Written numerically, the number you stated is written as:
5,004,300
Standard Form (Scientific Notation) numbers are written as a decimal (n.m) such that there is [normally] only one digit (n) to the left of the decimal point, and the rest of the number (m) to the right (with a possible limit). If there is only a single digit in the whole number then the value on the right is a zero (n.0).
To show that the value shown has shifted its decimal point position, the number of decimal positions shifted is shown after the number as a 10’s multiplied to the power of the number of positions shifted.
e.g., 102 = 102.0 = 1.02 (decimal point shifted left two places. If smallest digit is 0 then ignore it UNLESS it’s the ONLY digit!!)
1.02*10^2(2 Decimal shift to the left)
1.02*10^2
So, for the number asked in the question:
5,004,300 = 5,004,300.0
Left shift 6 times to largest digit = 5.004300
5.004300 (show the shift multiplier: 10^6)
5.004300*10^6
Matrix A order is 4x8.
Matrix B order is 4x16.
AB cannot result because an (m x n) can only be multiplied by an (n x p) matrix and orders don't match the requirements.
BA has the same negative result.
Answer:
A.) 1018 square inches
Step-by-step explanation:
The largest sphere will have a diameter equaling the length of the cube (see picture).
If the side length of the cube is 18 inches, the diameter of the sphere is also 18 inches. Use the surface area formula for a circle:
For this formula, we need the radius of the sphere. Divide the diameter by 2:
The radius is 9 inches. Plug this into the equation:
Simplify the equation:
Round the result to the nearest whole number:
→ 
The surface area is 1,018 inches².
:Done
Picture:
In a 2D version, we can clearly see that if the circle fits snuggly inside of the square, the diameter of a sphere is the same as the length of a side of the cube.
now, if we take 2000 to be the 100%, what is 2200? well, 2200 is just 100% + 10%, namely 110%, and if we change that percent format to a decimal, we simply divide it by 100, thus 
.
so, 1.1 is the decimal number we multiply a term to get the next term, namely 1.1 is the common ratio.
![\bf \qquad \qquad \textit{sum of a finite geometric sequence}\\\\S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad \begin{cases}n=n^{th}\ term\\a_1=\textit{first term's value}\\r=\textit{common ratio}\\----------\\a_1=2000\\r=1.1\\n=4\end{cases}\\\\\\S_4=2000\left[ \cfrac{1-(1.1)^4}{1-1.1} \right]\implies S_4=2000\left(\cfrac{-0.4641}{-0.1} \right)\\\\\\S_4=2000(4.641)\implies S_4=9282](https://tex.z-dn.net/?f=%20%5Cbf%20%5Cqquad%20%5Cqquad%20%5Ctextit%7Bsum%20of%20a%20finite%20geometric%20sequence%7D%5C%5C%5C%5CS_n%3D%5Csum%5Climits_%7Bi%3D1%7D%5E%7Bn%7D%5C%20a_1%5Ccdot%20r%5E%7Bi-1%7D%5Cimplies%20S_n%3Da_1%5Cleft%28%20%5Ccfrac%7B1-r%5En%7D%7B1-r%7D%20%5Cright%29%5Cquad%20%5Cbegin%7Bcases%7Dn%3Dn%5E%7Bth%7D%5C%20term%5C%5Ca_1%3D%5Ctextit%7Bfirst%20term%27s%20value%7D%5C%5Cr%3D%5Ctextit%7Bcommon%20ratio%7D%5C%5C----------%5C%5Ca_1%3D2000%5C%5Cr%3D1.1%5C%5Cn%3D4%5Cend%7Bcases%7D%5C%5C%5C%5C%5C%5CS_4%3D2000%5Cleft%5B%20%5Ccfrac%7B1-%281.1%29%5E4%7D%7B1-1.1%7D%20%5Cright%5D%5Cimplies%20S_4%3D2000%5Cleft%28%5Ccfrac%7B-0.4641%7D%7B-0.1%7D%20%20%5Cright%29%5C%5C%5C%5C%5C%5CS_4%3D2000%284.641%29%5Cimplies%20S_4%3D9282%20)
