Answer:The expression can be simplified in following steps;
Step-by-step explanation:
(16+a)+15=0
16+15+a=0
31+a=0
a=-31
Answer:
t=r/(sq)
Step-by-step explanation:
r=sqt
=> r/(sq)=t
Answer:
Yes, it is invertible
Step-by-step explanation:
We need to find in the matrix determinant is different from zero, since iif it is, that the matrix is invertible.
Let's use co-factor expansion to find the determinant of this 4x4 matrix, using the column that has more zeroes in it as the co-factor, so we reduce the number of determinant calculations for the obtained sub-matrices.We pick the first column for that since it has three zeros!
Then the determinant of this matrix becomes:
![4\,*Det\left[\begin{array}{ccc}1&4&6\\0&3&8\\0&0&1\end{array}\right] +0+0+0](https://tex.z-dn.net/?f=4%5C%2C%2ADet%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%264%266%5C%5C0%263%268%5C%5C0%260%261%5Cend%7Barray%7D%5Cright%5D%20%2B0%2B0%2B0)
And the determinant of these 3x3 matrix is very simple because most of the cross multiplications render zero:
![Det\left[\begin{array}{ccc}1&4&6\\0&3&8\\0&0&1\end{array}\right] =1 \,(3\,*\,1-0)+4\,(0-0)+6\,(0-0)=3](https://tex.z-dn.net/?f=Det%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%264%266%5C%5C0%263%268%5C%5C0%260%261%5Cend%7Barray%7D%5Cright%5D%20%3D1%20%5C%2C%283%5C%2C%2A%5C%2C1-0%29%2B4%5C%2C%280-0%29%2B6%5C%2C%280-0%29%3D3)
Therefore, the Det of the initial matrix is : 4 * 3 = 12
and then the matrix is invertible
Answer:
y= x^2 - 4x + 2
I did the test I got it right!
Hope it helps.
X=-1/8y^2
If the Vertex is at the origin: V=(h,k)=(0,0)→h=0, k=0
y^2=4px
(-8)(x=-1/8y^2)
-8x=y^2
y^2=-8x
4p=-8
Solving for p:
4p/4=-8/4
p=-2<0 (negative)→The parabola opens to the left
Focus: F=(p,0)→F=(-2,0)
Drectrix: x=-p=-(-2)→Directrix: x=2
Answers:
Focus=(-2,0)
Directrix: x=2