Ms. Cassidy instructed Miguel to change one sign of the graph of y < 2x – 4 so that point (2, 3) can be included in the solution set.
To check which of the given options might Miguel write we check the inequality that holds true for the point (2,3).Substituting x=2 ,y=3 we have:
1) y < 2x – 1
3<2(2)-1
3<3 Not True.
2)y ≤ 2x – 4
3≤ 2(2) -4
3≤ 0 .Not true.
3) y > 2x – 4
3> 2(2)-4
3> 0 True.
4) y < 2x + 4
3<2(2)+4
3<8 True
5) .y < 3.5x – 4
3< 3.5(2)-4
3<3 Not true
6) y < 4x – 4
3<4(2)-4
3<4 True.
Options 3 ,4 ,6 holds true for the point (2,3)
The relevant formula that we can use in this situation is:
v = v0 + at
where v is the final velocity, v0 is the initial velocity
= 20 km/hr, a is acceleration at 3.4 m/s^2 and t is time at 20 seconds
First convert v0 to m/s, v0 = 5.56 m/s
v = 5.56 m/s + 3.4 m/s^2 * 20 s
v = 73.56 m/s
Then convert back to km/hr:
<span>v = 264.82 km/hr</span>
Answer:
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Step-by-step explanation:
<h2>Nice</h2>
5x(4x^3 + 3x - 2)
There’s a 5 and a x common.
