A)
x = -y -2
y = -x -2
B)
0.5x + y = 1
0.5x + (-x -2) = 1
-0.5x -2 = 1
-0.5x = 3
x = -6
Answer:
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Step-by-step explanation:
If you use this equation then you say that the ground is h=0 and solve as a quadratic.
The quadratic formula is (-b±<span>√(b^2-4ac))/2a when an equation is in the form ax^2 + bx + c
So the equation you have been given would be -16t^2-15t-151 = 0
This equation has no real roots which leads me to believe it is incorrect.
This is probably where your difficulty is coming from, it's a mistake.
The equation is formed from S=ut+(1/2)at^2+(So) where (So) is the initial height and S is the height that you want to find.
In this case you want S = 0.
If the initial height is +151 and the initial velocity and acceleration are downwards (negative) and the initial velocity (u) is -15 and the initial acceleration is -32 then you get the equation S=-15t-16t^2+151
Solving this using the quadratic formula gives you t = 2.64 or t = -3.58
Obviously -3.58s can't be the answer so you're left with 2.64 seconds.
Hope this makes sense.
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Answer: g(x)=(9x+1)
Step by step explanation:
It's a box and whiskers plot, and the rightmost whisker end is the maximum. It goes A: minimum, B: 25th percentile, C: median, D: 75th percentile, E maximum.
Answer: E
