1- all five girls
2- four girls and one boy
3. three girls and two boys
4. two girls and three boys.
4 combinations
Answer:

Step-by-step explanation:
see the attached figure to better understand the problem
In this problem Triangles BAE and DAE are congruent by SAS postulate
so
BE=DE
substitute the given values

solve for x




which means the intersection of the parabolic cylinder

and paraboloid

is a circle of radius 2 centered at the origin.
So the integral can be represented in Cartesian coordinates by

where

is the region between the two surfaces.
Converting to cylindrical coordinates will make this slightly easier to compute.


Letting

denote the same region in cylindrical coordinates, you have

In either case the integral reduces to 0.
Step-by-step explanation:
Hey there!
(-32)^-2 <em>Into</em><em> </em><em>posi</em><em>tive</em><em> </em><em>expo</em><em>nent</em><em>s</em><em>.</em>
<em>We</em><em> </em><em>generally</em><em> </em><em>law</em><em> </em><em>of</em><em> </em><em>nega</em><em>tive</em><em> </em><em>index</em><em>.</em>
<em><u>Exa</u></em><em><u>mple</u></em><em><u>;</u></em>
<em><u>
</u></em>
<em><u>So</u></em><em><u>,</u></em><em> </em><em>let's </em><em>make</em><em> </em><em>it</em><em> </em><em>same</em><em> </em><em>as</em><em> </em><em>this</em><em>:</em>
<em>
</em>
<em><u>Hope</u></em><em><u> </u></em><em><u>it</u></em><em><u> helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>
Answer:

And the width for this case is:

And we know that the perimeter is given by:

And replacing we got:

And symplifying we got:

Step-by-step explanation:
For this problem we know that the lenght of the rectangle is given by:

And the width for this case is:

And we know that the perimeter is given by:

And replacing we got:

And symplifying we got:
