the system of inequalities?

Taya2010 [7]

Answer:

[see below]

Step-by-step explanation:

I used a graphing program to graph the given system. I also graphed the points given.

Options B's choice lies in the double shaded area, so it would be a solution to the system.

See the graph attached!

Hope this helps you!

3 0

3 years ago

If i am making $4 billion / sec then how much money would i have in 10 minutes? if i am making $6 billion / sec then how much mo

Ghella [55]

If you are making 4 billion a second do 4,000,000,000 x 60 = 240,000,000,000
Then times that by 10 which equals. 2,400,000,000,000

Is the answer for the second one
6,000,000,000 x 60 x 60 x 2 = 1,296,000,000,000,000

3 0

3 years ago