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:
150 cm2
Step-by-step explanation:
double of triangle abc is 200
this is in middle of both
If Paul mows the lawn once, he earns 1*25.50. If he mows it twice, he earns 2*25.50, and so on. So, he earns L*25.50. But, he spends 3.50. So the equation is:
Symmetrical functions can be about the x and y axis. Essentially, if we reflect the graph across the y or x axis, we get the same graph. Some other graphs can be reflected across both the x and y axis at the same time and be symmetrical. These can be classified as odd and even functions. You can test this by replacing x and y with -x and -y and simplify the equation. If the results comes out to be the same as the original, it is symmetrical across the origin.
Best of Luck!
100 times (6+5)
100 times 11
100 groups of 11=1100
Hope I helped:)
