A+ny = n
a = n-ny
a = (1-y)n
a/(1-y) = n
<span>n = a/(1-y)</span>
Answer:
1st picture: (0,4)
The lines intersect at point (0,4).
2nd picture: Graph D
2x ≥ y - 1
2x - 5y ≤ 10
Set these inequalities up in standard form.
y ≤ 2x + 1
-5y ≤ 10 - 2x → y ≥ -2 + 2/5x → y ≥ 2/5x - 2
When you divide by a negative number, you switch the inequality sign.
Now you have:
y ≤ 2x + 1
y ≥ 2/5x - 2
Looking at the graphs, you first want to find the lines that intersect the y-axis at (0, 1) and (0, -2).
In this case, it is all of them.
Next, you would look at the shaded regions.
The first inequality says the values are less than or equal to. So you look for a shaded region below a line. The second inequality says the values are greater than or equal to. So you look for a shaded region above a line.
That would mean Graph B or D.
Then you look at the specific lines. You can see that the lower line is y ≥ 2/5x - 2. You need a shaded region above this line. You can see the above line is y ≤ 2x + 1. You need a shaded region below this line. That is Graph D.
Answer:
Solution given
tan θ=opposite/adjacent=y/x is a required answer.
Answer:
g(x)= (x-(1+2i)) * (x-(1-2i)) * (x-(3-i)) * (x-(3+i))
Please see attached image
Step-by-step explanation:
We can easily solve this equation by using a technical solver or a programming language such as Octave.
The linear factorization of the equation can be obtained directly by finding the zeros of g(x)
Please see attached images for the answer to your problem
g(x)= (x-(1+2i)) * (x-(1-2i)) * (x-(3-i)) * (x-(3+i))
