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3 years ago

A movie theater has a seating capacity of 387. The theater charges $5.00 for children, $7.00 for students, and $12.00 of

Mathematics

1 answer:

Rainbow [258]3 years ago

8 0

Answer:

The attendance was 198 children, 90 students and 99 adults.

Step-by-step explanation:

We define:

c: children attendance

s: students attendance

a: adult attendance

The equation that describes the total ticket sales is:

$5c+7s+12a=2808$

We also know that the children attendance doubles the adult attendance:

$c=2a$

The third equation is the seating capacity, which we assume is full:

$c+s+a=387$

We start by replacing variables in two of the equations:

$c=2a\\\\s=387-c-a=387-2a-a=387-3a$

Then, we solve the remaining equation for a:

$5c+7s+12a=2808\\\\5(2a)+7(387-3a)+12a=2808\\\\10a+(2709-21a)+12a=2808\\\\10a+12a-21a=2808-2709\\\\a=99$

Then, we solve for the other two equations:

$c=2a=2*99=198\\\\s=387-3a=387-3*99=387-297=90$

The attendance was 198 children, 90 students and 99 adults.

You might be interested in

Two buses leave towns 304 miles apart at the same time and travel toward each other. One bus travels 14 mih slower than the othe

8090 [49]

Answer:

The faster bus moves at 83mi/h and the slower one moves at 69mi/h.

Step-by-step explanation:

Let's define:

R₁ = rate of bus 1, this is the faster one.

R₂ = rate of bus 2, this is the slower one.

We know that one bus travels 14mi/h slower, then:

R₂ = R₁ - 14mi/h.

Now we know that:

Distance = Speed*Time.

If we add the distances that both busses travel in 2 hours, it should be equal to the initial distance between the buses, then:

R₁*2h + R₂*2h = 304 mi

Then we have the two equations:

R₂ = R₁ - 14mi/h

R₁*2h + R₂*2h = 304 mi

The first step is to replace the first equation in the second one, to get:

R₁*2h + (R₁ - 14mi/h)*2h = 304 mi

And now we can solve this for R₁.

R₁*2h + R₁*2h - 14mi/h*2h = 304 mi

R₁*4h - 28mi = 304mi

R₁*4h = 304mi + 28mi = 332mi

R₁ = 332mi/4h = 83mi/h

The faster bus moves at 83mi/h

And we know that the slower one moves at 14mi/h slower than this, then:

R₂ = R₁ - 14mi/h = 83mi/h - 14mi/h = 69 mi/h

7 0

2 years ago

How can you determine if a triangle is a right triangle given the 3 side lengths?

sveta [45]

Answer:

To know that a triangle is right angled triangle we will have to apply pythagoras theorem to verify it

Like length 3, 4 and 5 are leghts of right angled triangle

5 0

2 years ago

what is the common number to be subtracted from each term of the ratio 11:8 to get the new ratio 2:1.?

JulijaS [17]

Answer:

5 should be subtracted from each term

Step-by-step explanation:

$\frac{11-x}{8-x}=\frac{2}{1}$

Cross multiply,

1 * (11 - x) = 2*(8-x)

11 -x = 2*8 - 2*x

11 - x = 16 - 2x

Subtract 11 from both sides,

-x = 16 - 2x - 11

-x = 5 - 2x

Add 2x to both sides

-x +2x = 5 - 2 + 2x

x = 5

7 0

2 years ago

2/3x-2=5/6x

Naddik [55]

Answer:

x = - 12

Step-by-step explanation:

2 5

--- x - 2 = --- x

3 6

2x 5

--- - 2 = ------ x

3 3 * 2

(2x) - 3 * 2 5x

---------------- = --------

3 2 * 3

x = - 12

5 0

3 years ago

Which value for the constant ccc makes w=5w=5w, equals, 5 an extraneous solution in the following equation? \sqrt{29+4w}=23-cw 2

miv72 [106K]

Answer:

c=6

Step-by-step explanation:

In order to solve the original equation, we would have to square both sides of the equation:

\begin{aligned}\sqrt{29+4w}&=23-cw\\\\ \left(\sqrt{29+4w}\right)^2&=(23-cw)^2\\\\ 29+4w&=(23-cw)^2\end{aligned}

29+4w

(

29+4w

)

29+4w

=23−cw

=(23−cw)

However, squaring both sides of an equation can create extraneous solutions! [Why?]

Hint #22 / 4

Let's plug \blueD w=\blueD{5}w=5start color #11accd, w, end color #11accd, equals, start color #11accd, 5, end color #11accd into the last equation we obtained:

\begin{aligned}29+4\blueD w&=(23-c\blueD{w})^2\\\\ 29+4(\blueD 5)&=(23-c(\blueD{5}))^2\\\\ 49&=(23-5c)^2\end{aligned}

29+4w

29+4(5)

=(23−cw)

=(23−c(5))

=(23−5c)

This equation is correct, both when 23-5c=723−5c=723, minus, 5, c, equals, 7 and when 23-5c=-723−5c=−723, minus, 5, c, equals, minus, 7.

However, the original equation is not correct for 23-5c=-723−5c=−723, minus, 5, c, equals, minus, 7, since this way we obtain \sqrt{49}=-7

=−7square root of, 49, end square root, equals, minus, 7.

Hint #33 / 4

Therefore, an extraneous solution is obtained for the ccc-value that makes 23-5c23−5c23, minus, 5, c equal -7−7minus, 7, which is c=6c=6c, equals, 6.

Substituting this back into the original equation gives \sqrt{29+4w}=23-6w

29+4w

=23−6wsquare root of, 29, plus, 4, w, end square root, equals, 23, minus, 6, w. You can now solve this for www and see for yourselves that w=5w=5w, equals, 5 is indeed extraneous.

Hint #44 / 4

The answer is:

c=6c=6

4 0

2 years ago