Iam so sorry! i am terrible at math!
suppose the length and the width of the sandbox are doubled. (look at the picture)
find the percent change in the perimeter
find the percent of change in the area

I have been learning this at school just add 6+6 and 10+10

Percent change=change/original times 100

perimiter
if length and width double then the perimiter doubles
because
P=2(L+W)
L=10
W=6
old one
P=2(10+6)
P=2(16)
P=32
new one
20 and 12
P=2(20+12)
P=2(32)
P=64

percent change=change/original times 100
change=64-32=32
original=32
percent change=32/32 times 100=100%

area
old are=legnth tiems wdith =6 times 10=60
nwe one
area=12 times 20=480
change is 240-60=180
percent change=180/60 times 100=3 times 100=300

perimiter is 100% change
area is 300% change

Answer: d. 39 cmin the diagram, the approximate length of the minor arc de is 39 centimetres.step-by-step explanation: the formula for the length of an arc in a circle is as displayed below.let the length of the minor arc de = xx = 2( pi )r × angle of sector / 360°from this, we will substitute the values given from the problem into the formula to obtain the length of the minor arc de.x = 2( pi )r × angle of sector / 360°r = radius = 25cmpi as an approximate value = 3.14angle of sector = 120°x = 2( 3.14 )( 25 ) × 120° / 360°x = 157 × 1 / 4x = 39.25x = 39 ( rounded to the nearest whole number as an approximate value )