DB/dt = k*B
dB/(kB) = dt
Integrating both sides:
1/k*ln|B| = t + C
B = C*e^(kt)
At time t = 0, B = 4000 and at time t = 3 (3 hours past 9 AM), B = 4500
4000 = C
4500 = 4000*e^(3k) => k = 1/3*ln(9/8)
At time t = 15 (Midnight), B = 4000*e^(1/3*ln(9/8)*15) = 4000*(9/8)^5 = 7208 bacteria
Step-by-step explanation:
solution
let,A(5,-1)
given
y=12x-1
or, y = 12 × x + (-1)
comparing it with y=mx+c
we get
x=1
y = c = -1
now
A(5,-1) and B(1,-1)
equation of line is,
(y-y1)/(x-x1) = (y2-y1)/(x2-x1)
or, (y+1)/(x-5) = (-1+1)/(1-5)
or, (y+1)/(x-5) = 0/-4
cross multiply
-4y-4 = 0
or, 0=4y+4
or, 4y+4=0 is the required eqn
I think this is the process and answer too.
I hope this helps you
Answer/Step-by-step explanation:
1. The figure is composed of a triangle and a rectangle.
Area of the triangle = ½*base*height
base = 4 ft
height = 12 - 8 = 4ft
Area of triangle = ½*4*4 = 8 ft²
Area of rectangle = length * width
Length = 8 ft
Width = 4 ft
Area of rectangle = 8*4 = 32 ft²
✔️Area of the figure = 8 + 32 = 40 ft²
2. The figure is composed of a semicircle and a triangle
Area of the semicircle = ½(πr²)
radius (r) = 3 cm
π = 3
Area = ½(3*3²) = 13.5 cm²
Area of triangle = ½*base*height
base = 3*2 = 6 cm
height = 6 cm
Area = ½*6*6 = 6 cm²
✔️Area of the figure = 13.5 + 6 = 19.5 cm²
