Answer:
1/3,-0.3,1.2,4/3,3/2
Step-by-step explanation:
-1/3,-0.3,1.2,4/3,3/2
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:(6, -1)
Step-by-step explanation:


notice... the dog's pen perimeter, does not include the side that's bordering the garden's, since that side will use the heavy duty fence, instead of the light one
so, the sum of both of those costs, will be the C(x)

so, just take the derivative of it, and set it to 0 to find the extremas, and do a first-derivative test for any minimum
Answer:
12.5%
Step-by-step explanation:
$4.50-$4/$4 x 100%
0.5/4x100%
50/4= 25/2 = 12.5%