Can someone help me with this
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<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
First, we need to find the length of the side of the square.
Use phytagorean theorem to find the length of the side. The side acts as hypotenuse, the distance of x and the distance of y acts as the perpendicular side of a right triangle. For clear understanding, see image attached.
General phytagorean theorem
c² = a² + b²
c represents hypotenuse, a and b are the side perpendicular to each other.
In this case, we could write it as
s² = Δx² + Δy²
s represents the length of the side, Δx represents distance of x, Δy represents distance of y
Plug in the numbers, use two of the vertices
I use (4,-1) and (7,3)
s² = Δx² + Δy²
s² = (7-4)² + (3 - (-1))²
s² = 3² + (3 + 1)²
s² = 3² + 4²
s² = 9 + 16
s² = 25
s = √25
s = 5
The length of the side is equal to 5 units length.
Second, find the area of the square
General area to find the area of a square
a = s²
Plug in the numbers
a = s²
a = 5²
a = 25
The area of the square is equal to 25 units area.
Use phytagorean theorem to find the length of the side. The side acts as hypotenuse, the distance of x and the distance of y acts as the perpendicular side of a right triangle. For clear understanding, see image attached.
General phytagorean theorem
c² = a² + b²
c represents hypotenuse, a and b are the side perpendicular to each other.
In this case, we could write it as
s² = Δx² + Δy²
s represents the length of the side, Δx represents distance of x, Δy represents distance of y
Plug in the numbers, use two of the vertices
I use (4,-1) and (7,3)
s² = Δx² + Δy²
s² = (7-4)² + (3 - (-1))²
s² = 3² + (3 + 1)²
s² = 3² + 4²
s² = 9 + 16
s² = 25
s = √25
s = 5
The length of the side is equal to 5 units length.
Second, find the area of the square
General area to find the area of a square
a = s²
Plug in the numbers
a = s²
a = 5²
a = 25
The area of the square is equal to 25 units area.