Anything raised do 0 is always 1
I hope that’s correct
Answer:
Therefore, the inverse of given matrix is

Step-by-step explanation:
The inverse of a square matrix
is
such that
where I is the identity matrix.
Consider, ![A = \left[\begin{array}{ccc}4&3\\3&6\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%263%5C%5C3%266%5Cend%7Barray%7D%5Cright%5D)








Therefore, the inverse of given matrix is

<span>Cone Lateral Area = (<span>π<span> • r •<span> slant height)
</span></span></span></span><span>Cone Lateral Area = (PI * 3 * 5)
</span>Cone Lateral Area =
<span>
<span>
<span>
47.1
</span>
</span>
</span>
square inches
Cone Lateral Area =
<span>
<span>
47</span></span> square inches (rounded to nearest whole number).
Step-by-step explanation:
A=1/2(B1+B2)h is a trapezoid
A=1/2bh is a triangle
A=bh is a parallelogram
And A=Pir2 is a circle