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alexandr402 [8]
3 years ago
7

Find the perimeter of a rectangular strip of board is 3.28m long and 7.5m

Mathematics
1 answer:
OLEGan [10]3 years ago
8 0

Answer:

21.56 m

Step-by-step explanation:

2(3.28) + 2(7.5)

(twos represent the sides)

6.56 + 15

= 21.56 m

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Arrange the decimal fractions in descending order.<br>1.3 0.13 3.1 0.31<br>​
SVEN [57.7K]

Answer:

0.13 0.13 3.1.3

Step-by-step explanation:

Hope it's answer you plz mark as Brainlist

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3 years ago
If a+b = 7, and a² +b²=25<br>then ab = ?​
Rama09 [41]

a2+b2=25

or, (a+b)^2-2ab=25

or, (7)^2-2ab=25 [a+b=7]

or, 49-25=2ab

or, 24/2=an

•°• ab=12....

3 0
3 years ago
Plz explain step by step, I'm taking Adv. Math (Algebra) It's confusing If can't explain, plz at least give right answer ✌
Mashcka [7]

Question 4:

OOOOOHHH a breakthrough! I was confused when I read this question but now I get it!

Okay when dividing this equation I would do this

5^4

25

Since 25 is 5 x 5 I would mark it as 5^2, and then I can subtract!

5^4

5^2 =

5^2

If you do it like this then it would be the first choice.



7 0
3 years ago
What is the area of a circle with a diameter of 9.2 centimeters? Round your answer ro the nearest hundredth.
nika2105 [10]
The answer the the problem is A.
3 0
3 years ago
Please Help !!!
solong [7]
Best Answer: 2 LiCl = 2 Li + Cl2
mass Li = 56.8 mL x 0.534 g/mL=30.3 g
moles Li = 30.3 g / 6.941 g/mol=4.37
the ratio between Li and LiCl is 2 : 2 ( or 1 : 1)

moles LiCl required = 4.37
mass LiCl = 4.37 mol x 42.394 g/mol=185.3 g


Cu + 2 AgNO3 = Cu/NO3)2 + 2 Ag
the ratio between Cu and AgNO3 is 1 : 2
moles AgNO3 required = 4.2 x 2 = 8.4 : but we have only 6.3 moles of AgNO3 so AgNO3 is the limiting reactant
moles Cu reacted = 6.3 / 2 = 3.15
moles Cu in excess = 4.2 - 3.15 =1.05

N2 + 3 H2 = 2 NH3
moles N2 = 42.5 g / 28.0134 g/mol=1.52
the ratio between N2 and H2 is 1 : 3
moles H2 required = 1.52 x 3 =4.56
actual moles H2 = 10.1 g / 2.016 g/mol= 5.00 so H2 is in excess and N2 is the limiting reactant
moles NH3 = 1.52 x 2 = 3.04
mass NH3 = 3.04 x 17.0337 g/mol=51.8 g
moles H2 in excess = 5.00 - 4.56 =0.44
mass H2 in excess = 0.44 mol x 2.016 g/mol=0.887 g
5 0
3 years ago
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