HEY CAN ANYONE HELP ME OUT IN DIS PLS

Three stamps can be attached to each other in various ways.how many ways might three stamps be attached?

ivanzaharov [21]

Let's call the stamps A, B, and C. They can each be used only once. I assume all 3 must be used in each possible arrangement.
There are two ways to solve this. We can list each possible arrangement of stamps, or we can plug in the numbers to a formula.
Let's find all possible arrangements first. We can easily start spouting out possible arrangements of the 3 stamps, but to make sure we find them all, let's go in alphabetical order. First, let's look at the arrangements that start with A:
ABC
ACB
There are no other ways to arrange 3 stamps with the first stamp being A. Let's look at the ways to arrange them starting with B:
BAC
BCA
Try finding the arrangements that start with C:
C_ _
C_ _
Or we can try a little formula; y×(y-1)×(y-2)×(y-3)...until the (y-x) = 1 where y=the number of items.
In this case there are 3 stamps, so y=3, and the formula looks like this: 3×(3-1)×(3-2).
Confused? Let me explain why it works.
There are 3 possibilities for the first stamp: A, B, or C.
There are 2 possibilities for the second space: The two stamps that are not in the first space.
There is 1 possibility for the third space: the stamp not used in the first or second space.
So the number of possibilities, in this case, is 3×2×1.
We can see that the number of ways that 3 stamps can be attached is the same regardless of method used.

8 0

3 years ago

Which of the following expressions is not equivalent to 12x + 18y?

Fed [463]

Answer:

3(4x + 18y)

..................

8 0

2 years ago

You have 100 yards of fencing available to create an enclosure for farm animals. As the

Sunny_sXe [5.5K]

1. area of a pentagon when 1 side is x is $\frac{x^2}{4}\sqrt{5(5+2\sqrt{5})}$
so if perimiter is 100 then one side is 100/5 or 20
area will be $\frac{20^2}{4}\sqrt{5(5+2\sqrt{5})}$ ≈<span>688.2 square yards

2. area of a hexagon when 1 side is length x is $\frac{3x^2\sqrt{3}}{2}$
so if permiter is 100 then length of one side is 100/6 or about 16.666666666666
the area will be $\frac{3(16.6666666)^2\sqrt{3}}{2}$ ≈721.7 square yards

3. area of a regular octagon with side lenghts x is $2(1+\sqrt{2})x^2$
so if 8 sides then side length of each is 100/8 or 12.5
area will be $2(1+\sqrt{2})12.5^2$ ≈754.4 square yards

octagon wil have most space
you can tell because the more sides it has, the more it gets to a circle and the more area it encloses with a given perimiter
</span>

3 0

3 years ago

A semi- circle has an area of 35cm . What is its diameter?

patriot [66]

Answer:

<h2>Area Semicircle = 1/2 πr² </h2>

where

r radius of semicircle

→ ½ * 22/7 * r² = 35

→ 11/7 * r² = 35

→ r² = 35*7/11

→ r² = 245/11

→ r² = 22.77

→ r = √22.77

→ r = 4.77

Therefore,

Diameter = 2*4.77

= 9.54 cm

5 0

2 years ago

Y is directly proportional to square root of x<br> If y=56 when x=49 find,<br> y when x=81

STALIN [3.7K]

$\qquad \qquad \textit{direct proportional variation} \\\\ \textit{\underline{y} varies directly with \underline{x}}\qquad \qquad \stackrel{\textit{constant of variation}}{y=\stackrel{\downarrow }{k}x~\hfill } \\\\ \textit{\underline{x} varies directly with }\underline{z^5}\qquad \qquad \stackrel{\textit{constant of variation}}{x=\stackrel{\downarrow }{k}z^5~\hfill } \\\\[-0.35em] \rule{34em}{0.25pt}$

$\stackrel{\begin{array}{llll} \textit{"y" directly}\\ \textit{proportional to }\sqrt{x} \end{array}}{y = k\sqrt{x}}\qquad \textit{we know that} \begin{cases} y = 56\\ x = 49 \end{cases}\implies 56=k\sqrt{49} \\\\\\ 56=7k\implies \cfrac{56}{7}=k\implies 8=k~\hfill \boxed{y=8\sqrt{x}} \\\\\\ \textit{when x = 81, what is "y"?}\hfill y=8\sqrt{81}\implies y=8(9)\implies y=72$

7 0

2 years ago