A triangle has vertices at (2, 4), (-3, -6), and (4, -2). Find its perimeter rounded to two decimal places.

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

A triangle has vertices at (2, 4), (-3, -6), and (4, -2). Find its perimeter rounded to two decimal places.

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

I would plot the points so it is easier for me to see which three distances to calculate.
1 Attachment
|dw:1437665201466:dw|

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

|dw:1437665261302:dw| we need to find all three side measurements to do that we can use the distance formula so for example let's go ahead and find the distance between (2,4) and (4,-2) you do that by doing: \[d_1=\sqrt{(2-4)^2+(4-(-2))^2} \\ d_1=\sqrt{(-2)^2+(6)^2} \\ d_1=\sqrt{4+36} \\ d_1=\sqrt{40}\] so we have that|dw:1437665356288:dw| we also need to find the other side lengths you try the other two legs and remember: \[\text{ The distance \between } (x_1,y_1) \text{ and } (x_2,y_2) \text{ is } \\ \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\]
once you find all the side lengths you add the side lengths to find the perimeter
ok
so have you computed the distance between (2,4) and (-3,-6) ?
no
|dw:1437665675706:dw| well you still need to find these two distances (these two side lengths)
I got 10
that is you need to find the distance between (2,4) and (-3,-6) you also need to find the distance between (-3,-6) and (4,-2)
the distance between (2,4) and (-3,-6) is given by: \[d_3=\sqrt{(2-(-3))^2+(4-(-6))^2}=\sqrt{5^2+(10)^2}\] see if you can finish simplifying this
39 and 125
do you mean sqrt(125) for d_3 ?
yes
|dw:1437665932861:dw| now you just need to find d_2
which is the distance between (4,-2) and (-3,-6)
use that formula above to find it
\[\text{ The distance between } (x_1,y_1) \text{ and } (x_2,y_2) \text{ is } \\ \sqrt{(x_1-x_2)^2+(y_1-y_2)^2} \]
I got 10
hmm... that is what you got for the other one... try again... you have (4,-2) and (-3,-6) replace x1 with 4 replace x2 with -3 replace y1 with -2 replace y2 with -6 ... \[d_2=\sqrt{(4-(-3))^2+(-2-(-6))^2}\]
11
4-(-3) is the same as 4+3 -2-(-6) is the same as -2+6
\[d_2=\sqrt{(4+3)^2+(-2+6)^2}\]
compute 4+3 and -2+6 then square both results
then add those results
i got 11
sounds like you are doing (4+3)+(-2+6) which means you are ignoring the squares
(4+3)^2=(7)^2 (-2+6)^2=(4)^2
\[d_2=\sqrt{(4+3)^2+(-2+6)^2} \\ d_2=\sqrt{(7)^2+(4)^2}\]
you square the numbers before adding them because that is what order of operations tells us to do
29 65
don't know where the 29 comes from but sqrt(49+16) is sqrt(65)
okay
|dw:1437666447857:dw| now add all the sides to find the perimeter
239
230
so you added sqrt(125) and sqrt(40) and sqrt(65)?
or did you add 125 and 40 and 65?
because those aren't the same numbers
i added them all
sqrt(125) is not the same as 125 sqrt(40) is not the same as 40 sqrt(65) is not the same as 65 so adding sqrt(125) and sqrt(40) and sqrt(65) is not equivalent to adding 125 and 40 and 65.
\[\sqrt{125}+\sqrt{40}+\sqrt{65}\] just enter this into the calculator to get the approximated perimeter
25.567
ok and you are suppose to round to 2 decimal places so that would actually be...
25.57
yes

Not the answer you are looking for?

Search for more explanations.

Ask your own question