The standard form of a parabola is y=ax²+bx+c
use the three given points to find the three unknown constants a, b, and c:
-2=a+b+c............1
-2=4a+2b+c......... 2
-4=9a+3b+c...........3
equation 2 minus equation 1: 3a+b=0..........4
equation 3 minus equation 2: 5a+b=-2.........5
equation 5 minus equation 4: 2a=-2, so a=-1
plug a=-1 in equation 4: -3+b=0, so b=3
Plug a=-1, b=3 in equation 1: -2=-1+3+c, so c=-4
the parabola is y=-x²+3x-4
double check: when x=1, y=-1+3-4=-2
when x=2, y=-4+6-4=-2
when x=3, y=-9+9-4=-4
Yes.
RS => y - 5 = (8 - 5)/(1 - (-1)) (x - (-1))
y - 5 = 3/2 (x + 1) => slope = 3/2
ST => y - 8 = (-2 - 8)/(7 - 1) (x - 1)
y - 8 = -10/6 (x - 1) = -5/3 (x - 1) => slope = -5/3
TU => y - (-2) = (0 - (-2))/(2 - 7) (x - 7)
y + 2 = 2/5(x - 7) => slope = 2/5
UR => y = 5/(-1 - 2) (x - 2)
y = -5/3 (x - 2) => slope = -5/3
The median is the line joining the midpoints of the non-parallel sides.
Midpoint of RS = ((-1 + 1)/2, (5 + 8)/2) = (0, 13/2)
Midpoint of TU = ((7 + 2)/2, -2/2) = (9/2, -1)
Equation of the line joining (0, 13/2) and (9/2, -1) is given by y - 13/2 = (-1 - 13/2)/(9/2) x
y - 13/2 = (-15/2)/(9/2) x
y - 13/2 = -15/9x
18y - 117 = -30x
30x + 18y = 117
Answer:
Min
Step-by-step explanation:
To check if it has a max or min, you must reference the a value. The a value in this case is 2, and that is positive. Positive parabolas point up, so they can only have a minimum point.
Answer:
Step-by-step explanation:
If solving for x it's 0
Let X be the random variable denoting the number of successful throws.
Here X~ Binomial Distribution with n = 5 and p = 0.80.
the probability of her missing 3 (or more) free throws out of 5
= P ( X ≤ 2)
= P (X= 0) + P(X= 1) + P(X= 2)
=0.00032 + 0.0064 + 0.0512
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= 0.05792
I hope my answer has come to your help. God bless and have a nice day ahead!
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