Ask question

Ask question

All categories

3 years ago

11

Suppose a movie theater sold 200 adult and student tickets for a showing with a revenue of $980. If the adult tickets were $6.50

and the student tickets were $4.50, how many of each type of ticket were sold? Help!!! I need this quick!!!

1 answer:

frosja888 [35]3 years ago

4 0

Answer:

Adult - 40

Student - 160

Step-by-step explanation:

You might be interested in

Find the absolute maximum and minimum values of the following function on the specified region R.

yan [13]

Answer:

Step-by-step explanation:

Since it is said that the region R is a semicircular disc, we asume that the boundaries of the region are given by $-2\leq x \leq 2, 0\leq y \leq \sqrt[]{4-x^2}$ . First, we must solve the optimization problem without any restrictions, and see if the points we get lay inside the region of interest. To do so, consider the given function F(x,y). We want to find the point for which it's gradient is equal to zero, that is

$\frac{dF}{dx}=9y=0$

$\frac{dF}{dy}=9x=0$

This implies that (x,y) = (0,0). This point lays inside the region R. We will use the Hessian criteria to check if its a minimum o r a maximum. To do so, we calculate the matrix of second derivates

$\frac{d^2F}{dx^2} = 0 = \frac{d^2F}{dy^2}$

$\frac{d^2F}{dxdy} = 9 = \frac{d^2F}{dydx}$

so we get the matrix

$\left[\begin{matrix} 0 & 9 \\ 9 & 0\end{matrix}\right]$

Note that the first determinant is 0, and the second determinant is -9. THis tell us that the point is a saddle point, hence not a minimum nor maximum.

Since the function is continous and the region R is closed and bound (hence compact) the maximum and minimum must be attained on the boundaries of R. REcall that when $-2\leq x \leq x$ and y=0 we have that F(x,0) = 0. So, we want to pay attention to the critical values over the circle, restricting that the values of y must be positive. To do so, consider the following function

$H(x,y, \lambda) = 9xy - \lambda(x^2+y^2-4)$ which consists of the original function and a function that describes the restriction (the circle x^2+y^2=4), we want that the gradient of H is 0.

Then,

$\frac{dH}{dx} = 9y-2\lambda x =0$

$\frac{dH}{dx} = 9x-2\lambda y =0$

$\frac{dH}{d\lambda} = x^2+y^2-4 =0$

From the first and second equation we get that

$\lambda = \frac{9y}{2x} = \frac{9x}{2y}$

which implies that $y^2=x^2$ . If we replace this in the restriction, we have that $x^2+x^2 = 2x^2 = 4$ which gives us that $x=\pm \sqrt[]{2}$ . Since we only care for the positive values of y, and that $y=\pm x$ , we have the following critical points $(\sqrt[]{2},\sqrt[]{2}), (-(\sqrt[]{2},\sqrt[]{2})$ . Note that for the first point, the value of the function is

$F(\sqrt[]{2},\sqrt[]{2}) = 9\cdot 2 =18$

as for the second point the value of the function is

$F(-\sqrt[]{2},\sqrt[]{2}) = 9\cdot -2 =-18$ .

Then, the point $(\sqrt[]{2},\sqrt[]{2})$ is a maximum and the point $(-(\sqrt[]{2},\sqrt[]{2})$ is a minimum.

6 0

3 years ago

Clear parentheses and combine like terms:

Nuetrik [128]

Answer:

2x - 6y + 16

3rd option

Step-by-step explanation:

Step 1: Write out expression

2(x - 3y + 8)

Step 2: Distribute

2x - 6y + 16

Since there is no other like terms to combine, that is our final answer.

4 0

3 years ago

What would be the radius of A= 530.66 square meter

alekssr [168]

530.66÷3.14=169
√169=√r²
13=r

4 0

3 years ago

Marking brainliest!!!

natka813 [3]

Answer:

its a right angle!

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Area of square i calculated by using formula (side)2., true or false

krok68 [10]

The calculate the area of a square multiply the base times the height.

5 0

3 years ago