Answer:
Yes.
Step-by-step explanation:
To see if h is a factor of f, we can use the factor theorem.
h(x)=x-3 has a zero at x=3 because h(3)=3-3=0.
So we want to see if the zero of h is a zero of f.
Is x=3 a zero of f?
You can check using synthetic division or just plug in 3.
Let's do both.
If you use synthetic you are trying to see if you get a remainder of 0.
If you plug it in you are trying to see if you get 0 as the output when plugging your input 3.
Let's do synthetic first:
Since we are checking to see if x=3 is a zero, that will go on the outside:
3 | 1 -1 -1 -15
| 3 6 15
- ------------------------
1 2 5 0
So yep the remainder is 0 so the answer is yes.
Let's plug it in:
f(3)=3^3-3^2-3-15
f(3)=27-9-3-15
f(3)=18-3-15
f(3)=15-15
f(3)=0
The result is 0 so the answer is yes.
You pick your favorite way here.
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, M, N and C are coplanar, because they are on the same plane.
Noncollinear, because they are not on the same line.
Hope this helps :)
Answer:
ok so we just use the pyagram theorem
25^2+65^2=c^2
4850=c^2
the square root of 4850 is 69(*snickers*).6419413859
so the height of the platform is 70 feet
