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Fofino [41]
3 years ago
5

Navy’s Works At A Cell phone Store Monday To friday . She Earns $80 Per day plus $8 per Phone Sold . On tuesday , Nayva Makes at

least $176 . Which Number Line Could be used to show the number of phones Navya Sold By The end Of her Shift On Tuesday
Answer Choices

Mathematics
1 answer:
rodikova [14]3 years ago
3 0

Answer:

D.

Step-by-step explanation:

If She earns 80$ PER day + 8$ per phone sold and makes AT LEAST 176$ on Tuesday, You need to figure out how many phones she has sold by Tuesday.

Okay so now you need to set up your problem.

8x + 80 ≥ 176

Take away the 80, then take 80 from 176.

176 - 80 = 96

96 ÷ 8

x ≥ 12

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Round inglés
Andrew [12]

Answer:

13. 200

14. 8,300

15. 500

16. 2,500

Step-by-step explanation:

13. It is more than 4

14. It is more than 4

15. It is less than 4

16. It is more than 4

5 0
3 years ago
2. Select the factor table with the appropriate
Alexandra [31]

Answer:

60

Step-by-step explanation:

5+5+5+5+5+5+5+5+5+5+5+5

4 0
3 years ago
Does this table represent an exponential function? x 1 2 3 4 y 4 16 64 256
4vir4ik [10]

Answer:

Yes, the given table represent an exponential function.

Step-by-step explanation:

Given table is :

x y

1 4

2 16

3 64

4 256

Now we need to identify if the given table represent an exponential function or not. To find that we need to check if we can write the numbers in y-column in form of exponential function y=ab^x.

We see that y-values are basically powers of 4 so we can write the related function as y=4^x.

Which is clearly in form of y=ab^x.

Hence yes, the given table represent an exponential function.

6 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
I need help please and thank you
grin007 [14]

Answer:30

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
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