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allsm [11]
1 year ago
14

For a school drama performance, student tickets cost $5 each and adult tickets cost $10 each. The sellers collected $3,570 from

512 tickets sold. If c is the number of student tickets sold, which equation can be used to find the amount of tickets sold of each type?
10(512 – c) + 5c = 512
10(512 – c) + 5c = 3,570
5(512 – c) + 10c = 3,570
5(512 – c) + 10c 512
Mathematics
2 answers:
Mumz [18]1 year ago
5 0

Answer: 10(512 – c) + 5c = 3,570

Step-by-step explanation:

Given : The number of student tickets sold is represented by c.

For a school drama performance, student tickets cost $5 each and adult tickets cost $10 each.

The sellers collected $3,570 from 512 tickets sold.

Since , the total tickets sold = 512

Therefore , the number of adults ticket = 512-c

Now, cost of total child tickets : 5c

Cost of total adult tickets = 10(512 – c)

Now, according to the cost of the tickets , we have the following equation :-

10(512 – c) + 5c = 3,570

olchik [2.2K]1 year ago
4 0
I think the right equation is 10(512 - c) plus 5c equals 3,570
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2 years ago
I WILL AWARD BRAINLIEST!! PLEASE HELP!!! The figure below shows the movement of a pedestrian from point B to point E. Using the
MissTica

A) The speed of the pedestrian BC is 5 km/h

    The speed of the pedestrian CD is 0 km/h

    The speed of the pedestrian DE is 5 km/h

B) He arrived E since the stop after 6 hours

C) The formula for section BC is d(t) = 40 - 5t

    The formula for section CD is d(t) = 20

    The formula for section DE is d(t) = 50 - 5t

Step-by-step explanation:

A)

In the time-distance graph the speed is the rate of change of distance

and that mean speed = Δd/Δt ⇒ (slope of the line)

In line BC:

1. Δd = 40 - 20 = 20 km

2. Δt = 4 - 0 = 4 hours

3. The speed = 20 ÷ 4 = 5 km/h

The speed of the pedestrian BC is 5 km/h

In line CD:

1. Δd = 20 - 20 = 0 km

2. Δt = 6 - 4 = 2 hours

3. The speed = 0 ÷ 2 = 0 km/h

The speed of the pedestrian CD is 0 km/h

In line DE:

1. Δd = 20 - 0 = 20 km

2. Δt = 10 - 6 = 4 hours

3. The speed = 20 ÷ 4 = 5 km/h

The speed of the pedestrian DE is 5 km/h

B)

∵ He stop at t = 4 hours

∵ He arrived at point E at t = 10 hours

∵ 10 - 4 = 6 hours

He arrived E since the stop after 6 hours

C)

The speeds are represented by lines

The form of the equation of a line is f(x) = mx + c, where m represents

the slope of the line and c is the y-intercept (y when x = 0)

1. f(x) is d(t)

2. m is the speed

3. x is t

4. You can find c by substitute d and t by any point on the line

Line BC

Line BC has negative slope because d decreases when t increases

∵ m = -5 and c = 40

∴ d(t) = 40 - 5t

The formula for section BC is d(t) = 40 - 5t

Line CD

Line CD is a horizontal line (equation any horizontal line is y = c)

∴ m = 0 and c = 20

∴ d(t) = 20

The formula for section CD is d(t) = 20

Line DE

Line DE has negative slope because d decreases when t increases

∵ m = -5

∴ d(t) = -5t + c

To find c substitute the coordinates of point D in the equation

∵ The coordinates of point D are (6 , 20)

∴ 20 = -5(6) + c

∴ 20 = -30 + c

Add 30 to both sides

∴ c = 50

∴ d(t) = 50 - 5t

The formula for section DE is d(t) = 50 - 5t

Learn more:

You can learn more about the distance, speed, and time in

brainly.com/question/5102020

#LearnwithBrainly

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Given that the random sample size is n=25

let x represent number of customers who purchase running shoes

Let "p" be the probability of customers in a sporting goods store purchase a pair of running shoes.

It is given that 70% of the customers in a sporting goods store purchase a pair of running shoes.

Thus p=\frac{70}{100}=0.7

Thus the Probability distribution of x is given by

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Seth is using the figure shown below to prove the Pythagorean Theorem using triangle similarity: In the given triangle PQR, angl
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Answer:

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Part B. All angles are same, so the triangles are similar.  

Part C. RP = 8

Step-by-step explanation:

We are given a right angled triangle \triangle RPQ with \angle P = 90^\circ.

PS is perpendicular to the hypotenuse RQ of \triangle RPQ and S lies on RQ.

Part A:

To identify the pair of similar triangles.

\triangle RPQ \sim \triangle RSP.

Part B:

To identify the type of similarity.

Kindly refer to the image attached in the answer area.

Let us consider the triangles \triangle RPQ \ and\ \triangle RSP.

\angle RSP =\angle RPQ =90^\circ

Also, \angle R is common to both the triangles under consideration.

Now, we can see that two angles of two triangles are equal.

So, third angle of the two triangles will also be same.

i.e. All three angles of two triangles \triangle RPQ \ and\ \triangle RSP are equal to each other.

So, by A-A-A (Angle - Angle - Angle) similarity, we can say that \triangle RPQ \sim \triangle RSP.

Part C:

RS = 4

RQ = 16, Find RP.

There is one property of similar triangles that:

The ratio of corresponding sides of two similar triangles is equal.

i.e.

\dfrac{RS}{RP} = \dfrac{RP}{RQ}\\\Rightarrow RP ^2 = RS \times RQ\\\Rightarrow RP ^2 = 4 \times 16\\\Rightarrow RP ^2 = 64\\\Rightarrow \bold{RP = 8\ units}

5 0
2 years ago
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