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Basile [38]
3 years ago
11

Evaluate the related series described 4 - 12 + 36 - 108 n=6

Mathematics
1 answer:
Arada [10]3 years ago
6 0

Answer:

11/54 is n did you want it in decimal form 0.2037 bar on top of 03

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When radioactive substances​ decay, the amount remaining will form a geometric sequence when measured over constant intervals of
svetoff [14.1K]
I am not sure because there is no info about what is rate of decay
I assume if there is rate of decay is constant every hour then 0-2 it is
It 1/4th
So for 1st hour it is 933 gram
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3 years ago
How many seconds are in 7 minutes?(feel free to show your work :)
Bumek [7]

Answer:

420 seconds

Step-by-step explanation:

60 seconds are in 1 minute

60 x 7 = 420

therefore there are 420 seconds in 7 minutes

5 0
3 years ago
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The half-life of uranium-238 is 4.5x10^9 years. The half-life of uranium-234 is 2.5x10^5 years. How many times greater is the ha
lara31 [8.8K]

<u>Answer:</u>

Half-life of uranium-238 18 \times 10^{3} \text { times greater } that of uranium-234

<u>Explanation:</u>

Half time of uranium-238 = 4.5\times 10^9  years

Half time of Uranium-234 = 2.5\times 10^5 years

To find how much times greater the half life of uranium-238 is from uranium-234

= \frac{\text { Half life of Uranium-238 }}{\text { Half time of Uranium - 234 }}

=\frac{4.5 \times 10^{9}}{2.5 \times 10^{5}}

=18 \times 10^{3} \text { times greater }

Hence Uranium-238 is  18 \times 10^{3} \text { times greater } than Uranium-234

4 0
3 years ago
Nobody haves answers
mote1985 [20]

Answer:

Nobody haves answersNobody haves answersNobody haves answersNobody haves answersNobody haves answersNobody haves answersNobody haves answersNobody haves answersNobody haves answersNobody haves answersNobody haves answers

Step-by-step explanation:

6 0
3 years ago
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Eduardo and Sarawong are selling wrapping paper for a school fundraiser. Eduardo sold 5 rolls of plain wrapping paper and 5 roll
Digiron [165]

Answer: A roll of plain wrapping paper costs $17, while a roll of holiday wrapping paper costs $20.

Step-by-step explanation: First let us represent a roll of plain wrapping paper with letter p and a roll of holiday wrapping paper would be represented by d. If Eduardo sold 5 rolls of plain wrapping paper and 5 rolls of holiday wrapping paper for $185, then we can express this as

5p+ 5d= 185

Also if Sarawong sold 14 rolls of plain wrapping paper and 5 rolls of holiday wrapping paper for $338, then this too can be expressed as

14p + 5d = 338

Now we have a pair of simultaneous equations which are,

5p + 5d = 185 ———(1)

14p + 5d = 338 ———(2)

Since all the variables have coefficients greater than 1, we shall use the elimination method. Note that the coefficients of the d variable are both 5, so straight away we subtract equation (1) from equation (2) and we now have;

(14p - 5p) + (5d - 5d) = 338 - 185

9p = 153

Divide both sides of the equation by 9

p = 17

Having calculated p, we can now substitute for the value of p into equation (1)

5p + 5d = 185

5(17) + 5d = 185

85 + 5d = 185

Subtract 85 from both sides of the equation

5d = 100

Divide both sides of the equation by 5

d = 20

Hence, the cost per roll of plain wrapping paper is $17, while the cost per roll of holiday wrapping paper is $20.

5 0
3 years ago
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