Answer:
p ∈ IR - {6}
Step-by-step explanation:
The set of all linear combination of two vectors ''u'' and ''v'' that belong to R2
is all R2 ⇔
And also u and v must be linearly independent.
In order to achieve the final condition, we can make a matrix that belongs to
using the vectors ''u'' and ''v'' to form its columns, and next calculate the determinant. Finally, we will need that this determinant must be different to zero.
Let's make the matrix :
![A=\left[\begin{array}{cc}3&1&p&2\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%261%26p%262%5Cend%7Barray%7D%5Cright%5D)
We used the first vector ''u'' as the first column of the matrix A
We used the second vector ''v'' as the second column of the matrix A
The determinant of the matrix ''A'' is

We need this determinant to be different to zero


The only restriction in order to the set of all linear combination of ''u'' and ''v'' to be R2 is that 
We can write : p ∈ IR - {6}
Notice that is
⇒


If we write
, the vectors ''u'' and ''v'' wouldn't be linearly independent and therefore the set of all linear combination of ''u'' and ''b'' wouldn't be R2.
Answer:
∠ CED = 58°
Step-by-step explanation:
∠ CED and ∠ AED are adjacent angles and sum to 180° , then
∠ CED + 122° = 180° ( subtract 122° from both sides )
∠ CED = 58°
X. 0,2,6,4,8,10
Y. 2,0,2,4,6,8
It’s how far the number is from zero
Answer:
D, 10.
Step-by-step explanation:
10+8=18 4/3x18=24