Given:
Volume of water in each container.
To find:
Difference in the rate of change.
Solution:
Take any two points on container 1.
Let the points are (10, 2) and (20, 4).
Rate of change for container 1 is 
.
Take any two points on container 2.
Let the points are (5, 2) and (10, 4).
Rate of change for container 2 is 
.
Difference = 
The difference in the rate of change between the two containers is gallon per minute.
Scott: 8,748 x 30% = 2,624.40 his share for health insurance.
2,624.40 / 52 weeks = 50.47
50.47 x 2 = 100.94 (biweekly pay) CHOICE D.
Jarvis: 41 y.o male ; 115,000 for 20 years.
115,000 / 1,000 = 115
115 x 13.68 = 1,573.20 annual premium
1,573.20 / 52 weeks = 30.25 weekly deduction CHOICE D.
I think the initial investment is 3,000 to have an answer among the given choices.
Marco: 3,000 (1.0425)^17 = 3,000 x 2.029 = 6,087 CHOICE D. about 17 years.
Stephen: 8,000 x 6.75% p.a. = 540 / 2 (semi annual) = 270 CHOICE A.
Skye: 36,192 annual salary / 52 weeks = 696 weekly salary
696 x 4% deduction = 27.84 skye's share
696 x 3% deduction = 20.88 employer's share
27.84 + 20.88 = 48.72 nearest to CHOICE D.
Ruth problem needs additional data.
Chanel:
1175 x 1.45% x 4 yrs = 68.15 Choice A.
Answer:
A car's velocity changes from 25 m/s to 20 m/s in 5 seconds.
Step-by-step explanation:
By definition, Acceleration is the rate of change of Velocity of an object.
I.e it is the( final velocity minus initial velocity ) ÷ by the time it took for the velocity to change.
If we observe the choices only one choice gives us all 3 components (final velocity, initial velocity, time taken for velocity to change), which is :
A car's velocity changes from 25 m/s to 20 m/s in 5 seconds.
The answer is 62.84 kg.
To calculate the mass of air (m) we need to use the volume (V) of the air (which is the same as the volume of the room) and the density (D) of the air.
D = m/V
The density of air is: D = 1.225 kg/m³
The volume of air is: V = 5.4m×3.8m×2.5m = 51.3 m³
D = m/V
m = D * V
m = 1.225 kg/m³ * 51.3 m³
m = 62.84 kg
The total if all 236 coins were nickels would be $11.80, which is $3.95 short of the actual amount.
Replacing a nickel with a dime adds $0.05 to the total value, so there must have been $3.95/$0.05 = 79 such replacements.
There are 79 dimes.
There are 236 -79 = 157 nickels.
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Using the given variables, the problem statement gives rise to two equations. One is the based on the number of coins. The other is based on their value.
- n + d = 236
- .05n +.10d = 15.75
Solving the first for n, we get
... n = 236 - d
Substituting that into the second equation, we have
... .05(236 -d) +.10d = 15.75
... .05d = 15.75 -236·.05 . . . . . collect terms, subtract .05·236
... d = 3.95/.05 . . . . . . . . . . . . . divide by .05
... d = 79
... n = 236-79 = 157
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The solution should look familiar, as it matches the verbal description at the beginning.
