Answer would be the third option.
Area of a triangle is half of the base times the height. A=(b·h)/2
In the third option the height is 8 ft. and base is 6 ft. which when multiplied together gets you 48. Half of that is 24.
Answer:
<em>The time it takes the ball to hit the ground is 3.05secs</em>
step - by - step explanation:
<em>In order to find height from where ball is dropped, you have to find height or h(t) when time or t is zero.So plug in t=0 into your quadratic equation:h(0) = -16.1(0^2) + 150h(0) = 0 +150h(0) = 150 ft is the height from where ball is dropped. When ball hits the ground, the height is zero. So plug in h(t) = 0 and solve for t.0 = -16.1t^2 + 15016.1 t^2 = 150t^2 = 150/16.1t = sqrt(150/16.1)t = ± 3.05Since time cannot be negative, your answer is positive solution i.e. t = 3.05 </em>
Let’s say that the other one is described as X. So if one piece is twice as long as the other one, it’ll be 2x.
2x + X =111
3x = 111
3x divided by 3= 111 divided by 3
X= 37
So 2x = 2(37) = 74
X = 37
74
+ 37
————
111
Answer:
The distance between A and D to the nearest tenth is;
Explanation:
Given the two points;
Applying the distance between two points formula;
![d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}](https://tex.z-dn.net/?f=d%3D%5Csqrt%5B%5D%7B%28x_2-x_1%29%5E2%2B%28y_2-y_1%29%5E2%7D)
substituting the given coordinates we have;
![\begin{gathered} AD=\sqrt[]{(-3-6)^2+(-2-2)^2} \\ AD=\sqrt[]{(-9)^2+(-4)^2} \\ AD=\sqrt[]{81+16} \\ AD=\sqrt[]{97} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20AD%3D%5Csqrt%5B%5D%7B%28-3-6%29%5E2%2B%28-2-2%29%5E2%7D%20%5C%5C%20AD%3D%5Csqrt%5B%5D%7B%28-9%29%5E2%2B%28-4%29%5E2%7D%20%5C%5C%20AD%3D%5Csqrt%5B%5D%7B81%2B16%7D%20%5C%5C%20AD%3D%5Csqrt%5B%5D%7B97%7D%20%5Cend%7Bgathered%7D)
Simplifying;
Therefore, the distance between A and D to the nearest tenth is;
2x + 21 = 35
hope this helps
