Which system of linear inequalities represented by the graph y_>2x-1 and y_<2x+2
2 answers:
1. Check picture 1.
2. By drawing 2 pairs of points,as shown in the picture, we can easily draw the lines y=2x-1 and y=2x+2
3. We notice that the lines are parallel. (we can notice this also from the slopes of the lines, that is the coefficients of x'es, both equal to 2, which means the lines are parallel)
4. Pick (1, 3), as shown in the picture to check which side of each line we will color. (any point can work.)
5. Check (1, 3) in inequality y_<2x+2:
3_<2*1+2=4 is actually true, so we color the part of the line containing (1, 3)
Check (1, 3) in the inequality y_>2x-1: 3_>2*1-1=1 which is true, so we color the part of the line containing the point:
6. So we color the part below the black line and the part above the blue line.
The solution is the part colored in both cases, which is the strip between the lines.
7. Answer: picture 2 is the graph of the system of inequalities
<span>y_>2x-1 and y_<2x+2</span>
2. By drawing 2 pairs of points,as shown in the picture, we can easily draw the lines y=2x-1 and y=2x+2
3. We notice that the lines are parallel. (we can notice this also from the slopes of the lines, that is the coefficients of x'es, both equal to 2, which means the lines are parallel)
4. Pick (1, 3), as shown in the picture to check which side of each line we will color. (any point can work.)
5. Check (1, 3) in inequality y_<2x+2:
3_<2*1+2=4 is actually true, so we color the part of the line containing (1, 3)
Check (1, 3) in the inequality y_>2x-1: 3_>2*1-1=1 which is true, so we color the part of the line containing the point:
6. So we color the part below the black line and the part above the blue line.
The solution is the part colored in both cases, which is the strip between the lines.
7. Answer: picture 2 is the graph of the system of inequalities
<span>y_>2x-1 and y_<2x+2</span>
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Answer:
Step-by-step explanation:
We have to take the derivatives for both functions and replace in the differential equation. Hence
for y=e^{-5x}:
for y=e^{6x}:
Now we replace in the differential equation y'' − y' − 30y = 0
for y=e^{-5x}:
for y=e^{6x}:
Now, to know if both function are linearly independent we calculate the Wronskian
I hope this is useful for you
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