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3 years ago

100 POINTS HELP ME!!!!!!!

Mathematics

2 answers:

trasher [3.6K]3 years ago

3 0

Given : A theater company has raised $930.75 by selling 25 floor seat tickets. Each ticket costs the same.

To Find : Write an equation with a variable that can be solved to correctly find the price of each ticket.

the price of each floor seat ticket.

Solution:

Let say price of each ticket = x $

then price of 25 tickets = 25x $

Price of 25 tickets = $930.75

Equating both

25x = 930.75

is required equation

25x = 930.75

Divide both sides by 25

=> x = 930.75/25

=> x = 37.23 $

price of each floor seat ticket. = 37.23 $

also, there was another question like this.

dimaraw [331]3 years ago

3 0

Given : A theater company has raised $930.75 by selling 25 floor seat tickets. Each ticket costs the same.

To Find : Write an equation with a variable that can be solved to correctly find the price of each ticket.

the price of each floor seat ticket.

Solution:

Let say price of each ticket = x $

then price of 25 tickets = 25x $

Price of 25 tickets = $930.75

Equating both

25x = 930.75

is required equation

25x = 930.75

Divide both sides by 25

=> x = 930.75/25

=> x = 37.23 $

price of each floor seat ticket. = 37.23 $

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If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar

erma4kov [3.2K]

Answer:

$\frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{120}$

Step-by-step explanation:

Given

5 tuples implies that:

$n = 5$

$(h,i,j,k,m)$ implies that:

$r = 5$

Required

How many 5-tuples of integers $(h, i, j, k,m)$ are there such that $n\ge h\ge i\ge j\ge k\ge m\ge 1$

From the question, the order of the integers h, i, j, k and m does not matter. This implies that, we make use of combination to solve this problem.

Also considering that repetition is allowed: This implies that, a number can be repeated in more than 1 location

So, there are n + 4 items to make selection from

The selection becomes:

$^{n}C_r => ^{n + 4}C_5$

$^{n + 4}C_5 = \frac{(n+4)!}{(n+4-5)!5!}$

$^{n + 4}C_5 = \frac{(n+4)!}{(n-1)!5!}$

Expand the numerator

$^{n + 4}C_5 = \frac{(n+4)!(n+3)*(n+2)*(n+1)*n*(n-1)!}{(n-1)!5!}$

$^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{5!}$

$^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{5*4*3*2*1}$

$^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{120}$

Solved

6 0

3 years ago

How do u guys solve this?

enyata [817]

Just graph it in the calculator

6 0

3 years ago

Reduce the fraction: x-y/x^2-1 times x-1/x^2-y^2

Dominik [7]

Answer:

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Molodets [167]

Answer:

x = (125,0) y = (0,275)

Step-by-step explanation:

I think this is right, good luck!

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3 years ago

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zvonat [6]

Answer:

5. m<1 = 114 degrees

m<2 = 66 degrees

6. 2x + 3x + 4x = 9x = 180, x = 20

7. 3

8. 1/3

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Step-by-step explanation:

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