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AlladinOne [14]

3 years ago

7

Theater m has 25 rows with 27 seats in each row. how many of the seats were occupiedduring a certain show?(1) during the show, t

here was an average (arithmetic mean) of 10 unoccupied seatsper row for the front 20 rows.(2) during the show, there was an average (arithmetic mean) of 20 unoccupied seatsper row for the back 15 rows.

1 answer:

Elis [28]3 years ago

5 0

1) 475
2)375
You need to multiply 25 by 27 and you get 675 then to get answer one you need to multiply 10 by 20 to get 200 the take 200 away from 675 to get 475 then do the same thing with the second problem 20x15 which is 300 the take 300 away from 675.

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Use set notation to write the members of the following set, or state that the set has no members. Odd numbers between 2 and 82 t

adell [148]

Let the set of all Odd multiples of 9 between 2 and 82 be denoted by D, then, using set-builder notation,

$D=\{ 18n+9 \mid n \in \mathbb{N}, 0\le n\le 4 \}$

The odd multiples of 9, $m$ , in the range $2\le m \le 82$ form the set

$\{9,27,45,63,81\}$

Each member of the set is a term of the arithmetic progression

$U_n=18n+9$

where the values of $n$ range from 0 to 4, or $0\le n\le 4$

Putting these facts together, we get the result

$D=\{ 18n+9 \mid n \in \mathbb{N}, 0\le n\le 4 \}$

Learn more about set-builder notation here: brainly.com/question/17238769

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Step-by-step explanation:

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3 years ago

Equation<br><img src="https://tex.z-dn.net/?f=%28z%20%2B%20%20%5Cfrac%7B6%7D%7B7%7D%20%29%20%2B%20%20%5Cfrac%7B2%7D%7B14%7D%20%2

ddd [48]

Given:

The equation is

$\left(z+\dfrac{6}{7}\right)+\dfrac{2}{14}=-1$

To find:

The solution of the given equation.

Solution:

We have,

$\left(z+\dfrac{6}{7}\right)+\dfrac{2}{14}=-1$

It can be written as

$\left(z+\dfrac{6}{7}\right)+\dfrac{1}{7}=-1$

$\left(z+\dfrac{6}{7}\right)=-1-\dfrac{1}{7}$

Multiply both sides by 7.

$7z+6=-7-1$

$7z+6=-8$

$7z=-8-6$

$7z=-14$

Divide both sides by 7.

$z=\dfrac{-14}{7}$

$z=-2$

Therefore, the value of z is -2.

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3 years ago

Question 3 of 10

katovenus [111]

Answer:

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Step-by-step explanation:

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4 years ago

Determine whether the system of linear equations has one and only one solution, infinitely many solutions, or no solution. 2x-y=

Bond [772]

Answer:

The system of linear equations has infinitely many solutions

Step-by-step explanation:

Let's modified the equations and find the answer.

Using the first equation:

$2x-y=5$ we can multiply by 2 in both sides, obtaining:

$2*(2x-y)=2*5$ which can by simplified as:

$4x-2y=10$ which is equal to:

$4x=2y+10$

Considering the second equation:

$=4x+ky=2$

Taking into account that from the first equation we know that: $4x=2y+10$ , we can express the second equation as:

$2y+10+ky=2$ , which can be simplified as:

$(2+k)y=2-10$

$(2+k)y=-8$

$y=-8/(2+k)$

Because (-8) is being divided by (2+k), then (2+k) can't be equal to 0, so:

$2+k=0$ if $k=-2$

This means that k can be any number different than -2, and for each of these solutions, there is a different solution for y, allowing also, different solutions for x.

For example, if k=0 then

$y=-8/(2+0)$ which give us y=-4, and, because:

$4x=2y+10$ if y=-4 then $x=(-8+10)/4=0.5$

Now let's try with k=-1, then:

$y=-8/(2-1)$ which give us y=-8, and, because:

$4x=2y+10$ if y=-8 then $x=(-16+10)/4=-1.5$ .

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3 years ago