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alisha [4.7K]
3 years ago
9

The admission fee at a movie theater is $5 for children and $9 for adults. If 3200 people go to the movies and $24000 is collect

ed, how many adults went to the movies?
Mathematics
1 answer:
Usimov [2.4K]3 years ago
6 0
For this problem, let x be the number of children and y for adults. Formulate the equations: 1st equation, x + y = 3,200 and 2nd equation 5x + 9y = 24,000. Re-arrange 1st equation into x = 3200 - y. Then, substitute into 2nd equation, 5(3,200-y) + 9y = 24,000. Then, solve for y. The 16,000 - 5y + 9y = 24000. Final answer is, y = 2000 adults went to watch the movie.
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Why are there no properties of subtraction or division?
tia_tia [17]
There are no properties because you can do anything with subtraction and division
4 0
3 years ago
Find all points on the portion of the plane x+y+z=5 in the first octant at which f(x, y, z) = xy2z2 has a maximum value.
irina1246 [14]
Lagrange multipliers:

L(x,y,z,\lambda)=xy^2z^2+\lambda(x+y+z-5)

L_x=y^2z^2+\lambda=0
L_y=2xyz^2+\lambda=0
L_z=2xy^2z+\lambda=0
L_\lambda=x+y+z-5=0

\lambda=-y^2z^2=-2xyz^2=-2xy^2z

-y^2z^2=-2xyz^2\implies y=2x (if y,z\neq0)

-y^2z^2=-2xy^2z\implies z=2x (if y,z\neq0)

-2xyz^2=-2xy^2z\implies z=y (if x,y,z\neq0)

In the first octant, we assume x,y,z>0, so we can ignore the caveats above. Now,

x+y+z=5\iff x+2x+2x=5x=5\implies x=1\implies y=z=2

so that the only critical point in the region of interest is (1, 2, 2), for which we get a maximum value of f(1,2,2)=16.

We also need to check the boundary of the region, i.e. the intersection of x+y+z=5 with the three coordinate axes. But in each case, we would end up setting at least one of the variables to 0, which would force f(x,y,z)=0, so the point we found is the only extremum.
4 0
3 years ago
Rename the number 120,000
jeka94

Answer:

One Hundred Twenty Thousand

Step-by-step explanation:

I hope this is waht your looking for. But you were a bit unclear.

4 0
3 years ago
Please help me with this question :)​
klio [65]

Answer:

y>5/6

Step-by-step explanation:

5y+3>-7y+13

5y+7y>13-3

12y/12>10/12

y>5/6

This is an inequality, which shows where the variable y does or does not exist

from this we can say

y: (-infinity,5/6) (5/6, infinity]

or, y does not exist from (-infinity,5/6)

and does exist from (5/6, infinity]

3 0
2 years ago
Is (2, 3) a solution to the inequality y > 6x + 3?
Sindrei [870]

Good morning

Answer:

NO

Step-by-step explanation:

Here we just substitute x and y by their values and see if the inequality still true or not

y = 3

6x + 3 = 6(2) + 3 = 15

since 3 < 15 then the inequality y > 6x + 3 is wrong.

______________________________________

:)

8 0
3 years ago
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