A quadratic pattern has a second term equal to 1,a third term equal to -16 and a fifth term equal to -14 1.1.1.calculate the second difference of this quadratic pattern

1.1.2.hence or otherwise, calculate the first term of term

1.1.1. The common difference is 12

1.1.2. The first term is 30

Step-by-step explanation:

1.1.1. The given quadratic pattern is presented as follows;

The 2nd term = 1

The 3rd term = -16

The 5th term = -14

Let 'd' represent the second difference, we have;

1 + 2d + a =  -16

2·d + a = -16 - 1 = -17

2·d + a = -17...(1)

-16 + 3d + a + 4d + a = -14

7·d + 2·a = -14 + 16 = 2

7·d + 2·a = 2...(2)

Making 'a' the subject of both equation (1) and (2) gives;

a = -17 - 2·d

a = 1 - 7/2·d

∴ -17 - 2·d = 1 - 7/2·d

(7/2)·d - 2·d = 1 + 17

(3/2)·d = 18

d = (2/3) × 18 = 12

The common difference, d = 12

a = -17 - 2·d = -17 - 2 × 12 = -41

1.1.2. Let 'x' represent the first term, we have;

x + a + d = 1

x = 1 - (a + d)

x = 1 - (-41 + 12) = 30

The first term, x = 30

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204

it is 570

step-by-step explanation: