Water is being pumped into an inverted conical tank at a rate of 9.4 cubic meters per min. The tank has height 13 meters and the diameter at the top is 5 meters. Find the rate at which the water level is rising when the height of the water is 1.5 meters.

dh/dt = 35.96 m/min

Step-by-step explanation:

Volume of a cone is given as;

V = (1/3)πr²h

Now, we need to find a relationship between dV/dt and dh/dt

Now, if we imagine the cone, at the height of 1.5m where the water level is rising, let's call the radius there r.

We are told the height of the cone is 13m and the radius of top is 5/2 =2.5m

Thus,by similar triangles;

r/h = 2.5/13

r = 2.5h/13

So,let's put this in the volume equation;

V = (1/3)π(2.5h/13)²h

V = (1/3)π(2.5²h³/169)

Since we need to find a relationship between dV/dt and dh/dt, thus;

dV/dt = π(2.5²h²/169) dh/dt

From the question, dV/dt = 9.4m³/min and h = 1.5m. so,

9.4 = π(2.5²•1.5²/169) dh/dt

9.4 = 0.2614 dh/dt

dh/dt = 9.4/0.2614 = 35.96 m/min

x =-.20752

step-by-step explanation:

4^(x+2)=12

4^(x+2)=12

take the log base 4 on each side

log 4 (4^(x+2) = log 4(12)

we know log b(a^y) = y log b (a)

(x+2) log 4(4) = log 4 (12)

x+2 = log 4 (12)

we know   logb c = loga c/loga b

we want to convert to base 10   so a = 10

by default, we do not write the 10   c = 4   and b = 10

log4 10 = log 12/log 4

log4 10 = log 12/ log 4

x+2 = log 12/ log 4

subtract 2 from each side

x +2 -2 = log 12/ log 4 -2

x = 1.79248125 -2

x =-.20752

i think its doubled. let me know

step-by-step explanation: