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Fittoniya [83]
3 years ago
7

Simplify completely 25 w to the sixth power over 10 w to the third power divided by 30 w squared over 5 w.

Mathematics
2 answers:
asambeis [7]3 years ago
5 0

5w^2 over 12 is your answer. hope it helped!

malfutka [58]3 years ago
4 0
\cfrac{25w^6}{10w^3}:  \cfrac{30w^2}{5w}=  \cfrac{5w^3}{2}*  \cfrac{5w}{30w^2}= \cfrac{w}{2}*  \cfrac{5w}{6}= \cfrac{5w^2}{12}
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I have two fair dice each numbered 1 to 6 I throw both dice and add the two numbers together. What is the probability that the n
liraira [26]

Answer:

7 / 36

Step-by-step explanation:

Using the sample space attached,

Required outcome = multiple of 5 from the sun of two rolled dice = 7

Total possible outcomes =(number of faces)² = 6² = 36

Therefore, the probability of obtaining a sum which is a multiple of 5 is :

P(sum which is a multiple of 5) :

Required outcome / Total possible outcomes

= 7 / 36

8 0
2 years ago
What is the product of the 2 solutions of the equation x^2+3x-21=0
Advocard [28]
X²+3x-21=0

1) we solve this square equation:
x=[-3⁺₋√(9+84)] / 2=(-3⁺₋√93)/2
We have two solutions:
x₁=(-3-√93)/2
x₂=(-3+√93)/2

2) we compute the product of the 2 solutions found.
[(-3-√93)/2][(-3+√93)/2] =(-3-√93)(-3+√93) / 4=
=(9-93)/4=-84/4=-21

Answer: the product of the 2 solutions of this equation is -21
3 0
3 years ago
Farmer Joe is planning to fence in a garden area in a portion of his field which is 75 feet wide. The farmer would like to creat
LenKa [72]
He could make his plot 300 feet long
5 0
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A recent tornado damaged Jack's house and left it a little distorted. A question on the insurance form asked, "What shape is the
Svetlanka [38]

Answer:

45

Step-by-step explanation:

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8 0
2 years ago
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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