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Hoochie [10]
3 years ago
5

Roger's Cafe has usual daily earnings of $800. Today, the cafe earned 120% of the usual daily earnings. How much did they earn t

oday
Mathematics
1 answer:
riadik2000 [5.3K]3 years ago
8 0

Answer:

960

Step-by-step explanation:

Over here, I'm going to do it in two ways, and either way works, one is just easier to do out:

1st way:

You can represent the percent 120 as a decimal which is 1.20, then you do 800*1.2.

2nd way:

You can use benchmark percents. You find 10 percent of 800 which is 80 and multiply 80 by 2 to get 160, so 20% is 160 dollars. Then you add 160 to 800 (800+160) and get 960.

Please tell me if you have any questions.

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Slope=4 Point=(1,6)
Y=Mx+b
6=4(1)+b
6=4+b
-4 -4
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2=b
Equation:4x+2
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3 years ago
(−8k+1)(−8k+1) standard form
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Answer:

64k^2 - 16k +1

Step-by-step explanation:

We can rewrite this as

(-8k+1) ^2

We know that (a+b)^2 = a^2 +2ab +b^2

Let a = -8k  and b = 1

(-8k+1) = (-8k)^2 +2*(-8k)(1) + 1^2

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Step-by-step explanation:

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2 years ago
Johanna receives a report that says her typing speed is 65 words per minute. She knows that she made 4 errors in the 5-minutes t
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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
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