The ordered pair which makes both inequalities true is: D. (3, 0).
<h3>How to determine ordered pair?</h3>
In Mathematics, an inequality can be used to show the relationship between two (2) or more integers and variables in an equation.
In order to determine ordered pair which makes both inequalities true, we would substitute the points into the inequalities as follows:
At (0, 0), we have:
y > -2x + 3
0 > -2(0) + 3
0 > 3 (false).
y < x – 2
0 < 0 - 2
0 < -2 (false)
At (0, -1), we have:
y > -2x + 3
-1 > -2(0) + 3
-1 > 3 (false).
y < x – 2
-1 < 0 - 2
-1 < -2 (false)
At (1, 1), we have:
y > -2x + 3
1 > -2(1) + 3
1 > -1 (true).
y < x – 2
1 < 1 - 2
1 < -1 (false)
At (3, 0), we have:
y > -2x + 3
0 > -2(3) + 3
0 > -3 (true).
y < x – 2
0 < 3 - 2
0 < 1 (true).
Read more on inequalities here: brainly.com/question/24372553
#SPJ1
Answer:
y=-4x+5
Step-by-step explanation:
y+3=-4(x-2)
y+3=-4x+8
-3. -3
y=-4x+5
Inflection point is the point where the second derivative of a graph is zero.
y = (x+1)arctan xy' = (x+1)(arctan x)' + (1)arctan xy' = (x+1)/(x^2+1) + arctan xy'' = (x+1)(1/(1+x^2))' + 1/(1+x^2) + 1/(1+x^2)y'' = (x+1)(-1/(1+x^2)^2)(2x)+2/(1+x^2)y'' = ((x+1)(-2x)+1+x^2)/(1+x^2)^2y'' = (-2x^2-2x+2+2x^2)/(1+x^2)^2y'' = (-2x+2)/(1+x^2)^2
Solving for point of inflection: y'' = 00 = (-2x+2)/(1+x^2)^20 = -2x+2x = 1y(1) = (1+1)arctan(1) = 2 * pi/4 = pi/2
Therefore, E(1, pi/2).
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
