Answer:
x = 6
Step-by-step explanation:
5x - 11 = 19
+11 = +11
5x = 30
5x/5 = 30/5
x = 6
Answer:
g(x)= 1
Step-by-step explanation:
![g(x) = \frac{1}{2} x - 4 \\g(10) = \frac{1}{2} (10) = 5 \\ 5 - 4 = 1](https://tex.z-dn.net/?f=g%28x%29%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20x%20-%204%20%5C%5Cg%2810%29%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%2810%29%20%3D%205%20%5C%5C%205%20-%204%20%3D%201)
Answer:
a) ![t = \sqrt{\frac{2 *10m}{9.8 m/s^2}}=1.43s](https://tex.z-dn.net/?f=%20t%20%3D%20%5Csqrt%7B%5Cfrac%7B2%20%2A10m%7D%7B9.8%20m%2Fs%5E2%7D%7D%3D1.43s%20)
b) ![4.9t^2 -2t -10 =0](https://tex.z-dn.net/?f=%204.9t%5E2%20-2t%20-10%20%3D0)
And we can use the quadratic formula to solve it:
![t = \frac{2 \pm \sqrt{(-2)^2 -4(4.9)(-10)}}{2*4.9}](https://tex.z-dn.net/?f=%20t%20%3D%20%5Cfrac%7B2%20%5Cpm%20%5Csqrt%7B%28-2%29%5E2%20-4%284.9%29%28-10%29%7D%7D%7B2%2A4.9%7D)
![t =\frac{2 \pm \sqrt{200}}{9.8}](https://tex.z-dn.net/?f=%20t%20%3D%5Cfrac%7B2%20%5Cpm%20%5Csqrt%7B200%7D%7D%7B9.8%7D)
![t_1 =1.65 s, t_2 =-1.24 s](https://tex.z-dn.net/?f=%20t_1%20%3D1.65%20s%2C%20t_2%20%3D-1.24%20s)
And since the time can't be negative the correct option would be ![t=1.65 s](https://tex.z-dn.net/?f=%20t%3D1.65%20s)
Step-by-step explanation:
For this case we can use the following kinematics formulas:
![y_f = y_o + V_o t + \frac{1}{2} a t^2](https://tex.z-dn.net/?f=%20y_f%20%3D%20y_o%20%2B%20V_o%20t%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20a%20t%5E2)
For this case we assume that the only acceleration is the gravity a = g =9.8 m/s^2. And for this case we can assume that the reference point is
and the final height would be
since is below the initial point.
Teh acceleration would be a=-g, since the gravity is acting dowward, we assume that the initial velocity is 0, so then we have everything to replace and we got:
![-10 m = 0m + (0m/s)t -\frac{1}{2} (9.8 m/s^2) t^2](https://tex.z-dn.net/?f=%20-10%20m%20%3D%200m%20%2B%20%280m%2Fs%29t%20-%5Cfrac%7B1%7D%7B2%7D%20%289.8%20m%2Fs%5E2%29%20t%5E2)
And solving for t we got:
![t = \sqrt{\frac{2 *10m}{9.8 m/s^2}}=1.43s](https://tex.z-dn.net/?f=%20t%20%3D%20%5Csqrt%7B%5Cfrac%7B2%20%2A10m%7D%7B9.8%20m%2Fs%5E2%7D%7D%3D1.43s%20)
For the second part assuming that We have an initial vertical speed of
we have the following equation:
![-10 m = 0m + (2m/s)t -\frac{1}{2} (9.8 m/s^2) t^2](https://tex.z-dn.net/?f=%20-10%20m%20%3D%200m%20%2B%20%282m%2Fs%29t%20-%5Cfrac%7B1%7D%7B2%7D%20%289.8%20m%2Fs%5E2%29%20t%5E2)
And we have this quadratic equation:
![-10 = 2t -4.9t^2](https://tex.z-dn.net/?f=%20-10%20%3D%202t%20-4.9t%5E2)
![4.9t^2 -2t -10 =0](https://tex.z-dn.net/?f=%204.9t%5E2%20-2t%20-10%20%3D0)
And we can use the quadratic formula to solve it:
![t = \frac{2 \pm \sqrt{(-2)^2 -4(4.9)(-10)}}{2*4.9}](https://tex.z-dn.net/?f=%20t%20%3D%20%5Cfrac%7B2%20%5Cpm%20%5Csqrt%7B%28-2%29%5E2%20-4%284.9%29%28-10%29%7D%7D%7B2%2A4.9%7D)
![t =\frac{2 \pm \sqrt{200}}{9.8}](https://tex.z-dn.net/?f=%20t%20%3D%5Cfrac%7B2%20%5Cpm%20%5Csqrt%7B200%7D%7D%7B9.8%7D)
![t_1 =1.65 s, t_2 =-1.24 s](https://tex.z-dn.net/?f=%20t_1%20%3D1.65%20s%2C%20t_2%20%3D-1.24%20s)
And since the time can't be negative the correct option would be ![t=1.65 s](https://tex.z-dn.net/?f=%20t%3D1.65%20s)