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if you had a brain you would know that im foxy the pirate so your comment is wrong

thank you so much! i have been stuck on this question for a few hours now.

the junior class is renting a laser tag facility with a capacity of 325 people. the cost for the facility is $1200. the party must have at least 13 adult chaperones

a. if every student who attends shares the facility cost equally, what function models the cost per student C with respect to the number of students N who attend?

Cost per student = $1200/N  

what is the domain of the function?

\(domain: \;\;1\le N \le (312-additional\;adult\;chaperones)\qquad N\in Z\)

how many students must attend to make the cost per student no more than $7.50?

\(\frac{1200}{N}\le7.50\\ \frac{1200}{7.5}\le N\\ N \ge \frac{1200}{7.5}\\ N \ge 160 \)

b. the class wants to promote the event by giving away 30 spots to students in a drawing. how does the model change? now how many paying students must attend so the cost for each is no more than $7.50?

It doesn't change the number of PAYING students who must attend. It will still need to be 160 or more paying students. 

BUT N is not paying students, it is all students so the model does change.

Cost per attending student that is actually paying will be

\(Cost\;per\;paying\;student=\frac{1200}{N-30}\)

The points are the same      x = -4  and   x= -2

the options are 

x = 12 and x = 0

x = -4 and x = 12

x = 12 and x = 12

x = -4 and x = -2

The solutions to f(x) = g(x) are any points that are on both of the graphs...the points where they intersect..   -2,0    -4, 12     

4 x 4 x 4 x 4 = 256

=4ba^(1/2)

I do not know what the rest of your question means.....

From the problem statement, we can make these two inequalities:

x + y  ≤  600            and            5x + 7y  ≥  3500

If  x = 330 , then...

330 + y  ≤  600

y  ≤  270

and

5(330) + 7y  ≥  3500

1650 + 7y  ≥  3500

7y  ≥  1850

y  ≥  264 + 2/7      To sell at least  264 and 2/7  tickets, we have to sell at least 265 tickets.

y  ≥  265

Is there a possible value of  y  such that   y  ≤  270   and   y  ≥  265   ?

yes, for example,  y = 265 , or  y  =  266 , or  y = 267

330 x $5 =$1,650 - money raised before the day of the show

$3,500 - $1,650 =$1,850 - money that must be raised on the day of the show

600 - 330 =270 tickets to be sold on the day of the show

270 x $7 =$1,890 money that CAN be raised on the day of the show.

$1,890 - $1,850 = $40 additional money that WILL be raised if ALL the remaining 270 tickets are sold on the day of the show.

So, the answer is "YES", on the condition that you can sell at least $1,850 / $7 =~ 265 tickets on the day of the show.

V=Van

B=Bus

8V + 8B = 240.........................(1)   

4V + B =54.............................(2)

From (2) above, we have:

B = 54 - 4V sub this into (1) above

8V + 8[54 - 4V] = 240

8V + 432 - 32V = 240

- 24V = - 192

V = 8 - Number of students in each Van

Sub this into (2) above:

[4x8] + B = 54

B = 22 - Number of students in each Bus.

i am

thank you

I hope you're doing well too Foxy.

And ignore the annoying guest.

x = 500mg  price

y= 250 mg price

1800x + 2200y =39200

and

2200x + 2200y =44000  

Subtract second from first equation

-400x = -4800     x = 12       

So 12 bucks for 500 mg bottle 

1800(12) + 2200y = 39200         y= 8     $ 8 for 250 mg tab bottle

V=Van

B=Bus

Solve the following system:

{6 B + V = 372 | (equation 1)

12 B + 4 V = 780 | (equation 2)

Swap equation 1 with equation 2:

{4 V + 12 B = 780 | (equation 1)

V + 6 B = 372 | (equation 2)

Subtract 1/4 × (equation 1) from equation 2:

{4 V + 12 B = 780 | (equation 1)

0 V+3 B = 177 | (equation 2)

Divide equation 1 by 4:

{V + 3 B = 195 | (equation 1)

0 V+3 B = 177 | (equation 2)

Divide equation 2 by 3:

{V + 3 B = 195 | (equation 1)

0 V+B = 59 | (equation 2)

Subtract 3 × (equation 2) from equation 1:

{V+0 B = 18 | (equation 1)

0 V+B = 59 | (equation 2)

Collect results:

V = 18 - Number of students in each Van

B = 59 - Number of students in each Bus.

x + y = 6

4x + 3.50y = 22, solve for x, y

Solve the following system:

{x + y = 6

4 x + 3.5 y = 22

In the first equation, look to solve for y:

{x + y = 6

4 x + 3.5 y = 22

Subtract x from both sides:

{y = 6 - x

4 x + 3.5 y = 22

Substitute y = 6 - x into the second equation:

{y = 6 - x

3.5 (6 - x) + 4 x = 22

3.5 (6 - x) + 4 x = 4 x + (21. - 3.5 x) = 0.5 x + 21.:

{y = 6 - x

0.5 x + 21. = 22

In the second equation, look to solve for x:

{y = 6 - x

0.5 x + 21. = 22

0.5 x + 21. = x/2 + 21:

x/2 + 21 = 22

Subtract 21 from both sides:

{y = 6 - x

x/2 = 1

x = 2 - Pounds of nuts

y = 4 - Pounds of dried fruit

1.   The square root of a negative number is not real.  -9/16  has no real square roots.

2.   \(\sqrt[3]{-0.000125} \,=\,\sqrt[3]{(-0.05)^3}\,=\,-0.05\)

Solve for y:

y^3 + 2 y^2 + 16 y + 32 = 0

The left hand side factors into a product with two terms:

(y + 2) (y^2 + 16) = 0

Split into two equations:

y + 2 = 0 or y^2 + 16 = 0

Subtract 2 from both sides:

y = -2 or y^2 + 16 = 0

Subtract 16 from both sides:

y = -2 or y^2 = -16

Take the square root of both sides:

y = -2    or    y = 4i    or    y = -4i

Solve for x:

x^2 + 20 = 0

Subtract 20 from both sides:

x^2 = -20

Take the square root of both sides:

x = (2i) sqrt(5)      or       x = (-2i) sqrt(5)