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3 years ago

6

Tickets to the college basketball game are $2.50 for students and $3 for general admission. If 54 people attended the last game

and the box office collected $149, how many of each type of ticket did they sell?

1 answer:

Yuki888 [10]3 years ago

8 0

Answer:

28 general admission
26 student

Step-by-step explanation:

Let g represent the number of general admission tickets sold. Then the revenue is ...

3g +2.50(54-g) = 149

.50g = 14 . . . . . . . . . . . subtract 135

g = 14/.5 = 28 . . general admission tickets sold

54-g = 26 . . . . . student tickets sold

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A particle moving in a planar force field has a position vector x that satisfies x'=Ax. The 2×2 matrix A has eigenvalues 4 and 2

andrey2020 [161]

Answer:

The required position of the particle at time t is: $x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

Step-by-step explanation:

Consider the provided matrix.

$v_1=\begin{bmatrix}-3\\1 \end{bmatrix}$

$v_2=\begin{bmatrix}-1\\1 \end{bmatrix}$

$\lambda_1=4, \lambda_2=2$

The general solution of the equation $x'=Ax$

$x(t)=c_1v_1e^{\lambda_1t}+c_2v_2e^{\lambda_2t}$

Substitute the respective values we get:

$x(t)=c_1\begin{bmatrix}-3\\1 \end{bmatrix}e^{4t}+c_2\begin{bmatrix}-1\\1 \end{bmatrix}e^{2t}$

$x(t)=\begin{bmatrix}-3c_1e^{4t}-c_2e^{2t}\\c_1e^{4t}+c_2e^{2t} \end{bmatrix}$

Substitute initial condition $x(0)=\begin{bmatrix}-6\\1 \end{bmatrix}$

$\begin{bmatrix}-3c_1-c_2\\c_1+c_2 \end{bmatrix}=\begin{bmatrix}-6\\1 \end{bmatrix}$

Reduce matrix to reduced row echelon form.

$\begin{bmatrix} 1& 0 & \frac{5}{2}\\ 0& 1 & \frac{-3}{2}\end{bmatrix}$

Therefore, $c_1=2.5,c_2=1.5$

Thus, the general solution of the equation $x'=Ax$

$x(t)=2.5\begin{bmatrix}-3\\1\end{bmatrix}e^{4t}-1.5\begin{bmatrix}-1\\1 \end{bmatrix}e^{2t}$

$x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

The required position of the particle at time t is: $x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

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zalisa [80]

Answer:

$360xy^2|xz^3|$

Step-by-step explanation:

Given expression:

$12x \sqrt{900x^2y^4z^6}$

$\textsf{Apply radical rule} \quad \sqrt{ab}=\sqrt{a}\sqrt{b}:$

$\implies 12x \sqrt{900}\sqrt{x^2}\sqrt{y^4}\sqrt{z^6}$

Replace 900 with 30² :

$\implies 12x \sqrt{30^2}\sqrt{x^2}\sqrt{y^4}\sqrt{z^6}$

$\textsf{Apply radical rule} \quad \sqrt{a^2}=a, \quad a \geq 0:$

$\implies 12x \cdot 30|x|\sqrt{y^4}\sqrt{z^6}$

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(We need to use the absolute value of √x² since the x term was originally to the power of 2, which means the value of x² is always positive since the exponent is even).

$\textsf{Apply exponent rule} \quad \sqrt{a^m}=a^{\frac{m}{2}}:$

$\implies 360x|x|\cdot y^{\frac{4}{2}}\cdot z^{\frac{6}{2}}$

Simplify:

$\implies 360xy^2|xz^3|$

(We need to use the absolute value of z³ since the z term was original to the power of 6, which means the value of z⁶ is always positive since the exponent is even).

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