Answer:
9
Step-by-step explanation:
Let the two perfect cubes be x and y where x > y.
According to the given conditions:
![{x}^{3} - {y}^{3} = 386...(1) \\ y = 7...(2) \\ plug \: y = 7 \: in \: equation \: (1) \\ {x}^{3} - {7}^{3} = 386 \\ {x}^{3} - 343 = 386 \\ {x}^{3} = 343 + 386 \\ {x}^{3} = 343 + 386 \\ {x}^{3} = 729 \\ x = \sqrt[3]{729} \\ x = 9](https://tex.z-dn.net/?f=%20%7Bx%7D%5E%7B3%7D%20%20-%20%20%7By%7D%5E%7B3%7D%20%20%3D%20386...%281%29%20%5C%5C%20y%20%3D%207...%282%29%20%5C%5C%20plug%20%5C%3A%20y%20%3D%207%20%5C%3A%20in%20%5C%3A%20equation%20%5C%3A%20%281%29%20%5C%5C%20%20%7Bx%7D%5E%7B3%7D%20%20-%20%20%7B7%7D%5E%7B3%7D%20%20%3D%20386%20%5C%5C%20%7Bx%7D%5E%7B3%7D%20%20-%20%20343%20%3D%20386%20%5C%5C%20%7Bx%7D%5E%7B3%7D%20%20%20%20%20%3D%20343%20%20%2B%20%20386%20%5C%5C%20%7Bx%7D%5E%7B3%7D%20%20%20%20%20%3D%20343%20%20%2B%20%20386%20%5C%5C%20%7Bx%7D%5E%7B3%7D%20%20%20%3D%20729%20%5C%5C%20x%20%3D%20%20%5Csqrt%5B3%5D%7B729%7D%20%20%5C%5C%20x%20%3D%209)
Thus the cube root of the larger number is 9.
It has become somewhat fashionable to have students derive the Quadratic Formula themselves; this is done by completing the square for the generic quadratic equation ax2 + bx + c = 0. While I can understand the impulse (showing students how the Formula was invented, and thereby providing a concrete example of the usefulness of abstract symbolic manipulation), the computations involved are often a bit beyond the average student at this point.
Answer:
I believe it is B
Step-by-step explanation:
3 out of 4 options to land on are less than 5
Answer:
The value of AB is
and it's not possible to multiply BA.
Step-by-step explanation:
Consider the provided matrices.
, ![B=\left[\begin{array}{ccc}3\\5\end{array}\right]](https://tex.z-dn.net/?f=B%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%5C%5C5%5Cend%7Barray%7D%5Cright%5D)
Two matrices can be multiplied if and only if first matrix has an order m × n and second matrix has an order n × v.
Multiply AB
Matrix A has order 2 × 2 and matrix B has order 2 × 1. So according to rule we can multiply both the matrix as shown:
![AB=\left[\begin{array}{ccc}2&3\\2&1\end{array}\right] \left[\begin{array}{ccc}3\\5\end{array}\right]](https://tex.z-dn.net/?f=AB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%263%5C%5C2%261%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%5C%5C5%5Cend%7Barray%7D%5Cright%5D)
![AB=\left[\begin{array}{ccc}2\times 3+3\times 5\\2\times 3+1\times 5\end{array}\right]](https://tex.z-dn.net/?f=AB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%5Ctimes%203%2B3%5Ctimes%205%5C%5C2%5Ctimes%203%2B1%5Ctimes%205%5Cend%7Barray%7D%5Cright%5D)
![AB=\left[\begin{array}{ccc}6+15\\6+5\end{array}\right]](https://tex.z-dn.net/?f=AB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D6%2B15%5C%5C6%2B5%5Cend%7Barray%7D%5Cright%5D)
![AB=\left[\begin{array}{ccc}21\\11\end{array}\right]](https://tex.z-dn.net/?f=AB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D21%5C%5C11%5Cend%7Barray%7D%5Cright%5D)
Hence, the value of AB is ![\left[\begin{array}{ccc}21\\11\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D21%5C%5C11%5Cend%7Barray%7D%5Cright%5D)
Now calculate the value of BA as shown:
Multiply BA
Matrix B has order 2 × 1 and matrix A has order 2 × 2. So according to rule we cannot multiply both the matrix.
We can multiply two matrix if first matrix has an order m × n and second matrix has an order n × v.
That means number of column of first matrix should be equal to the number of rows of second matrix.
Hence, it's not possible to multiply BA.
Answer: 20 pages
Step-by-step explanation: 100/5 is 20