Drag the tiles to the boxes to form correct pairs. Not all tiles will be used. The height (in feet) of a rocket launched from the ground is given by the function f(t) = -16t2 + 160t. Match each value of time elapsed (in seconds) after the rocket’s launch to the rocket's corresponding instantaneous velocity (in feet/second).

The pairs are;

t,         v

2,       96

4,       32

5,       0

6,      -32

7,      -64

9,     -128

Step-by-step explanation:

The given equation is f(t) = -16·t² + 160·t

We have, the velocity, v = d(f(t))/dt = d(-16·t² + 160·t)/dt = -32·t + 160

Which gives;

t,                                 v  

0,    -32×(0) + 160 =   160

1,     -32×(1) + 160 =    128

2,    -32×(2) + 160 =    96

3,    -32×(3) + 160 =    64

4,    -32×(4) + 160 =     32

5,    -32×(5) + 160 =     0

6,    -32×(6) + 160 =   -32

7,    -32×(7) + 160 =    -64

8,    -32×(8) + 160 =    -96

9,    -32×(9) + 160 =   -128

The given velocity values are;

96, -64, 32, 0, -128, -32  which correspond to 2, 7, 4, 5, 9, 6

The pairs are;

t,         v

2,       96

4,       32

5,       0

6,      -32

7,      -64

9,     -128

so firstly, we need to isolate the y variables to be able to solve these inequalities. firstly, add 0.5x on both sides of the first inequality and subtract 1.5x on both sides of the second inequality:

now since the slope is positive for the first inequality, this means that the line going upwards belongs to the first inequality, and the line going downwards belongs to the second inequality.

next, since y ≥ in the first inequality, this means that the shaded region will be above the first inequality's line, thus shading regions a and b.

next, since y ≤ in the second inequality, this means that the shaded region is going to be below this line, thus shading regions a and c.

since both lines shade region a, this means that region a is the solution.

step-by-step explanation:

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