12, 18 , 24, 30, 36, 42….
Answer:
t as a function of height h is t = √600 - h/16
The time to reach a height of 50 feet is 5.86 minutes
Step-by-step explanation:
Function for height is h(t) = 600 - 16t²
where t = time lapsed in seconds after an object is dropped from height of 600 feet
t as a function of height h
replacing the function with variable h
h = 600 - 16t²
Solving for t
Subtracting 600 from both side
h - 600 = -16t²
Divide through by -16
600 - h/ 16 = t²
Take square root of both sides
√600 - h/16 = t
Therefore, t = √600 - h/16
Time to reach height 50 feet
t = √600 - h/16
substituting h = 50 in the equation
t = √600 - 50/16
t = √550/16
t= 34.375
t = 5.86 minutes
-7.
5x-3 = -15
-15-4\4= -16
-16 + 9 = -7
<h3>Answer: Choice D</h3>
Divide both sides of the first equation by 7, then add the result to the second equation
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Explanation:
We can multiply both sides of an equation by the same number and the equation will be equivalent to the original.
For example, if we had x = 5, then we could get 2x = 10 after multiplying both sides by 2. The reason they are equivalent equations is because the same x value is the solution for both equations.
We can also divide both sides of an equation by the same number and the two equations would be equivalent. We can go from 2x = 10 back to x = 5 when we divide both sides by 2.
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If we divide both sides of 7x - 21y = 14 by 7, then we end up with x - 3y = 2. Simply divide each term (7x, -21y, and 14) by 7.
Because 7x-21y=14 and x-3y=2 are equivalent, this means we can replace the "7x-21y=14" with "x-3y=2"
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The new system would be
x-3y = 2
2x+3y = 11
From here you add straight down. Doing so will have the y terms add to -3y+3y = 0y = 0. After this point the y terms are eliminated and you can solve for x just like with any other equation of one variable.
The nearest tenth of how fast a rover will hit Mars' surface after a bounce of 15 ft in height is 20.7ft/s.
<h3>What is the approximation about?</h3>
From the question:
Mars: F(x) = 2/3
Therefore, If x = 15
Then:
f (15) = 2/3 ![\sqrt[8]{15}](https://tex.z-dn.net/?f=%5Csqrt%5B8%5D%7B15%7D)
= 16/3
= 20.7ft/s
Hence, The nearest tenth of how fast a rover will hit Mars' surface after a bounce of 15 ft in height is 20.7ft/s.
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