Across-section of an airplane wing is shown. measurements of the thickness of the wing, in centimeters, at 25-centimeter intervals are 6.1, 19.8, 27.2, 29.0, 27.5, 26.9, 23.7, 20.4, 16.3, 9.1, and 2.9. use the midpoint rule with n = 5 to estimate the area of the wing's cross-section if a = 250. (assume the thickness of the edges is nonzero.)

The thicknesses in centimeters at the midpoints of 5 equal-width intervals are 19.8, 29.0, 26.9, 20.4, and 9.1. The width of each of the 5 intervals is 50 cm.

An estimate of the area of the cross-section using the midpoint rule.

According to the midpoint rule, the area estimate is the sum of the midpoint values multiplied by the width of the interval. That is ...

... Area ≈ (50 cm)×((19.8 +29.0 +26.9 +20.4 +9.1) cm) = 5,260 cm²

a

step-by-step explanation:

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