The answer is 60.60dollars
Answer:
2
Step-by-step explanation:
Show that there do not exist scalars c1, c2, and c3 such that c1(1, 0, 1, 0) + c2(1, 0, -2, 1) + c3(2, 0, 1, 2) = (1, -2, 2, 3)
Aloiza [94]
Write the system in augmented-matrix form:

![\iff\left[\begin{array}{ccc|c}1&1&2&1\\0&0&0&-2\\1&-2&1&2\\0&1&2&3\end{array}\right]](https://tex.z-dn.net/?f=%5Ciff%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%261%262%261%5C%5C0%260%260%26-2%5C%5C1%26-2%261%262%5C%5C0%261%262%263%5Cend%7Barray%7D%5Cright%5D)
Row reduce this matrix:
![\left[\begin{array}{ccc|c}1&1&2&1\\0&0&0&-2\\0&-3&-1&1\\0&1&2&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%261%262%261%5C%5C0%260%260%26-2%5C%5C0%26-3%26-1%261%5C%5C0%261%262%263%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc|c}1&1&2&1\\0&0&0&-2\\0&0&5&10\\0&1&2&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%261%262%261%5C%5C0%260%260%26-2%5C%5C0%260%265%2610%5C%5C0%261%262%263%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc|c}1&1&2&1\\0&0&0&-2\\0&0&1&2\\0&1&2&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%261%262%261%5C%5C0%260%260%26-2%5C%5C0%260%261%262%5C%5C0%261%262%263%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc|c}1&1&2&1\\0&0&0&-2\\0&0&1&2\\0&1&0&-1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%261%262%261%5C%5C0%260%260%26-2%5C%5C0%260%261%262%5C%5C0%261%260%26-1%5Cend%7Barray%7D%5Cright%5D)
- Add -2(row 3) and -1(row 4) to row 1:
![\left[\begin{array}{ccc|c}1&0&0&-2\\0&0&0&-2\\0&0&1&2\\0&1&0&-1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%260%260%26-2%5C%5C0%260%260%26-2%5C%5C0%260%261%262%5C%5C0%261%260%26-1%5Cend%7Barray%7D%5Cright%5D)
This matrix tells us that
,
, and
, but clearly
, so there is no solution.
Answer:
p=25 degrees
Step-by-step explanation:
70 and 20+2p are verticle angles, so they are equal
70=20+2p
-20-20
50=2p
/2 /2
25=p