<h3>Option B</h3><h3>At 2 second and 1.75 second, the object be at a height of 56 feet</h3>
<em><u>Solution:</u></em>
Given that,
<em><u>The height h(t) in feet of an object t seconds after it is propelled straight up from the ground with an initial velocity of 60 feet per second is modeled by the equation:</u></em>

<em><u>At what times will the object be at a height of 56 feet</u></em>
<em><u>Substitute h = 56</u></em>

Solve the above equation by quadratic formula


Thus, at 2 second and 1.75 second, the object be at a height of 56 feet
Solutions
The terminal side of an angle in standard position is a ray that has been rotated from the positive (x) <span>axis.
</span>If the angle measure is positive <span>the rotation is counterclockwise
</span>If the angle measure is negative the <span>rotation is counterclockwise.
</span>
The angle measure is positive, so the rotation is 135° is <span>counterclockwise.
</span>
Rotate the ray 135° counterclockwise. This rotation puts the terminal side in Quadrant <span>II.</span>
The first option
letter a 0,2, 4
Answer: 378.57mi²
Step-by-step explanation:
Formula = a²+2a√{(a²/4)+h²}
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