Ya, calculus and related rates, such fun!
everything is changing with respect to t
altitude rate will be dh/dt and that is 1cm/min
dh/dt=1cm/min
area will be da/dt which is increasing at 2cm²/min
da/dt=2cm²/min
base=db/dt
alright
area=1/2bh
take dervitivie of both sides
da/dt=1/2((db/dt)(h)+(dh/dt)(b))
solve for db/dt
distribute
da/dt=1/2(db/dt)(h)+1/2(dh/dt)(b)
move
da/dt-1/2(dh/dt)(b)=1/2(db/dt)(h)
times 2 both sides
2da/dt-(dh/dt)(b)=(db/dt)(h)
divide by h
(2da/dt-(dh/dt)(b))/h=db/dt
ok
we know
height=10
area=100
so
a=1/2bh
100=1/2b10
100=5b
20=b
so
h=10
b=20
da/dt=2cm²/min
dh/dt=1cm/min
therefor
(2(2cm²/min)-(1cm/min)(20cm))/10cm=db/dt
(4cm²/min-20cm²/min)/10cm=db/dt
(-16cm²/min)/10cm=db/dt
-1.6cm/min=db/dt
the base is decreasing at 1.6cm/min
Looks like you have most of the details already, but you're missing one crucial piece.
is parameterized by
for 
and 
, and a normal vector to this surface is
with norm
So the integral of 
is
Answer:
300ml
Step-by-step explanation:
As values are missing in your question, the complete question is:
In the lab, Leila has two solutions that contain alcohol and is mixing them with each other. Solution A is 6% alcohol and Solution B is 20% alcohol. She uses 400 milliliters of Solution A. How many milliliters of Solution B does she use, if the resulting mixture is a 12% alcohol solution?
Considering x=ml of 20% solution B
Therefore, 400+x=ml of resulting 12% solution
Solution A alcohol+ solution B alcohol= Alcohol solution
6% (400)+ 20%x = 12%(400+x) ->(converting percentage into decimal)
.06*400+.2x=.12(400+x)
24+.2x=48+.12x
.08x=24
x=24/.08=300 ml
she used 300 milliliters of 20%Solution B in resulting mixture.
Answer:
Step-by-step explanation:
∡x = v0t + 1/2 at²
∡x / v0t = 1/2 at²
2∡x / v0t =at²
a = 2∡x / v0t³
that's your answer:)
