You should expect 28 defected cars.
We write the equation in the form of directional.
y -1 = 6x ⇔ y = 6x + 1
y - 1 = 3x ⇔ y = 3x + 1
y - 7 = 2x - 6 ⇔ y = 2x - 6 + 7
y = 2x + 1
y - 7 = x - 2 ⇔ y = x - 2 + 7
y = x + 5
Equations cleverly arranged .
Point Q = (0,1)
b factor , not only fits the last equation
In the drawing have engraved points Q and R are tangent linear function appropriate to that point . This graphics solution . y = 3x + 1
Answer b
We check choice by the system of equations , where substitute wartoćsi points Q and R to the model equations linear function
The result of equations confirmed our choice Answer b
That is your answer if it is wrong i will try to fix it
I believe the term is "constant".
Answer:
y = -4x² + 32x - 48
Step-by-step explanation:
The standard form of a quadratic equation is
y = ax² + bx + c
We must find the equation that passes through the points:
(2, 0), (6,0), and (3, 12)
We can substitute these values and get three equations in three unknowns.
0 = a(2²) + b(2) + c
0 = a(6²) + b(6) + c
12 = a(3²) + b(3) + c
We can simplify these to get the system of equations:
(1) 0 = 4a + 2b + c
(2) 0 = 36a + 6b + c
(3) 12 = 9a + 3b + c
Eliminate c from equations (1) and (2). Subtract (1) from (2).
(4) 0 = 32a + 4b
Eliminate c from equations (2) and (3). Subtract (3) from (2).
(5) -12 = 27a - 3b
Simplify equations (4) and (5).
(6) 0 = 8a + b
(7) -4 = 9a - b
Eliminate b by adding equations (6) and (7).
(8) a = -4
Substitute (4) into (6).
0 = -32 + b
(9) b = 32
Substitute a and b into (1)
0 = 4(-4) + 2(32) + c
0 = -16 + 64 + c
0 = 48 + c
c = -48
The coefficients are
a= -4, b = 32, c = -48
The quadratic equation is
y = -4x² + 32x - 48
The diagram below shows the graph of your quadratic equation and the three points through which it passes.