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GuDViN [60]
3 years ago
8

An electrolyte solution has an average current density of 111 ampere per square decimeter \left( \dfrac{\text{A}}{\text{dm}^2}\r

ight)(
dm

2


A

​

)left parenthesis, start fraction, start text, A, end text, divided by, start text, d, m, end text, squared, end fraction, right parenthesis.

What is the current density of the solution in \dfrac{\text{A}}{\text{m}^2}

m

2


A

​

start fraction, start text, A, end text, divided by, start text, m, end text, squared, end fraction?
Mathematics
1 answer:
Ksenya-84 [330]3 years ago
6 0

Answer:

The current density of the solution in m^2 is 11100A/m^2

Step-by-step explanation:

<em>One square decimeter is equal to 0.01 square meter,</em> Thus the current density in square meters is

\frac{111A}{dm^2} =\frac{111A}{0.01m^2} =\frac{111}{0.01}*\frac{A}{m^2}=\boxed{11100A/m^2}

11100 Amperes per square meter.

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When you find 36 is 42% of what number, what are you looking for?​
arsen [322]

Answer:

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Step-by-step explanation:

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5 0
3 years ago
A shopkeeper had a profit of ₹ 50 on Monday , a loss of ₹ 20 on Tuesday and of loss of
Oliga [24]

Answer:

12

Step-by-step explanation:

Remark

I can't use a symbol for the currency. I'll just get the answer in numbers. You can add the symbol.

Solution

First day: 50

Second day: - 20

Third day: - 18

50 - 20 - 18 = 12

4 0
3 years ago
My question got deleted so I'll post these into two separate
Agata [3.3K]

Answer:

see below

Step-by-step explanation:

1 million

1,000,000 seconds * 1 hour/ 3600 seconds * 1 day/ 24 hours * 1 year / 365 days

.031709792 years

Rounding to 3 decimal places

.032 years

50 years * 365 days/ 1 year * 24 hours/ 1 day * 3600 second / 1 hour    

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8 0
3 years ago
Hello Pls help and thanks
Anna11 [10]

Answer:

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6 0
2 years ago
The third term of an arithmetic sequence is 14 and the 9th term is -1. Find the first four terms of the sequence.​
Liula [17]

Answer:

The first four terms of the sequence are :  19, 16.5 , 14 , 11.5

Step-by-step explanation:

In the given sequence:

a(3) =  14, a(9)  = -1

The general term of a sequence in Arithmetic Progression is:

a(n) = a + (n-1)d

a(3) = a + (3 -1) d = a + 2 d

and a(9) = a + (9- 1 ) d = a  + 8 d

⇒   a + 2 d =  14      .........  (1)

and a + 8 d  = -1    ...........  (2)

Now, solving the given system of equation, we get:

From (1), a  = 14 - 2 d

Put in (2), we get:

a + 8 d  = -1                ⇒   14 - 2 d + 8d = -1

⇒  14 +   6d   = -1  

or,  6d = -1 -14 = -15

⇒  d = -15/6 = -2.5

or, d  = -2.5

Then a  = 14 - 2 d = 14 - 2(-2.5)  =14 + 5  = 19, or a  = 19

Now, first four terms of the sequence is:

a = 19

a(2) = a + 4 = 19 - 2.5 = 16.5

a(3) = a + 2d = 19 + 2(-2.5) = 19 - 5 = 14

a(4) = a  + 3d = 19 + 3(-2.5)  = 19  - 7.5  = 11.5

Hence, the first four terms of the sequence are :  19, 16.5 , 14 , 11.5

4 0
3 years ago
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