Answer:
a = 2, b = -9, c = 3
Step-by-step explanation:
Replacing x, y values of the points in the equation y = a*x^2 + b*x +c give the following:
(-1,14)
14 = a*(-1)^2 + b*(-1) + c
(2,-7)
-7 = a*2^2 + b*2 + c
(5, 8)
8 = a*5^2 + b*5 + c
Rearranging:
a - b + c = 14
4*a + 2*b + c = -7
25*a + 5*b + c = 8
This is a linear system of equations with 3 equations and 3 unknows. In matrix notation the system is A*x = b whith:
A =
1 -1 1
4 2 1
25 5 1
x =
a
b
c
b =
14
-7
8
Solving A*x = b gives x = Inv(A)*b, where Inv(A) is the inverse matrix of A. From calculation software (I used Excel) you get:
inv(A) =
0.055555556 -0.111111111 0.055555556
-0.388888889 0.444444444 -0.055555556
0.555555556 0.555555556 -0.111111111
inv(A)*b
2
-9
3
So, a = 2, b = -9, c = 3
Answer:
total change in time when you used 5 clues is 10.
Step-by-step explanation:
if you have 3 free clues, then 5-3 =2
That means you used 2 additional clues. 2 clues means 5*2 min reduced which is 10 mins reduced. The total time you have is x and the total time left is x-10 after you used the two clues.
Does this help?
"A. 4x^4 is the answer so try that."
The commutative property of addition means we can add two integers in any order. So yes, It would still apply to two negative integers (for example, -2 + -3 and -3 + -2 both equal five)
