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Vlad1618 [11]
3 years ago
10

Marco's painting has an area of 320 square inches. If the width of the painting is 16 inches, what is the length of the length o

f the painted?
Mathematics
2 answers:
Tamiku [17]3 years ago
6 0
If you would like to know what is the length of the painting, you can calculate this using the following steps:

an area = width * length
320 = 16 * length
length = 320 / 16
length = 20 inches

The correct result would be 20 inches.
Novay_Z [31]3 years ago
5 0
Area = length * width 320 = l * 16 320/16 16/16 320/16 = 20 Length = 20 inches
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The height of the tree is 42.

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The fair spinner shown in the diagram above is spun.
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Unable to provide a definitive answer due to lack of diagram.

Answer: 1/4

Step-by-step explanation:

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Multiples of 6 are numbers obtained when 6 is multiplied by an integer. Such as

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Probability is calculated by finding the ratio of the required outcome and total possible outcomes.

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P(multiple of 6) = number of multiples of 6 / total number of possible outcomes

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A = 1011 + 337 + 337/2 +1011/10 + 337/5 + ... + 1/2021
egoroff_w [7]

The sum of the given series can be found by simplification of the number

of terms in the series.

  • A is approximately <u>2020.022</u>

Reasons:

The given sequence is presented as follows;

A = 1011 + 337 + 337/2 + 1011/10 + 337/5 + ... + 1/2021

Therefore;

  • \displaystyle A = \mathbf{1011 + \frac{1011}{3} + \frac{1011}{6} + \frac{1011}{10} + \frac{1011}{15} + ...+\frac{1}{2021}}

The n + 1 th term of the sequence, 1, 3, 6, 10, 15, ..., 2021 is given as follows;

  • \displaystyle a_{n+1} = \mathbf{\frac{n^2 + 3 \cdot n + 2}{2}}

Therefore, for the last term we have;

  • \displaystyle 2043231= \frac{n^2 + 3 \cdot n + 2}{2}

2 × 2043231 = n² + 3·n + 2

Which gives;

n² + 3·n + 2 - 2 × 2043231 = n² + 3·n - 4086460 = 0

Which gives, the number of terms, n = 2020

\displaystyle \frac{A}{2}  = \mathbf{ 1011 \cdot  \left(\frac{1}{2} +\frac{1}{6} + \frac{1}{12}+...+\frac{1}{4086460}  \right)}

\displaystyle \frac{A}{2}  = 1011 \cdot  \left(1 - \frac{1}{2} +\frac{1}{2} -  \frac{1}{3} + \frac{1}{3}- \frac{1}{4} +...+\frac{1}{2021}-\frac{1}{2022}  \right)

Which gives;

\displaystyle \frac{A}{2}  = 1011 \cdot  \left(1 - \frac{1}{2022}  \right)

\displaystyle  A = 2 \times 1011 \cdot  \left(1 - \frac{1}{2022}  \right) = \frac{1032231}{511} \approx \mathbf{2020.022}

  • A ≈ <u>2020.022</u>

Learn more about the sum of a series here:

brainly.com/question/190295

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Step-by-step explanation:

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