Let width = w
Let length = l
Let area = A
3w+2l=1200
2l=1200-3w
l=1200-3/2
A=w*l
A=w*(1200-3w)/2
A=600w-(3/2)*w^2
If I set A=0 to find the roots, the maximum will be at wmax=-b/2a which is exactly 1/2 way between the roots-(3/2)*w^2+600w=0
-b=-600
2a=-3
-b/2a=-600/-3
-600/-3=200
w=200
And, since 3w+2l=1200
3*200+2l=1200
2l = 600
l = 300
The dimensions of the largest enclosure willbe when width = 200 ft and length = 300 ft
check answer:
3w+2l=1200
3*200+2*300=1200
600+600=1200
1200=1200
and A=w*l
A=200*300
A=60000 ft2
To see if this is max area change w and l slightly but still make 3w+2l=1200 true, like
w=200.1
l=299.85
A=299.85*200.1
A=59999.985
Answer:
12x^2+9x^2-25 (quadratic equation)
a=12, b=9, c=-25
put this in quadratic formula
Step-by-step explanation:
Cobalt has an atomic number (Z) of 27, which means the nuclei of all its isotopes have 27 protons. Cobalt 60 has an atomic mass of 60, so it has 60-27 = 33 neutrons.
The mass of 27 isolated protons plus the mass of 33 isolated neutrons would be:
27*(1.007825 u) + 33*(1.008665 u) = 60.497220 u
The actual mass of the nucleus of 60-Co is 59.933820 u.
Mass defect: 60.497220 u - 59.933820 u = 0.563400 u
The mass defect is equal to the binding energy of a nucleus.
using the fact that 1 u = 931.5 MeV/c^2
(0.563400 u)*(931.5 MeV/u) = 524.807 MeV
Answer:
1/12
Step-by-step explanation:
Given that
30/360 to lowest fraction equivalent
Now, we can write 30 as 3×10
Also, we can write 360 as 36×10
Then, we have
(3×10)/(36×10)
Then, 10 cancel 10, we are left with
3/36
Also we can write 36 as 12×3
Then, we have
3/(12×3)
Also, 3 cancel 3, we are left with
1/12
Then the lowest fraction is 1 / 12
1/12
The answer is D
To solve the answer you would have needed to multiply the reciprocal of 
but in this case, Unvi multiplied with the reciprocal of 14.
