What is the solution to the following system of equations?
1 answer:
You might be interested in
The general form of the equation of a line is represented by the Affine function y = mx + b, in which m represents the slope, b the y-intercept, y and x the coordinates of a point.
By hypothesis, we got the slope (m = 4), and the coordinates of a point (in which x = 0 and y = -9). So all we need to do is to find the y-intercept b.
Let's replace the values we have in the equation y = mx + b:
-9 = 4*0 + b
-9 = 0 + b
-9 = b.
We got now the y-intercept. So the final equation of the line is:
y = 4x - 9
If we subtract 4x from each side, we get:
y - 4x = -9
And if we multiply both sides by -1, we get:
4x - y = 9
So the equation of the line, based on your choices, is 4x - y = 9.
I've added a picture of the graph of the equation under the answer.
Hope this Helps! :)
By hypothesis, we got the slope (m = 4), and the coordinates of a point (in which x = 0 and y = -9). So all we need to do is to find the y-intercept b.
Let's replace the values we have in the equation y = mx + b:
-9 = 4*0 + b
-9 = 0 + b
-9 = b.
We got now the y-intercept. So the final equation of the line is:
y = 4x - 9
If we subtract 4x from each side, we get:
y - 4x = -9
And if we multiply both sides by -1, we get:
4x - y = 9
So the equation of the line, based on your choices, is 4x - y = 9.
I've added a picture of the graph of the equation under the answer.
Hope this Helps! :)
This question is incomplete, the complete question is;
Determine if the described set is a subspace. Assume a, b, and c are real numbers.
The subset of R³ consisting of vectors of the form , where abc = 0
- The set is a subspace
- The set is not a subspace
Answer:
Therefore; The set is not a subspace
Step-by-step explanation:
Given the data the question;
the subset R³;
S = { , where abc = 0 }
we know that a subset of R³ is a subspace if it stratifies the following properties;
- it contains additive identity
- it is closed under addition
- it is closed under scales multiplication
Looking at the properties, we can say that it is not a subspace
As;
u = ∈ S and v =
∈ S
As 1×1×0=0 0×1×1=0
But u+v =
∉ S as 1×2×1 ≠ 0
Hence, it is not closed under addition.
Therefore; The set is not a subspace