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SSSSS [86.1K]
2 years ago
9

PRE-FINAL REVIEW

Mathematics
1 answer:
densk [106]2 years ago
5 0

The equation that can be used to represent total tickets sales is 2170 = 5s + 2f + 10a

<h3>Equation</h3>

let

  • Number of students tickets = s
  • Number of faculty tickets = f
  • Number of alumni tickets = a

Expression for number of students tickets sold;

s = f + 15

Expression for number of faculty tickets sold;

f = 2a

Expression for number of alumni tickets sold;

f = 2a

a = f/2

  • Cost of students tickets = $5
  • Cost of faculty tickets = $2
  • Cost of Alumni tickets= $10
  • Total revenue = $2170

2170 = 5s + 2f + 10a

Learn more about equation:

brainly.com/question/4344214

#SPJ1

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Whats the square route of 49?
Vlad [161]

Answer:

The square root of 49 is 7.

Step-by-step explanation:

To find the square root of 49, you would need to look at the possible multiplication equations. For example:

1 x 49 = 49

7 x 7 = 49

49 x 1 = 49

As you can see, 7 x 7 would have the square root of 49, which is 7.

7^2=49

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In conclusion, the square root of 49 is 7

5 0
4 years ago
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stepan [7]
The original price would be $92.38
.25(73.90)=18.475
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3 years ago
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B. Pb/a = pb) is the answer
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3 years ago
An arts academy requires there to be 6 teachers for every 108 students and 5 tutors for every 45 students. How many students doe
prohojiy [21]

Answer:

a) 18 students per teacher

b) 9 students per tutor

c) 7 tutors

Step-by-step explanation:

An arts academy requires there to be 6 teachers for every 108 students and 5 tutors for every 45 students.

a) How many students does the academy have per​ teacher?

6 teachers = 108 students

1 teacher = x

Cross Multiply.

6x = 108

x = 108/6

x = 18 students per teacher

b) Per​ tutor?

5 tutor = 45 students

1 tutor = x

Cross Multiply

5x = 45

x = 45/5

x = 9 students

= 9 students per tutor

c) How many tutors does the academy need if it has 63 students​?

45 students = 5 tutors

63 students = x

Cross Multiply

45x = 63 × 5

x = 63 × 5/45

x = 7 tutors

8 0
3 years ago
When x = 2e, <img src="https://tex.z-dn.net/?f=%5Clim_%7Bh%20%5Cto%200%7D%20%5Cfrac%7Bln%28x%20%2B%20h%29%20-%20ln%28x%29%7D%7Bh
dybincka [34]

Supposing you know about the derivative, notice that

\displaystyle\lim_{h\to0}\frac{\ln(x+h)-\ln x}h=\dfrac{\mathrm d(\ln x)}{\mathrm dx}=\dfrac1x

so that when x=2e, the limit is equal to \dfrac1{2e} and the answer is A.

4 0
4 years ago
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