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Novosadov [1.4K]
2 years ago
5

Given Circle B where r = 2, determine the circumference.

Mathematics
1 answer:
kvv77 [185]2 years ago
5 0

Answer:

12.56

Step-by-step explanation:

Double the radius to get the diameter, 2 * 2 = 4

Multiply diameter x pi, 4 * 3.14 = 12.56

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the half-life of uranium -238 is 4.5 × 10 to the power of 9 years. The half-life of uranium-238 than that of uranium-234
Anika [276]

Answer:

Uranium-234 is an isotope of uranium. In natural uranium and in uranium ore, U-234 occurs as an indirect decay product of uranium-238, but it makes up only 0.0055% (55 parts per million) of the raw uranium because its half-life of just 245,500 years is only about 1/18,000 as long as that of U-238.

Step-by-step explanation:


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3 years ago
PLEASE HELP WITH THIS QUESTION​
Karolina [17]

Answer:

a) 0.8( 8/10 gives 0.8)

b)0.7( if it was 7/100 it would have been

0.07)

c)3.5( by converting mixed fractions to improper fractions, that is, 3×2+1 and having the same denominator 2… we get 7/2… for the purpose of converting it to decimal, 7×5/2*5= 35/10… by moving one decimal place to the left, we get 3.5)

D) 2.5 ( the same way as c) 5*5/2*5: 25/10 or you can make it to be 5*50/2*50: 250/100 which still gives 2.5)

7 0
3 years ago
Read 2 more answers
$13,957 is invested, part at 7% and the rest at 6%. If the interest earned from the amount invested at 7% exceeds the interest e
ch4aika [34]

Answer:

The Amount invested at 7% interest is $12,855

The Amount invested at 6% interest = $1,102  

Step-by-step explanation:

Given as :

The Total money invested = $13,957

Let The money invested at 7% = p_1  = $A

And The money invested at 6% = p_2 = $13957 - $A

Let The interest earn at 7% = I_1

And The interest earn at 6% = I_2

I_1 -  I_2 = $833.73

Let The time period = 1 year

Now,<u> From Simple Interest method</u>

Simple Interest = \dfrac{\textrm principal\times \textrm rate\times \textrm time}{100}

Or,  I_1 = \dfrac{\textrm p_1\times \textrm 7\times \textrm 1}{100}

Or,  I_1 = \dfrac{\textrm A\times \textrm 7\times \textrm 1}{100}

And

I_2 = \dfrac{\textrm p_2\times \textrm 6\times \textrm 1}{100}

Or,  I_2 = \dfrac{\textrm (13,957 - A)\times \textrm 6\times \textrm 1}{100}

∵  I_1 -  I_2 = $833.73

So, \dfrac{\textrm A\times \textrm 7\times \textrm 1}{100} -  \dfrac{\textrm (13,957 - A)\times \textrm 6\times \textrm 1}{100} = $833.73

Or, 7 A - 6 (13,957 - A) = $833.73 × 100

Or, 7 A - $83,742 + 6 A = $83373

Or, 13 A = $83373 + $83742

Or, 13 A = $167,115

∴ A = \dfrac{167115}{13}

i.e A = $12,855

So, The Amount invested at 7% interest = A = $12,855

And The Amount invested at 6% interest = ($13,957 - A) = $13,957 - $12,855

I.e The Amount invested at 6% interest = $1,102

Hence,The Amount invested at 7% interest is $12,855

And The Amount invested at 6% interest = $1,102   . Answer

8 0
3 years ago
In 2010, the average cost of electricity was 11.54 cents per kilowatt hour. In 2014, the average cost of electricity
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Answer:

B

Step-by-step explanation:

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2 years ago
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Evgesh-ka [11]

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2.) CM=MD                                                                          2.) Definition of midpoint

3.) 5x-2=3x+2                                                                       3.) Substitution

4.) 2x-2=2                                                                             4.) Subtraction Property of Equality

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4 0
3 years ago
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