I believe your answer is 6 seconds.
Because the height of the ball = -16(x² - 5x - 6), when the height of the ball is 0, which is when it is on the ground, we can set -16(x² - 5x - 6) equal to 0. This also allows us to divide by -16, and then we can solve the equation:
x² - 5x - 6 = 0
(x - 6)(x + 1) = 0
So x = 6 or x = -1, and because a quantity of time cannot be negative, x would have to be 6, which means it takes 6 seconds for the ball to reach 0 feet.
I hope this helps!
Formula for Perimeter of Rectangle:
P = 2(L + W)
Plug in 160:
160 = 2(L + W)
L = 4W
So we can plug in '4W' for 'L' in the first equation.
<span>160 = 2(L + W)
160 = 2(4W + W)
Combine like terms:
160 = 2(5W)
160 = 10W
Divide 10 to both sides:
W = 16
Now we can plug this back into any of the two equations to find the length.
L = 4W
L = 4(16)
L = 64
So the width is 16, and the length is 64.</span>
Distribute: 3x+6+4=5x+7
Simplify: 3x+10=5x+7
Subtract 3x from both sides: 10=2x + 7
Subtract 7 from both sides: 3=2x
Divide by 2: x=3/2
x=3/2
Answer: The percent error is -2.1352% of Jocelyn's estimate.
The area of square is
square units
<h3><u>Solution:</u></h3>
Given that square has side length (x+5) units
To find: area of square
<em><u>The area of square is given as:</u></em>
![\text {Area of square }=\mathrm{a}^{2}](https://tex.z-dn.net/?f=%5Ctext%20%7BArea%20of%20square%20%7D%3D%5Cmathrm%7Ba%7D%5E%7B2%7D)
Where "a" is the length of side
From question, length of each side "a" = x + 5 units
Substituting the value in above formula,
![\text {Area of square }=(x+5)^{2}](https://tex.z-dn.net/?f=%5Ctext%20%7BArea%20of%20square%20%7D%3D%28x%2B5%29%5E%7B2%7D)
![{\text {Expanding }(x+5)^{2} \text { using the algebraic identity: }} \\\\ {(a+b)^{2}=a^{2}+2 a b+b^{2}}\end{array}](https://tex.z-dn.net/?f=%7B%5Ctext%20%7BExpanding%20%7D%28x%2B5%29%5E%7B2%7D%20%5Ctext%20%7B%20using%20the%20algebraic%20identity%3A%20%7D%7D%20%5C%5C%5C%5C%20%7B%28a%2Bb%29%5E%7B2%7D%3Da%5E%7B2%7D%2B2%20a%20b%2Bb%5E%7B2%7D%7D%5Cend%7Barray%7D)
![\begin{array}{l}{\text {Area of square }=x^{2}+2(x)(5)+5^{2}} \\\\ {\text {Area of square }=x^{2}+10 x+25}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bl%7D%7B%5Ctext%20%7BArea%20of%20square%20%7D%3Dx%5E%7B2%7D%2B2%28x%29%285%29%2B5%5E%7B2%7D%7D%20%5C%5C%5C%5C%20%7B%5Ctext%20%7BArea%20of%20square%20%7D%3Dx%5E%7B2%7D%2B10%20x%2B25%7D%5Cend%7Barray%7D)
Thus the area of square is
square units