The greatest perimiter
let's experiment weather we want to have sides with as close to each other as possible, or with as far apart form each other
so if we have lenth times width, peimiter=legnth+legnth+widht+width
1 times 39:perimiter=80
3 times 13:perimiter=52
so we want one side as small as possible
0.1 times 360:perimiter=720.2
let's see how far we can go
0.00000000000000000001 times 3600000000000000000000=3,600,000,000,000,000,000,000.00000000000000000001
you get the idea
that is a really big perimiter
if you can be that exact
so just make a really small decimal
if you haven't learned decimals yet and only know whole numbers then
80 ft is the greatest perimiter made (with sides of 39,1,39,1)
Given that the perimeter of rhombus ABCD is 20 cm, the length of the sides will be:
length=20/4=5 cm
the ratio of the diagonals is 4:3, hence suppose the common factor on the diagonals is x such that:
AC=4x and BD=3x
using Pythagorean theorem, the length of one side of the rhombus will be:
c^2=a^2+b^2
substituting our values we get:
5²=(2x)²+(1.5x)²
25=4x²+2.25x²
25=6.25x²
x²=4
x=2
hence the length of the diagonals will be:
AC=4x=4×2=8 cm
BD=3x=3×2=6 cm
Hence the area of the rhombus wll be:
Area=1/2(AC×BD)
=1/2×8×6
=24 cm²
Answer:
(14n - 42)*pi
Step-by-step explanation:
Circumference is 2 * pi * r. Since r in this case is 7n - 21, you can multiply r by 2 and pi, giving you (14n - 42)*pi
A. Western Beach is reducing was width by 10 feet every 5 years for the first 10 years, then the pattern became less constant. Dunes Beach experienced a stable and fast increase in width, 25 feet every 5 years of 5 feet per year.
B. Somewhere between years 11 and 12 they had the same width.
C. You can place the values on a graph and connect the points, and look at the intersections to determine points in time where they were of the same width.
First, lets look at the data. Western Beach (or WB for short) decreases its width by 10 feet every 5 years from year 0 to year 10. Between year 11 and year 15 the pattern shifts and becomes less changing since there is barely any change between years 12 and 15. Dunes Beach (or DB) increases steadily by 25 feet every 5 years or 5 feet per year.
Assuming the changes in width happen over time and gradually, at some point between year 10 and 11, both beaches would have likely had the same width, somewhere between 70 and 75.
To determine the exact point in time where they meet we would need to draw a graph, with the width in feet on the X axis and the year on the Y axis. Then we place all the pairs in the graph by their coordinates, and connect the points that correspond to each beach. We then see where the lines intersect and use mathematics to determine the values of X and Y, giving us the time and width when the two beaches were equal.
please mark me brainliest it helps to become an genius
