Im pretty sure is teh 3rd one
Answer: Hm, this sounds a bit out of order I hope you use better grammar and please ask a question not a comment
Hello!
I believe that the answer would be 40²
I hope this helps!
Answer:
2x + 3y ≥ 5
Step-by-step explanation:
See the graph attached.
The bold straight line passes through the points (1,1) and (4,-1).
Therefore, the equation of the straight line will be
⇒ 3(y + 1) = - 2(x - 4)
⇒ 3y + 3 = - 2x + 8
⇒ 2x + 3y = 5 ............. (1)
Now, the shaded region i.e. the solution to the inequality does not include the origin(0,0).
So, putting x = 0 and y = 0 in the equation (1) we get, 0 < 5
Therefore, the inequality equation is 2x + 3y ≥ 5 (Answer)
The computation shows that the placw on the hill where the cannonball land is 3.75m.
<h3>How to illustrate the information?</h3>
To find where on the hill the cannonball lands
So 0.15x = 2 + 0.12x - 0.002x²
Taking the LHS expression to the right and rearranging we have:
-0.002x² + 0.12x -.0.15x + 2 = 0.
So we have -0.002x²- 0.03x + 2 = 0
I'll multiply through by -1 so we have
0.002x² + 0.03x -2 = 0.
This is a quadratic equation with two solutions x1 = 25 and x2 = -40 since x cannot be negative x = 25.
The second solution y = 0.15 * 25 = 3.75
Learn more about computations on:
brainly.com/question/4658834
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Complete question:
The flight of a cannonball toward a hill is described by the parabola y = 2 + 0.12x - 0.002x 2 . the hill slopes upward along a path given by y = 0.15x. where on the hill does the cannonball land?