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lubasha [3.4K]
3 years ago
12

Five times the sum of 173 and 32

Mathematics
2 answers:
Sladkaya [172]3 years ago
4 0
173+32=205 , so 205 times 5 would equal 1,025 . so the answer would be 1,025 .
kow [346]3 years ago
3 0

Answer:  The required value is 1025.

Step-by-step explanation:  We are given to find the following value :

"five times the sum of 173 and 32".

Let x represents the quantity equal to five times the sum of 173 and 32.

Then, we get the value of x as follows :

x\\\\=5(173+32)\\\\=5\times 205\\\\=1025.

Thus, the required value is 1025.

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Help me with this,i really forgot how to do these type of questions
Montano1993 [528]
Work=speed times time (my equation to help me figure it out)

ok, 10 men take 84 days to complete a job
therefor, each man completes 1/10 of the job in 84 days
each man does 1/840 of the job per 1 day

so, if we had 5 more, or 15 men

15 times 1/840 times xdays=1 job complete
15/840 times xdays=1job
times both sides by 840/15
xdays=840/15
xdays=56
56 days it would take
5 0
3 years ago
State the solution to the equations f(x) and g(x).
Digiron [165]

Answer:

x = -1

Step-by-step explanation:

f(x) = g(x) when x = -1

3 0
2 years ago
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What is the x-intercept for the equation -2x + 5y = -10
Zielflug [23.3K]

Answer:

The x-intercept is at the point (5,0).

Step-by-step explanation:

-2x + 5y = -10

At the x intercept y = 0 so we substitute y = 0 into the given equation:

-2x + 5(0) = -10

-2x = -10

x = 5.



5 0
3 years ago
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Given log34≈1.262 and log37≈1.771, what is the value of log3(167)?
Arada [10]

Answer:

0.753

Step-by-step explanation:

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2 years ago
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1. How much heat is absorbed by a 50g iron skillet when its temperature rises from 10oC to 124oC? Joules
hoa [83]
<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
</span>
7 0
3 years ago
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