Answer:
2 square cm
Step-by-step explanation:
Given :
A square is inscribed in a circle whose radius is r = 1 cm
Therefore, the diameter of the circle is 2 r = 2 x 1
= 2 cm.
So the diagonal of the square is 2r.
Using the Pythagoras theorem, we find each of the side of the triangle is
.
Therefore, the area of the square is given by 
= 



Hence the area of the largest square that is contained by a circle of radius 1 cm is 2 cm square.
Answer:converge at 
Step-by-step explanation:
Given
Improper Integral I is given as

integration of
is -
![I=\left [ -\frac{1}{x}\right ]^{\infty}_3](https://tex.z-dn.net/?f=I%3D%5Cleft%20%5B%20-%5Cfrac%7B1%7D%7Bx%7D%5Cright%20%5D%5E%7B%5Cinfty%7D_3)
substituting value
![I=-\left [ \frac{1}{\infty }-\frac{1}{3}\right ]](https://tex.z-dn.net/?f=I%3D-%5Cleft%20%5B%20%5Cfrac%7B1%7D%7B%5Cinfty%20%7D-%5Cfrac%7B1%7D%7B3%7D%5Cright%20%5D)
![I=-\left [ 0-\frac{1}{3}\right ]](https://tex.z-dn.net/?f=I%3D-%5Cleft%20%5B%200-%5Cfrac%7B1%7D%7B3%7D%5Cright%20%5D)

so the value of integral converges at 
Well the thing is you didn't tell us which number but if it is 12 then i strongly believe that the answer is 3/9
Answer:
x
=
26
Step-by-step explanation:
Isolate the variable by dividing each side by factors that don't contain the variable.
We know that
the distance from the centroid of the triangle to one of the vertices is the radius of the circle <span>required to inscribe an equilateral triangle.
[distance </span>centroid of the triangle to one of the vertices]=(2/3)*h
h=the <span>altitude of the equilateral triangle-----> 5.196 in
so
</span>[distance centroid of the triangle to one of the vertices]=(2/3)*5.196
[distance centroid of the triangle to one of the vertices]=3.464 in----> 3.5 in
the radius is equal to the distance of the centroid of the triangle to one of the vertices
hence
the radius is 3.5 in
the answer is
the radius is 3.5 in