Answer:
2160 cm³/hour
Step-by-step explanation:
By default, we know that the volume of a cube is given as s³
Thus, the Volume function, V = s³
When we differentiate with respect to time we have
dV/dt = 3s² (ds/dt), where ds/dt = 0.2
Then we go ahead and substitute all the given parameters
dV/dt = 3 x 60 x 60 x 0.2
dV/dt = 10800 * 0.2
dV/dt = 2160 cm³/hour
This means that the volume decreases by a rate of 2160 cm³/hour at the instant its edge is 60 cm
Answer:
I think it's 25/30.
Step-by-step explanation:
Answer:
solution given:
let's see only in a right-angled triangle Δ ACD.
AC=3 units
AD=5 units
since Δ ACD is a right-angled triangle. It satisfies Pythagoras law
25=9+
=25-9=16
Now
Area of rectangle Δ ACD=
similarly,
since AB=CE=2.4 units
Δ ACE is a right-angled triangle. It satisfies Pythagoras law.
now
area of trapezoid=area ofΔACD+Area of ΔABC
=6+
First you need to figure out what 5.4% of 3600 is. So take 3,600 x 0.054. This equals 194.4. You then want to add this to 3,600. This will equal 3,794.4. You then want to do the process again, so the interest this year comes from 3,794.4 x 0.054. This equals about 204.9 (if you can it is 204.8976). Then you want to add that back into 3,794.4 so it will equal about 4,000 (or 3,999.2976) Well now this is a long process, but its the one way I remember how to do it.
Answer: option A. 9.2 gal/year
Justification:
The average rate of change from 1960 to 1970 is the change on the variable gallons consumed per passenger car divided by the period
average rate of change = Δg / Δt
Δg = 760 gallons - 668 gallons
Δt = 1970 year - 1960 year
760 gal - 668 gal 92 gal
average rate of change = -------------------------------- = ---------- = 9.2 gal / year
1970 year - 1960 year 10 year
Answer: 9.2 gal / year
As you see for this specific period, 1960 to 1970, the vehicles increased their consumption of gas per passenge, meaning that they became less efficient.
