X=length
y=width
Area (rectangle)=length x width
we suggest this system of equations:
y=x/5
xy=9/20
solve this system of equations by substitution method:
x(x/5)=9/20
x²/5=9/20
Least common multiple=20
4x²=9
x²=9/4
x=⁺₋√(9/4)
we have two solutions:
x₁=-3/2 it does not validate
x₂=3/2 ⇒y=x/5=(3/2)/5=3/10
The dimensions of the lake are:
lengh=3/2 miles
width=3/10 miles
to check:
Area=3/2 miles x 3/10 miles=9/20 miles².
width is 1/5 the length of the lake ⇒ 3/10 miles=(3/2) /5 miles
What is this type of math called please and what year/grade if you don’t mind me asking so I can be prepared
hi it me and I will be there
1.
15 × 12× 20
3600
2.
2×2×3
12
3.
24.3×8.5×9.7
2003.535
Answer: 2.11X=11.41
X=9.3
or
211x=1141
930
Step-by-step explanation:
