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3 years ago

A line intersects the points (-2,5) and (6,5) What is the slope-intercept equation for the slope

Mathematics

1 answer:

avanturin [10]3 years ago

8 0

Answer:

Step-by-step explanation:

(y2-y1)/(x2-x1)

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The cost of renting a moving truck is $15 plus an additional $0.10 for each mile driven. Write an expression to represent the co

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Answer:

m(0.10)+15

Step-by-step explanation:

X is representing the miles driven. Then we're adding $15 for the rent fee.

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2 years ago

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Lubov Fominskaja [6]

Noy completwly possitive what yhe quesrion is asking for but The number that shows up most often is 55 but the mean (average) is 50.06

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3 years ago

Why is it important to reduce radicals

OleMash [197]

Simplifying radical expressions expression is important before addition or subtraction because it you need to which like terms can be added or subtracted. If we hadn't simplified the radical expressions, we would not have come to this solution. In a way, this is similar to what would be done for polynomial expression.

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2 years ago

What is the quotient of the division below?<br><br> 96 divided by 3 =

egoroff_w [7]

Answer:

Step-by-step explanation:

96 divided by 3

A quick way to do it:

9 divided by 3 = 3

6 divided by 3 = 2

answer: 32

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3 years ago

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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:

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