there is a sandbox with a width of 6ft and a length of 10ft, suppose the length and the width are doubled.
a. find the percent of change in the perimeter
b. find the percent of change in the area
i think the answer is 66.67% for both of them
14 answers
just realized i put my name in the subject and the subject in my name, oops!
don't understand how you came up with that answer.
old sandbox : 6 by 10
so perimeter = 32 ft
area = 60 ft^2
new sandbox : 12 by 20
new perimeter = 64 --> it doubled, so a 100% increase, the multiplication factor is 2
new area = 240 ft^2 --> it quadrupled
so the increase is 300% , the multiplication factor was 4
in general, since perimeter is a linear measurement, the perimeter would be the same multiplication factor
and the area would be the (multiplication factor)^2
e.g. If we had tripled the dimensions, the area would be 9 times as large
old sandbox : 6 by 10
so perimeter = 32 ft
area = 60 ft^2
new sandbox : 12 by 20
new perimeter = 64 --> it doubled, so a 100% increase, the multiplication factor is 2
new area = 240 ft^2 --> it quadrupled
so the increase is 300% , the multiplication factor was 4
in general, since perimeter is a linear measurement, the perimeter would be the same multiplication factor
and the area would be the (multiplication factor)^2
e.g. If we had tripled the dimensions, the area would be 9 times as large
What about the perimeter?
Oh wait I see it now sorry
ummm no... that didnt help
not correct
Imagine giving the wrong answer to everyone who actually needs help. Couldn't be ruby
Didnt work
didnt work bout to post this on my insta and no one is ever going to use this website again
Fdfdbtfhtrhtryhtyhthytyht
Its actually correct dummys
Lol at least Reiny actually gave the correct answer oh my flipping god
Whats the proportion
nobody is helping
2 answers