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Fed [463]
3 years ago
10

A Chinese restaurant offers buffet takeout for $4.99 per pound. How much does your takeout meal cost?

Mathematics
1 answer:
Degger [83]3 years ago
4 0

Answer:

$7.68

Step-by-step explanation:

4.99/0.65 = $7.68

Hope it helps!

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Riley's dentist has an aquarium in his office with the dimensions 16 in. By 8.5 in. By 10.5 in. What is the volume of the aquari
ad-work [718]

Answer:

1428

Step-by-step explanation:

I did 16 times 8.5 times 10.5. Good Luck!

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Answer: Gear 14 or 14.3

Step-by-step explanation:

8 0
3 years ago
M. Score: 0 of 1 pt
beks73 [17]

Answer:

  3×5×53

Step-by-step explanation:

You can use divisibility rules to find the small prime factors.

The number ends in 5, so is divisible by 5.

  795/5 = 159

The sum of digits is 1+5+9 = 15; 1+5 = 6, a number divisible by 3, so 3 is a factor.

  159/3 = 53 . . . . . a prime number,* so we're done.

795 = 3×5×53

_____

* If this were not prime, it would be divisible by a prime less than its square root. √53 ≈ 7.3. We know it is not divisible by 2, 3, or 5. We also know the closest multiples of 7 are 49 and 56, so it is not divisible by 7. Hence 53 is prime.

5 0
3 years ago
1. Evaluate: <br>0.234 x 10^2<br><br>​
stellarik [79]

Answer: 23.4 is the answer

3 0
3 years ago
Read 2 more answers
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
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