Prove that if four numbers are chosen from the set {l, 2, 3, 4.5.6}, at least one pair of the selected numbers must add up to 7

Step-by-step explanation:

GIven that:

The set is :  {l, 2, 3, 4.5.6}

To Prove that:

If four numbers are to be chosen , at least one pair of the selected numbers must add up to 7

From the data set given, this set has three pairs of number which can add up to 7. These are:

{1, 6}  i.e 1 + 6 = 7

{2,5}  i.e  2+ 5 = 7

{3,4}  i.e  3 + 4 = 7

So, if four number are chosen from the set that contains six numbers, then that the possibility for such to occur is either by selecting two pairs out of the three pairs or one pair and two numbers.

Thus, in both cases, at least one pair of the selected numbers must add up to 7.

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.