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3 years ago

The end points of one side of a regular pentagon are (-1,4) and (2,3). What is the perimeter of the pentagon?

Mathematics

1 answer:

marissa [1.9K]3 years ago

8 0

To determine the perimeter of the pentagon, you must first calculate a side length of it. Let's name the coordinates A(-1,4) and B(2,3).

To figure out how far the points are from each other, you have to use the distance formula:
$D_{AB} = \sqrt{( x_{2}-x_{1})^2+{(y_{2}-y_{1})^2}
$
$x_{1} =1, x_{2} =-2, y_{1} =2, y_{2} =3$
$D_{AB}= \sqrt{(2--1)^2+{(3-4)^2}$
D_{AB}= \sqrt{(2--1)^2+{(3-4)^2}
$D_{AB}= \sqrt{(2+1)^2+{(3-4)^2}$
$D_{AB}= \sqrt{(3)^2+(-1)^2}$
$D_{AB}= \sqrt{9+1}$
$D_{AB}= \sqrt{10}$

Now, the formula for the perimeter of a pentagon is
P = 5×side length

So...
Perimeter = 5× $\sqrt{10}$

The answer is (2)

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Examine the following expression.

svetoff [14.1K]

Answer:

True expressions:

The constants, -3 and -8, are like terms.

The terms 3 p and p are like terms.

The terms in the expression are p squared, negative 3, 3 p, negative 8, p, p cubed.

The expression contains six terms.

Like terms have the same variables raised to the same powers.

Step-by-step explanation:

The expression is:

p² - 3 + 3p - 8 + p + p³

False expressions:

The terms p squared, 3 p, p, and p cubed have variables, so they are like terms. (They don't have the same exponents)

The terms p squared and p cubed are like terms. (They don't have the same exponents)

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3 years ago

Read 2 more answers

The slope is -3/2, the y intercept is 1. Write an equation for the linear function, f(x)

Delicious77 [7]

Answer:

f(x)=-3/2x+1

Step-by-step explanation:

y=mx+b

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b=y intercept

y=f(x)

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3 years ago

A bag contains two six-sided dice: one red, one green. The red die has faces numbered 1, 2, 3, 4, 5, and 6. The green die has fa

gayaneshka [121]

Answer:

the probability the die chosen was green is 0.9

Step-by-step explanation:

Given that:

A bag contains two six-sided dice: one red, one green.

The red die has faces numbered 1, 2, 3, 4, 5, and 6.

The green die has faces numbered 1, 2, 3, 4, 4, and 4.

From above, the probability of obtaining 4 in a single throw of a fair die is:

P (4 | red dice) = $\dfrac{1}{6}$

P (4 | green dice) = $\dfrac{3}{6}$ = $\dfrac{1}{2}$

A die is selected at random and rolled four times.

As the die is selected randomly; the probability of the first die must be equal to the probability of the second die = $\dfrac{1}{2}$

The probability of two 1's and two 4's in the first dice can be calculated as:

= $\begin {pmatrix} \left \begin{array}{c}4\\2\\ \end{array} \right \end {pmatrix} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^4$

= $\dfrac{4!}{2!(4-2)!} ( \dfrac{1}{6})^4$

= $\dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^4$

= $6 \times ( \dfrac{1}{6})^4$

= $(\dfrac{1}{6})^3$

= $\dfrac{1}{216}$

The probability of two 1's and two 4's in the second dice can be calculated as:

= $\begin {pmatrix} \left \begin{array}{c}4\\2\\ \end{array} \right \end {pmatrix} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^2 \times \begin {pmatrix} \dfrac{3}{6} \end {pmatrix} ^2$

= $\dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^2 \times ( \dfrac{3}{6})^2$

= $6 \times ( \dfrac{1}{6})^2 \times ( \dfrac{3}{6})^2$

= $( \dfrac{1}{6}) \times ( \dfrac{3}{6})^2$

= $\dfrac{9}{216}$

∴

The probability of two 1's and two 4's in both dies = P( two 1s and two 4s | first dice ) P( first dice ) + P( two 1s and two 4s | second dice ) P( second dice )

The probability of two 1's and two 4's in both die = $\dfrac{1}{216} \times \dfrac{1}{2} + \dfrac{9}{216} \times \dfrac{1}{2}$

The probability of two 1's and two 4's in both die = $\dfrac{1}{432} + \dfrac{1}{48}$

The probability of two 1's and two 4's in both die = $\dfrac{5}{216}$

By applying Bayes Theorem; the probability that the die was green can be calculated as:

P(second die (green) | two 1's and two 4's ) = The probability of two 1's and two 4's | second dice)P (second die) ÷ P(two 1's and two 4's in both die)

P(second die (green) | two 1's and two 4's ) = $\dfrac{\dfrac{1}{2} \times \dfrac{9}{216}}{\dfrac{5}{216}}$

P(second die (green) | two 1's and two 4's ) = $\dfrac{0.5 \times 0.04166666667}{0.02314814815}$

P(second die (green) | two 1's and two 4's ) = 0.9

Thus; the probability the die chosen was green is 0.9

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3 years ago

Travis has a budget of $300 that he can spend on perennial flowers and at least 6 annual flowers. Perennial flowers are 18 dolla

boyakko [2]

Answer:

18x + 15y ≤ 300
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Step-by-step explanation:

Variables are defined in the problem statement.

18x +15y ≤ 300 . . . . total budget

y ≥ 6 . . . . . . . . . . . . minimum number of annuals

x ≥ 0 . . . . . number of perennials cannot be negative

This system of inequalities describes the situation.

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3 years ago

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Answer:

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