Westonci.ca is the best place to get answers to your questions, provided by a community of experienced and knowledgeable experts. Discover a wealth of knowledge from professionals across various disciplines on our user-friendly Q&A platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
Sagot :
The equation for for perimeter of a rectangle is
2w + 2L = 154 meters
and we know that
w = .4L
so all we have to do is plug in what we know
2(.4L) + 2L = 154
.8L + 2L = 154
2.8L = 154
L = 55
now we know L we can solve for w with our first equation
2w + 2(55) = 154
2w = 44
w = 22
so the length is 55 and our width is 22.
Let check
55 *.4 = 22 so we are correct.
2w + 2L = 154 meters
and we know that
w = .4L
so all we have to do is plug in what we know
2(.4L) + 2L = 154
.8L + 2L = 154
2.8L = 154
L = 55
now we know L we can solve for w with our first equation
2w + 2(55) = 154
2w = 44
w = 22
so the length is 55 and our width is 22.
Let check
55 *.4 = 22 so we are correct.
Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.