Answer:
<h2> 40in^3</h2>
Step-by-step explanation:
Step one:
given data
dimension of the box
lengh= 5 in
width= 2 in
height =4 in
Required
the volume of the box
the expression for the volume is
V=L*W*H
V=5*2*4
V=40 in^3
Hence the volume of sand the box can take is 40in^3
X=width of a coccer pitch
2x-20=length of a soccer pitch
Area (rectangule)=length x width
We suggest this equation:
x(2x-20)=6000
2x²-20x=6000
2x²-20x-6000=0
x²-10x-3000=0
We solve this quadratic equation:
x=[10⁺₋√(100-4*1*-3000)]/2=[10⁺₋√(100+12000)]/2=
=(10⁺₋110)/2
we have two solutions:
x₁=(10-110)/2=-50, invalid solution.
x₂=(10+110)/2=60
x=60
2x-20=2(60)-20=120-20=100
Solution: the length is 100 m, and the width is 60 m.
To check:
Area=100 m*60 m=6000 m²
The twice of width is =2(60 m)=120 m,
20 m less than twice its width is: 120 m-20 m=100 m=the length.
Answer:
f(3) = 9
Step-by-step explanation:
Wherever you see an x on the right, put a 3 in for that x
f(x) = 2x^2 + x - 12
f(3) = 2*(3)^2 + 3 - 12
f(3) = 2*9 + 3 - 12
f(3) = 18 + 3 - 12
f(3) = 9
1 mile = 5,280 feet
Car + space = 13.5 + 3 = 16.5 feet
5,280 / 16.5 = 320
Answer: 320 cars
