|1-3x| + 6 = 29 Justify Each Step With An Algebraic Property Please show work =)

1 answer:

8 0

X= -23/3, 23/3 calculate absolute value move constant to the right subtract those two numbers divide both by 3 then you get the answer i put up there

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How to find the measure of angles?

nevsk [136]

Answer:

The best way to measure an angle is to use a protractor. To do this, you'll start by lining up one ray along the 0-degree line on the protractor. Then, line up the vertex with the midpoint of the protractor. Follow the second ray to determine the angle's measurement to the nearest degree.

Step-by-step explanation:

7 0

3 years ago

It costs $5.15 to rent a kayak for 1 hour at a local state park. The price per hour stays the same for up to 5 hours of rental.

iren [92.7K]

5.15 x 5 = 25.75

25.75 + 3.75 = 29.5

So your answer is $29.50

Hope it helps! :D

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2 years ago

Angle 1 and Angle 2 form a linear pair. If the measure of angle is 113°, find the measure of angle 2.

LuckyWell [14K]

Answer:

<2 =67

Step-by-step explanation:

A linear pair adds to 180 degrees

<1 + <2 =180

113+ <2 = 180

<2 = 180-113

<2 =67

5 0

2 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:

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