Answer:
The answer is "
f and g are arbitrary".
Step-by-step explanation:
The matrix of the device is increased
![\left[\begin{array}{ccc}1&3&f\\ c&d&g\\ \end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%263%26f%5C%5C%20c%26d%26g%5C%5C%20%5Cend%7Barray%7D%5Cright%5D)
Reduce the echelon row matrix
![\left[\begin{array}{ccc}1&3&f\\ c&d&g\\ \end{array}\right] \\\\R_1 \leftrightarrow R_2 \\\\\frac{R_2 -1}{C R_1 \to R_2} \sim \left[\begin{array}{ccc} c&d&g\\ 0 & \frac{3c-d}{c}& \frac{cf-g}{c}\\ \end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%263%26f%5C%5C%20c%26d%26g%5C%5C%20%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5CR_1%20%5Cleftrightarrow%20%20R_2%20%5C%5C%5C%5C%5Cfrac%7BR_2%20-1%7D%7BC%20R_1%20%5Cto%20R_2%7D%20%20%20%5Csim%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%20c%26d%26g%5C%5C%200%20%26%20%5Cfrac%7B3c-d%7D%7Bc%7D%26%20%5Cfrac%7Bcf-g%7D%7Bc%7D%5C%5C%20%5Cend%7Barray%7D%5Cright%5D)
Therefore, if 3c
0 is d 3c, the device is valid. Therefore d are arbitrary 3c, g and f.
Answer:
Side a is 35/5=7 inches long
Side b is 30/5=6 inches long
Side c is 15/5=3 inches long
I believe that the correct newer would actually be the last option. By this Eddie person finding the length of the cube, the small cube, as also the cube which is the bigger one, if we were just to find the length and just add them all up, we would then find how many would then fit.
The mixed number is a fraction so the question is not able to be solved.
Hope this helps.
Answer:
the ask a question about the question isnt working
Step-by-step explanation:
but can you zoom in its really glitchy and blurred from far away
