Answer:
C and E are not on the line.
Step-by-step explanation:
Just substitute for the x and the y and see which ordered pair fits and which does not:
If Left side of the equation = right side then the point fits.
A. y = -3
-2x+5 = -2(4) + 5 = -3 So A is on the line
B. y = 9
-2(-2) + 5 = 9 So this point does lie on the line.
C. y = 1
-2(3) + 5 = -1 No.
D. y = 7
-1(-1) + 5 = 7 YES.
E. y = 1
-2(1) + 5 = 3 NO.
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:
156
Step-by-step explanation:
Write 26% as
26
100
Since, finding the fraction of a number is same as multiplying the fraction with the number, we have
26
100
of 600 =
26
100
× 600
Therefore, the answer is 156
If you are using a calculator, simply enter 26÷100×600 which will give you 156 as the answer.
The coordinates would be (2,7) you just have to move 1 left 4 up
The hundredths place value was rounded.
