Of all rectangles with a fixed perimeter of p, which one has the maximum area? (give the dimensions in terms of p.)

1 answer:

8 0

<span>For all rectangles with a fixed perimeter of p the rectangle with the maximum area is one where the length is p/4 and the width is p/4, in other words a square. The area of such a rectangle would be (p*p) /16.</span>

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(a) Estimate the area under the graph of f(x) = 2 cos(x) from x = 0 to x = π/2 using four approximating rectangles and right end

nirvana33 [79]

<span>first, we are going to define variables as the following:

a = 0
a = π/2
n = 4 rectangles

Δx = [ b - a ] / n ------>Δx = [ π/2 - 0 ] / 4 = π/8

right endpoints :
sum( seq( 4 cos(x) * π/8 , x , 0+π/8 , π/2 , π/8 ) ) = 3.163065 underestimate

left endpoints:
sum( seq( 4 cos(x) * π/8 , x , 0 , π/2-(π/8) , π/8 ) ) = 4.733861 overestimate

the reason because the actual estimate by integral as the following:
π/2
∫ 4cos(x) dx = 4
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8 0

3 years ago

Put together, Dulcina and Tremaine have 129 total matchbooks. Tremaine's collection has 39 fewer matchbooks in it than Dulcina's

Anit [1.1K]

Answer:

Let the Dulcina's collection be 'x'

Let the Tremaine collection be 'x-39'

x + x - 39 =129

2x = 129 +39

2x = 168

x = 168/2

x = 84

Dulcina's collection = x = 84

Tremaine's collection = x - 39 = 84 - 39 = 45

7 0

3 years ago

How will the solution of the system y > 2x + and y < 2x + change if the inequality sign on both inequalities is reversed t

Gala2k [10]

Answer:

The System of inequality is ,

1. y > 2 x + p

2. y < 2 x + p

Suppose we assign some values to p and q and draw its graph

And, then the inequality sign on both inequalities is reversed

3. y < 2 x + p

4. y > 2 x + p

And , then draw it's graph

it has been found that, the solution set of both the inequality remains same.That is there is no point or set of points , which satisfy both the system of inequality.

The system has no solution.