Answer: (x2 – 1) – 5(x – 1)
V = (pi)r² h
v = (pi)5² h
v = 25(pi) h
take derivative with respect to h
dv/dh = 25(pi)
(dv/dt)*(dt/dh) = dv/dh
dv/dt is given to be 5 cm³
solved for dv/dh = 25(pi)
5(dt/dh) = 25(pi)
dt/dh = 25(pi)/5
dt/dh = 5(pi)
then the recipricol of dt/dh; we want to find dh/dt = 1/(5(pi))
Gradient of line,m=(17-6)/(20-0)=11/20
finding equation of line using a point (20,17)
y-17=11/20(x-20)
y-17=(11/20x)-11
y=11/20x+(17-11)
y=(11/20)*x+6
Hi there!

The two trapezoids are similar, so we can determine a common scale factor:
OL/UR = NM/TS
9/3 = 6/2
3 = 3
Trapezoid ONML is 3x larger than UTSR, so:
RS = 4, LM = y
3RS = LM
3 · 4 = y = 12.
Find x using the same method:
3UT = ON
3(2x+1) = 4x + 9
6x + 3 = 4x + 9
2x = 6
x = 3.