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3 years ago

It has 2 fewer sides than a heptagon

2 answers:

5 0

2 fewer sides than a heptagon.

A heptagon has 7 sides.

2 fewer than 7 is 5.

A 5 sided shape is called a Pentagon.

It is a Pentagon.

marta [7]3 years ago

4 0

A pentagon because 7-2=5

You might be interested in

Paul is filling the rectangular prism shown with cubes that measure 4 inch along each edge.

kumpel [21]

Awnser:

You can fit up to 64 cubes in the rectangular prism.

Step-by-step explanation:

Well, we know that each side measures 4 inch each.

So, 4(L) x 4(W) x 4(H) = 64.

8 0

3 years ago

24 : 18<br> In its simplest form

beks73 [17]

Answer:

4 : 3

Step-by-step explanation:

Because when you divide 24 by 6 and divide 18 by 6. It would be 4 : 3

7 0

3 years ago

Read 2 more answers

Arrange these numbers from least to greatest:

zmey [24]

First let's put them all over 100.
- $\frac{5}{8}$ would change to - $\frac{62.5}{100}$

- $\frac{7}{9}$ to - $\frac{77.7}{100}$

- $\frac{4}{5}$ to - $\frac{80}{100}$

- $\frac{3}{7}$ to - $\frac{42.8}{100}$

Then if we put them from least to greatest it would be from highest to lowest (since they are negatives)

- $\frac{4}{5}$ , - $\frac{7}{9}$ , - $\frac{5}{8}$ , - $\frac{3}{7}$

The answer would be B

8 0

4 years ago

If y varies directly with x and y = 16 when x = 4, find x when y = 8.

VladimirAG [237]

Answer: X = 2

Step-by-step explanation:

Y & X

Y = KX

K = Y/X = 16/4 = 4

Y= 8

K =4

X=?

X = Y/K

X = 8/4 = 2

7 0

4 years ago

Read 2 more answers

Pigeon Hole Principle :

yaroslaw [1]

Answer:

A proof can be as follows:

Step-by-step explanation:

Let $S=\{a_{1},a_{2},...,a_{n},a_{n+1}\}$ be a set of $n+1$ integers. By the division algorithm the possible remainders when we divide by $n$ are $0,1,2,....,n-1$ . Then, each integer $a_{i}\in S$ can be written as:

$a_{i}=np_{i}+r_{i},\,\,0\leq r_{i}$

Observe that the set of remainders $\{r_{1},r_{2},...,r_{n+1}\}$ has $n+1$ elements and each element has $n$ possible values. By the Pigenhole principle at least two remainders have the same value. Suppose that this two elements are $a_{i}, a_{j}$ . Then,

$\begin{array}{c}a_{i}=np_{i}+r\\a_{j}=np_{j}+r\end{array}$

Where $r_{i}=r_{j}=r$ . Then, $a_{i}-a_{j}=np_{i}-np_{i}=n(p_{i}-p_{j})$ . Then we have that $n$ divides $a_{i}-a_{j}$ .

5 0

4 years ago