Answer: her call lasted for 26 minutes.
Step-by-step explanation:
Let x represent the number of minutes for which her call lasted.
Rachel purchased a prepaid phone card for $30. Long distance calls cost 6 cents a minute using this card. Converting 6 cents to dollars, it becomes 6/100 = $0.06
This means that the cost of x minutes of long distance call is
0.06 × x = $0.06x.
If the remaining credit on her card is $28.44, it means that
0.06x + 28.44 = 30
0.06x = 30 - 28.44
0.06x = 1.56
x = 1.56/0.06
x = 26
For question 4 the Y needs to be by itself. So instead of it being x+2y>/-12 it would be y>/-6+x. You would have to subtract x on both sides and then divide 2 on both sides as well. whatever you do on one side, you have to do it on the other side. Hope this helps. After you redo all the equations, you graph them. If you need anymore help , let me know!
The Correct Answer is : 4x - 4 * 2
Hope This Helps :)
If it does please give brainliest answer xD
Solution :
Given initial velocity, v= 48 ft/s
Acceleration due to gravity, g = ![$12\ ft/s^2$](https://tex.z-dn.net/?f=%2412%5C%20ft%2Fs%5E2%24)
a). Therefore the maximum height he can jump on Mars is
![$H_{max}=\frac{v^2}{2g}$](https://tex.z-dn.net/?f=%24H_%7Bmax%7D%3D%5Cfrac%7Bv%5E2%7D%7B2g%7D%24)
![$H_{max} = \frac{(48)^2}{2 \times 12}$](https://tex.z-dn.net/?f=%24H_%7Bmax%7D%20%3D%20%5Cfrac%7B%2848%29%5E2%7D%7B2%20%5Ctimes%2012%7D%24)
= 96 ft
b). Time he can stay in the air before hitting the ground is
![$T=\frac{2v}{g}$](https://tex.z-dn.net/?f=%24T%3D%5Cfrac%7B2v%7D%7Bg%7D%24)
![$T=\frac{2 \times 48}{12}$](https://tex.z-dn.net/?f=%24T%3D%5Cfrac%7B2%20%5Ctimes%2048%7D%7B12%7D%24)
= 8 seconds
c). Considering upward motion as positive direction.
v = u + at
We find the time taken to reach the maximum height by taking v = 0.
v = u + at
0 = 16 + (12) t
![$t=\frac{16}{12}$](https://tex.z-dn.net/?f=%24t%3D%5Cfrac%7B16%7D%7B12%7D%24)
![$=\frac{4}{3} \ s$](https://tex.z-dn.net/?f=%24%3D%5Cfrac%7B4%7D%7B3%7D%20%5C%20s%24)
We know that, ![$S=ut + \frac{1}{2}at^2$](https://tex.z-dn.net/?f=%24S%3Dut%20%2B%20%5Cfrac%7B1%7D%7B2%7Dat%5E2%24)
Taking t =
, we get
![$S=16 \times\frac{4}{3} + \frac{1}{2}\times(-12) \times \left(\frac{4}{3}\right)^2$](https://tex.z-dn.net/?f=%24S%3D16%20%5Ctimes%5Cfrac%7B4%7D%7B3%7D%20%2B%20%5Cfrac%7B1%7D%7B2%7D%5Ctimes%28-12%29%20%5Ctimes%20%5Cleft%28%5Cfrac%7B4%7D%7B3%7D%5Cright%29%5E2%24)
feet
Thus he can't reach to 100 ft as it is shown in the movie.
d). For any jump whose final landing position will be same of the take off level, the final velocity will be the initial velocity.
Therefore final velocity is = -16 ft/s
You have to multiply the 2 numbers together.
0.3 x 0.4
0.12