Eighth grade
> t3 ara and perimeter, word problems 6my
a rectangular sheet of metal has an area of 247 square inches. its perimeter is 64 inches. what are
the dimensions of the sheet?

The length and width of the rectangle are 19 inches and 13 inches respectively.

Solution:

Given that , A rectangular sheet of metal has an area of 247 square inches. Its perimeter is 64 inches.

Let the length of rectangle be a and width of rectangle be b.  

The area of rectangle is given as:

The perimeter of rectangle is given as:

Perimeter = 2(length + width)

Then 2(a + b) = 64

a + b = 32 ⇒ (1)

a – b = 6 ---- (2)

Now solve (1) and (2)

2a = 38

a = 19

Then, 19 + b = 32  

b = 32 – 19 = 13  

hence, the length and width of the rectangle are 19 inches and 13 inches respectively.

.632

step-by-step explanation:

6 * 1/10 = 3/5

3*0.01 = 0.03

3/5 + 0.03 = .63

2*0.002 = .002

.63 + 0.002 = .632

answer: ratio of volume of cone (with 3 times the height and equal diameter) to volume of cylinder is 2 to 1. the value of π is not used.

step-by-step explanation:

radius is diameter/2 = 4.

area of circle is π r^2.

volume of cylinder is height × area of base = hπr^2 = 3(4^2) π = 48π cubic inches.

not knowing the formula for volume of cone, i calculate it using calculus (which is for calculating things).

volume of cone is integral 0 to h area(y) dy. area(y) is π r(y)^2. r(y) runs r to zero as y runs zero to h, r(0) = r, r(h) = 0.

r(y) = r - r/h y

area(y) = π (r - r/h y)^2 = π (r^2 - 2r^2y/h + r^2y^2/h^2) = π r^2(1-2y/h+y^2/h^2)

antiderivative area(y) = π r^2 (y-y^2/h+y^3/(3h^2))

integral 0 to h = π r^2(h-h^2/h+h^3/(3h^2) - (0)) = π r^2(h - h + h/3) = π r^2(h/3) = π/3 r^2 h

volume of this cone is π/3(4^2)18 cubic inches.

= 6 16 π = 96π cubic inches.

cone has twice the volume, 96/48.

cone height is 6 times the cylinder.

ratio of volume formula: π r^2 cancels.

cone over cylinder ratio is 6h/3 to h, 2 to 1.