All categories

2 years ago

How are the side lengths of a polygon and its perimeter related? Use examples to explain

1 answer:

8 0

The perimeter of a polygon or any shape is just the sum of all of its sides. If you are asked to find the perimeter of a polygon, you would simply just add all the sides together. For example, if a square has 4 sides with the length 2m, the perimeter of that square would be 2m + 2m + 2m + 2m = 8m. Perimeter just means add all the sides.

You might be interested in

2/3 divided by 4/6 =

const2013 [10]

1 ? I checked on the calculator and it said 1

4 0

2 years ago

What is .07 times 1.22

lidiya [134]

.07×1.22 = ?
first take away the decimals in the question, 7×122 =854.
now add the decimals back to the answer, we had 4 decimal places in the questions so add 4 decimal places to 854 = 0.0854

answer= 0.0854

7 0

2 years ago

I will marke brainliest. what is If 1=3

pogonyaev

Answer:

Step-by-step explanation:

<u>Count the number of letters</u>

1 = one ⇒ 3 letters
2 = two ⇒ 3 letters
3 = three ⇒ 5 letters
4 = four ⇒ 4 letters
5 = five ⇒ 4 letters
6 = six ⇒ 3 letters

3 0

1 year ago

Read 2 more answers

Two streams flow into a reservoir. Let X and Y be two continuous random variables representing the flow of each stream with join

zlopas [31]

Answer:

c = 0.165

Step-by-step explanation:

Given:

f(x, y) = cx y(1 + y) for 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3,

f(x, y) = 0 otherwise.

Required:

The value of c

To find the value of c, we make use of the property of a joint probability distribution function which states that

$\int\limits^a_b \int\limits^a_b {f(x,y)} \, dy \, dx = 1$

where a and b represent -infinity to +infinity (in other words, the bound of the distribution)

By substituting cx y(1 + y) for f(x, y) and replacing a and b with their respective values, we have

$\int\limits^3_0 \int\limits^3_0 {cxy(1+y)} \, dy \, dx = 1$

Since c is a constant, we can bring it out of the integral sign; to give us

$c\int\limits^3_0 \int\limits^3_0 {xy(1+y)} \, dy \, dx = 1$

Open the bracket

$c\int\limits^3_0 \int\limits^3_0 {xy+xy^{2} } \, dy \, dx = 1$

Integrate with respect to y

$c\int\limits^3_0 {\frac{xy^{2}}{2} +\frac{xy^{3}}{3} } \, dx (0,3}) = 1$

Substitute 0 and 3 for y

$c\int\limits^3_0 {(\frac{x* 3^{2}}{2} +\frac{x * 3^{3}}{3} ) - (\frac{x* 0^{2}}{2} +\frac{x * 0^{3}}{3})} \, dx = 1$

$c\int\limits^3_0 {(\frac{x* 9}{2} +\frac{x * 27}{3} ) - (0 +0) \, dx = 1$

$c\int\limits^3_0 {(\frac{9x}{2} +\frac{27x}{3} ) \, dx = 1$

Add fraction

$c\int\limits^3_0 {(\frac{27x + 54x}{6}) \, dx = 1$

$c\int\limits^3_0 {\frac{81x}{6} \, dx = 1$

Rewrite;

$c\int\limits^3_0 (81x * \frac{1}{6}) \, dx = 1$

The $\frac{1}{6}$ is a constant, so it can be removed from the integral sign to give

$c * \frac{1}{6}\int\limits^3_0 (81x ) \, dx = 1$

$\frac{c}{6}\int\limits^3_0 (81x ) \, dx = 1$

Integrate with respect to x

$\frac{c}{6} * \frac{81x^{2}}{2} (0,3) = 1$

Substitute 0 and 3 for x

$\frac{c}{6} * \frac{81 * 3^{2} - 81 * 0^{2}}{2} = 1$

$\frac{c}{6} * \frac{81 * 9 - 0}{2} = 1$

$\frac{c}{6} * \frac{729}{2} = 1$

$\frac{729c}{12} = 1$

Multiply both sides by $\frac{12}{729}$

$c = \frac{12}{729}$

$c = 0.0165 (Approximately)$

8 0

3 years ago

Expand and simplify x(x + 1)(x + 2)

zimovet [89]

Answer:

$x^3$ + 3 $x^2$ + 2x

Step-by-step explanation:

x(x^2 + x + 2x + 2)

x ( x^2 + 3x + 2)

$x^3$ + 3 $x^2$ + 2x

6 0

2 years ago