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gladu [14]
3 years ago
13

What is 12%expressed as a fraction? 3/25 1/8 6/25 1/4

Mathematics
1 answer:
Alexxx [7]3 years ago
3 0
3/25 because it's 12% expressed as a fraction
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The circumference of a circle of diameter 3,5 cm is 11 cm. What is the circumference of a circle of diameter 42 cm? Answer ... c
andreyandreev [35.5K]

Answer:

131.95 cm

Step-by-step explanation:

The Formula for the circumference of a circle is πd where d is the diameter. Therefore you simply multiply 42 with PI to get the circumference

5 0
3 years ago
Ian often buys in large quantities. A few months ago he bought several cans of frozen orange juice for $24. The next time Ian pu
ANEK [815]
For the answer to the question above asking, w<span>hat was the price per can and the numbers of cans purchased each time?
let x be the number of cans he bought
the let us go to the 2nd statement which is t</span><span>he next time Ian purchased frozen orange juice, the price had increased by $0.10 per can and he bought 1 less can for the same total price.
The equation for this is .10(x-1) = 24
So now let's solve, 
</span> .10(x-1) = 24
.10x - .10 = 24
.10x = 24+ .10
.10x = 24.10
Then divide both sides by .10
So the answer for this question is 
241 cans of juice


4 0
3 years ago
Jay went on a diet and lost 5 pounds each month. Find Jay's net weight change in 6 months.
Goryan [66]

Answer:

a 30 pound weight change in 1 month

Step-by-step explanation:

5 pounds

--------------- times 6 months  results in a 30 pound weight change in 1 month

 1 month

7 0
3 years ago
Use any of the methods to determine whether the series converges or diverges. Give reasons for your answer.
Aleks [24]

Answer:

It means \sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6} also converges.

Step-by-step explanation:

The actual Series is::

\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}

The method we are going to use is comparison method:

According to comparison method, we have:

\sum_{n=1}^{inf}a_n\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n

If series one converges, the second converges and if second diverges series, one diverges

Now Simplify the given series:

Taking"n^2"common from numerator and "n^6"from denominator.

=\frac{n^2[7-\frac{4}{n}+\frac{3}{n^2}]}{n^6[\frac{12}{n^6}+2]} \\\\=\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{n^4[\frac{12}{n^6}+2]}

\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n=\sum_{n=1}^{inf} \frac{1}{n^4}

Now:

\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\ \\\lim_{n \to \infty} a_n = \lim_{n \to \infty}  \frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\=\frac{7-\frac{4}{inf}+\frac{3}{inf}}{\frac{12}{inf}+2}\\\\=\frac{7}{2}

So a_n is finite, so it converges.

Similarly b_n converges according to p-test.

P-test:

General form:

\sum_{n=1}^{inf}\frac{1}{n^p}

if p>1 then series converges. In oue case we have:

\sum_{n=1}^{inf}b_n=\frac{1}{n^4}

p=4 >1, so b_n also converges.

According to comparison test if both series converges, the final series also converges.

It means \sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6} also converges.

5 0
3 years ago
What is the first step to solve this equation
shepuryov [24]

Answer:

First step, add 11 to 44.

Step-by-step explanation:

After that, divide both sides by negative 3.

7 0
3 years ago
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