Well it depends on what the model is but if it's an IRA or whatever so if you make your own model that's easy
14 + {-2 + 3[1 + 3(-6-2)]}
14 + { -2 + 3[1 + 3(-8)]}
14 + { -2 + 3[1 - 24]}
14 + { -2 + 3[-23]}
14 + {-2 - 69}
14 + {-71}
14 - 71
- 57 <==
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Step-by-step explanation:
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Y = -2(x-5)^2 + 4
Note: vertex form is 0, f(x), y, or nothing= a(x-h)^2 + k
The opposite of the x value of the vertex is h
The y value value of the vertex is k.
Find a. Plug in (6,-10)
-10 = a(6-5)^2 + 4
-10 = 1a + 4
-10 = 5a
a = -2
Answer:
Step-by-step explanation:
Let the solution to
2x^2 + x -1 =0
x^2+ (1/2)x -(1/2)
are a and b
Hence a + b = -(1/2) ( minus the coefficient of x )
ab = -1/2 (the constant)
A. We want to have an equation where the roots are a +5 and b+5.
Therefore the sum of the roots is (a+5) + (b+5) = a+ b +10 =(-1/2) + 10 =19/2.
The product is (a+5)(b+5) =ab + 5(a+b) + 25 = (-1/2) + 5(-1/2) + 25 = 22.
So the equation is
x^2-(19/2)x + 22 =0
2x^2-19x + 44 =0
B. We want the roots to be 3a and 3b.
Hence (3a) + (3b) = 3(a+b) = 3(-1/2) =-3/2 and
(3a)(3b) = 9(ab) =9(-1/2)=-9/2.
So the equation is
x^2 +(3/2) x -9/2 = 0
2x^2 + 3x -9 =0.
