Answer:
4020 grams
Step-by-step explanation:
4kg=4000g
2 dag= 20g
Total: 4000g + 20g = 4020g
P = 2a + b
10= 2a + 3
2a = 10-3
a = 7/2 = 3.5
Therefore, numerical value of a is 3.5
The limit does not exist. Why? Because the left hand limit DOES NOT equal the right hand limit. Let’s double check:
We could use -0.000001 to represent the left hand limit. This is less than 0. We plug in 5x - 8
5(-0.000001) - 8
-0.000005 - 8
-8.000005
If we would continue the limit (extend the zeros to infinity), we would get exactly
-8
That is our left hand limit.
Our right hand limit will be represented by 0.000001. This is greater than 0. We plug in abs(-4 - x)
abs(-4 - (0.000001))
abs(-4.000001)
4.000001
If we would continue the limit (extend the zeros to infinity), we would get exactly
4
4 does not equal -8, therefore
The limit does not exist
If you would like to know in which step did the student first make an error and what is the correct step, you can calculate this using the following steps:
3(2x - 4) = 8 + 2x + 6
6x - 12 = 8 + 2x + 6
6x - 12 = 14 + 2x ... Step 2
6x - 2x = 14 + 12
4x = 26
x = 26/4 = 13/2
The correct result would be Step 2; <span>6x - 12 = 14 + 2x.</span>
<span>1. How much heat is absorbed by a 50g iron skillet when its temperature rises from 10oC to 124oC? Joules
Formula: Heat = mass * specific heat * ΔT
Data:
mass = 50 g = 0.050 kg
specific heat of iron = 450 J/ kg °C
ΔT = 124°C - 10°C ¿ 114 °C
=> heat = 0.050kg * 450 J / kg°C * 114°C ≈ 2.6 J
2. If a refrigerator is a heat pump that follows the first law of
thermodynamics, how much heat was removed from food inside of the
refrigerator if it released 492J of energy to the room? Joules
The firs law of thermodynamics is conservation of energy => energy removed from inside of the refrigerator = energy released to the room
=> Answer = 492 J
3. How much heat is needed to raise the temperature of 45g of water by 63oC? Joules
Formula: heat = mass * specific heat * ΔT
specific heat of water = 4186 J / Kg °C
heat = 0.045 kg * 4186 J/kg °C * 63°C = 11,867.31 J
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