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.
If you would like to find the decimal number which is equivalent to 1/10, you can do this using the following step:
1/10 = 0.1
The correct result would be B. 0.1.
Answer:
x= 1 , y=4
Step-by-step explanation:
x-y= -3 => Equation 1
x+5y= 21 => Equation 2
<u>Substitut</u><u>ion</u><u> </u><u>Method</u><u>:</u>
<u>Substitu</u><u>te</u><u> </u><u>Equation</u><u> </u><u>1</u>=>
x=y-3 <= Equation 3
Put x=y-3 in Equation 2:
x+5y=21
( y-3)+5y=21
y-3+5y=21
6y-3=21
6y=21+3
6y=24
y=24÷6
y=4
Put y=4 in Equation 1:
x-y= -3
x-4=-3
x=4-3
x=1
Hope this helps :)
Answer:
width is 2 and length is 5
Step-by-step explanation:
Answer:
1,860 per minute and 111,600 per hour
