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2 years ago

Find the ninth partial sum of infinite sumnation 3i-4

1 answer:

3 0

$\bf \sum\limits_{i=1}^9\ 3i-4\implies \sum\limits_{i=1}^9\ 3i-\sum\limits_{i=1}^9\ 4\implies 3\sum\limits_{i=1}^9\ i-\sum\limits_{i=1}^9\ 4
\\\\\\
3\cdot \cfrac{9(9+1)}{2}-(9\cdot 4)$

and surely you know how much that is.

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7 (n+4) = 21<br> Please help me solve this

RoseWind [281]

Answer is n= -1 Explanation first thing you need to do is to divide the 7 and 21 to get n+4=3. After that move the constant 4 and switch it with 3. N+3=4. Then calculate n+3=4 to get n= -1

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2 years ago

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The product of two fractions that are each between 0 and 1 is also between 0 and 1.

Len [333]

They would be any fraction between 0 and 1.

Example- 1/3 * 1/2 = 1/6 (They all are less than one)

1/5 * 1/8 = 1/40

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2 years ago

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Find the perimeter of a quadrilateral with vertices at C (−2, 1), D (2, 4), E (5, 0), and F (1, −3).

antoniya [11.8K]

The perimeter of the quadrilateral with vertices at C (−2, 1), D (2, 4), E (5, 0), and F (1, −3) is 20units.

Option C) is the correct answer.

<h3>What a quadrilateral?</h3>

A quadrilateral is simply a polygon with four sides, four angles, and four vertices.

To get the perimeter, we simply add the values of the four side.

Given that;

The vertices are at C (−2, 1), D (2, 4), E (5, 0), and F (1, −3).

To get the dimension between the given coordinates, we use;

d = √((x2 -x1)² +(y2 - y1)²)

For length CD, DE, EF and FC

CD = √((2 - (-2))² + (4 - 1)²) = √( 16+9) = √25 = 5

DE = √((5 - 2)² + (0 - 4)²) = √( 9+16) = √25 = 5

EF = √((1 - 5)² + (-3 - 0)²) = √( 16+9) = √25 = 5

FC = √((-2 - 1)² + ( 1 - (-3))²) = √( 9+16) = √25 = 5

Perimeter of the quadrilateral = CD + DE + EF + FC

Perimeter of the quadrilateral = 5 + 5 + 5 + 5

Perimeter of the quadrilateral = 20units

Therefore, the perimeter of the quadrilateral with vertices at C (−2, 1), D (2, 4), E (5, 0), and F (1, −3) is 20units.

Option C) is the correct answer.

Learn more about area of rectangle here: brainly.com/question/27612962

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1 year ago

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What is the output, or dependent variable of quantity?<br><br> 1: x<br> 2:f(x)<br> 3:y

Semenov [28]

The answer would be y

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2 years ago

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There are eight students in a class. Only one of them has passed Exam P/1 and only one of them has passed Exam FM/2. No student

Svetach [21]

Answer: There is probability of 0.57 chances that exactly three students from a group of four students have not passed Exam P/1 or Exam FM/2.

Step-by-step explanation:

Total number of students = 8

Number of student who has passed Exam P/1 = 1

Number of student who has passed Exam FM/2 = 1

No student has passed more than one exam.

According to question, exactly three students from a randomly chose group of four students have not passed Exam P/1 or Exam FM/2.

So, Probability will be

$\frac{^6C_3\times ^2C_1}{^8C_4}\\\\=\frac{20\times 2}{70}\\\\=\frac{4}{7}\\\\=0.57$

Hence, there is probability of 0.57 chances that exactly three students from a group of four students have not passed Exam P/1 or Exam FM/2.

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2 years ago