Question

50 points + brainliest

Answer 1

Answer:

17.9 is your answer

Hope this helps

Step-by-step explanation:

Answer 2

Solving this problem involves repeated application of the distance formula. In order to figure out which vertices we need to connect to another vertex, we should first plot the points on the coordinate plane to get an idea of what the polygon looks like. To form the sides of this polygon (which is, in our case, a pentagon), we'll need to connect the points in the following pairs:

(-2, -2) and (3, -3)
(3, -3) and (4, -6)
(4, -6) and (1, -6)
(1, -6) and (-2, -4)
(-2, -4) and (-2, -2)

In case you forgot, the distance formula is simply an application of the Pythagorean Theorem that treats the x-distance and y-distance between two points as the "legs" of a right triangle, and the shortest distance between them as the "hypotenuse."

If a and b are the legs of a right triangle, and c is the hypotenuse, the Pythagorean Theorem can be written as:

$a^2+b^2=c^2$

Or, if we're just looking for the value of c:

$c=\sqrt{a^2+b^2}$

Since the hypotenuse in our case represents distance, it's more descriptive to rename that variable d. Also, the "legs" a and b in this problem represent the distances between the x and y components of the two points. If we take any two points $(x_1,y_1)$ and $(x_2,y_2)$, the distance between the x components of those points would be their difference, $x_2-x_1$, and the distance between the y components would be $y_2-y_1$. Substituting that all in, the distance formula becomes:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2$

All that's left to do now is substitute our specific points into the formula for each side of the polygon:

(-2, -2) and (3, -3):
$d=\sqrt{(3-(-2))^2+(-3-(-2))^2}\\ d=\sqrt{(3+2)^2+(-3+2)^2}\\ d=\sqrt{5^2+(-1)^2}\\ d=\sqrt{25+1}\\ d=\sqrt{26}\\ d\approx5.1$

(3, -3) and (4, -6)
$d=\sqrt{(4-3)^2+(-6-(-3))^2}\\ d=\sqrt{1^2+(-6+3)^2}\\ d=\sqrt{1+(-3)^2}\\d=\sqrt{1+9}\\ d=\sqrt{10}\\ d\approx3.2$

(4, -6) and (1, -6)
$d= \sqrt{(1-4)^2+(-6-(-6))^2} \\ d= \sqrt{(-3)^2+0^2} \\ d= \sqrt{9} \\ d=3$

(1, -6) and (2, -4)
$d= \sqrt{(2-1)^2+(-4-(-6))^2}\\ d= \sqrt{1^2+(-4+6)^2}\\ d= \sqrt{1+2^2}\\ d= \sqrt{1+4} \\ d= \sqrt{5}\\ d\approx2.2$

(2, -4) and (-2, -2)
$d= \sqrt{(-2-2)^2+(-2-(-4))^2}\\ d= \sqrt{(-4)^2+(-2+4)^2} \\ d= \sqrt{16+2^2}\\ d= \sqrt{16+4}\\ d= \sqrt{20} \\ d\approx4.5$

Rounding beforehand and adding up all of the distances gives us a perimeter of 18 units, which is remarkably close to the more precise approximation of 17.96 units. Given your options, 17.9 units would be the closest to the result we obtained here.