The polynomial 3x ^ 3 - 16x ^ 2 + 31x - 20 the area of a trapezoidal desktop . Of the bases of the x ^ 3 - 5x the height of the trapezoid? Hins: Use long division , what is trapezoid represented by the expressions trapezoid

Step-by-step explanation:

Given:

Area = 3x^3 - 16x^2 + 31x - 20

Base:

x^3 - 5x

Area of trapezoid, S = 1/2 × (A + B) × h

Using long division,

(2 × (3x^3 - 16x^2 + 31x - 20))/x^3 - 5x

= (6x^3 - 32x^2 + 62x - 40))/x^3 - 5x = 6 - (32x^2 - 92x + 40)/x^3 - 5x = 2S/Bh - Ah/Bh

= 2S/Bh - A/B

= (2S/B × 1/h) - A/B

Since, x^3 - 5x = B

Comparing the above,

A = 32x^2 - 92x + 40

2S/B = 6

Therefore, h = 1

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

a rotation about point l

step-by-step explanation:

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