find the inverse of the following matrix
\\begin{bmatrix}
6 & -5 \\
3 & -2
\end{bmatrix}\
select one:
$\circ$ a. $\begin{bmatrix} -6 & 5 \\ -3 & 2 \end{bmatrix}$
$\circ$ b. $\begin{bmatrix} -2 & 5 \\ -3 & 6 \end{bmatrix}$
$\circ$ c. $\begin{bmatrix} -\frac{2}{3} & \frac{5}{3} \\ -1 & 2 \end{bmatrix}$
$\circ$ d. undefined

a frog jumps off the bank of a creek with an initial velocity of 2.5 ft/sec at an angle of 45° with the horizontal. the frog is 1 ft above the base of the creek when he jumps off. write the parametric equations that model this situation.
select one:
$\circ$ a. $x = 2.5\cos45^\circ t$ and $y = 2.5\sin45^\circ t - 16t^2$
$\circ$ b. $x = 2.5\cos45^\circ t$ and $y = 2.5\sin45^\circ t - 16t^2 + 1$
$\circ$ c. $x = 2.5\cos45^\circ t$ and $y = 2.5\sin45^\circ t + 16t^2 - 1$
$\circ$ d. $x = 2.5\sin45^\circ t$ and $y = 2.5\cos45^\circ t - 16t^2 + 1$

Question

find the inverse of the following matrix
\\begin{bmatrix}
6 & -5 \\
3 & -2
\end{bmatrix}\
select one:
$\circ$ a. $\begin{bmatrix} -6 & 5 \\ -3 & 2 \end{bmatrix}$
$\circ$ b. $\begin{bmatrix} -2 & 5 \\ -3 & 6 \end{bmatrix}$
$\circ$ c. $\begin{bmatrix} -\frac{2}{3} & \frac{5}{3} \\ -1 & 2 \end{bmatrix}$
$\circ$ d. undefined

a frog jumps off the bank of a creek with an initial velocity of 2.5 ft/sec at an angle of 45° with the horizontal. the frog is 1 ft above the base of the creek when he jumps off. write the parametric equations that model this situation.
select one:
$\circ$ a. $x = 2.5\cos45^\circ t$ and $y = 2.5\sin45^\circ t - 16t^2$
$\circ$ b. $x = 2.5\cos45^\circ t$ and $y = 2.5\sin45^\circ t - 16t^2 + 1$
$\circ$ c. $x = 2.5\cos45^\circ t$ and $y = 2.5\sin45^\circ t + 16t^2 - 1$
$\circ$ d. $x = 2.5\sin45^\circ t$ and $y = 2.5\cos45^\circ t - 16t^2 + 1$

Accepted Answer

### First Question (Matrix Inverse)
# Answer:
b. $\begin{bmatrix}-2&5\-3&6\end{bmatrix}$
# Explanation:
## Step1: Recall inverse formula
For a $2	imes2$ matrix $\begin{bmatrix}a&b\c&d\end{bmatrix}$, the inverse is $\frac{1}{ad - bc}\begin{bmatrix}d&-b\-c&a\end{bmatrix}$, where $ad - bc$ is the determinant.
## Step2: Calculate determinant
For $\begin{bmatrix}6&-5\3&-2\end{bmatrix}$, determinant $= (6)(-2)-(-5)(3)= -12 + 15 = 3$.
## Step3: Apply inverse formula
Inverse $=\frac{1}{3}\begin{bmatrix}-2&5\-3&6\end{bmatrix}$? Wait, no, wait: Wait, the formula is $\frac{1}{\det}\begin{bmatrix}d&-b\-c&a\end{bmatrix}$. So $a = 6, b=-5, c = 3, d=-2$. So inverse is $\frac{1}{(6)(-2)-(-5)(3)}\begin{bmatrix}-2&5\-3&6\end{bmatrix}=\frac{1}{-12 + 15}\begin{bmatrix}-2&5\-3&6\end{bmatrix}=\frac{1}{3}\begin{bmatrix}-2&5\-3&6\end{bmatrix}$? Wait, but option b is $\begin{bmatrix}-2&5\-3&6\end{bmatrix}$. Wait, maybe I made a mistake. Wait, no, wait: Wait, the determinant is $6	imes(-2)-(-5)	imes3 = -12 + 15 = 3$. Then the inverse is $\frac{1}{3}\begin{bmatrix}-2&5\-3&6\end{bmatrix}$? But option b is $\begin{bmatrix}-2&5\-3&6\end{bmatrix}$. Wait, maybe the problem has a typo, or maybe I miscalculated. Wait, no, wait: Wait, let's check the options. Option b is $\begin{bmatrix}-2&5\-3&6\end{bmatrix}$. Let's multiply the original matrix with option b: $\begin{bmatrix}6&-5\3&-2\end{bmatrix}\begin{bmatrix}-2&5\-3&6\end{bmatrix}=\begin{bmatrix}6	imes(-2)+(-5)	imes(-3)&6	imes5+(-5)	imes6\3	imes(-2)+(-2)	imes(-3)&3	imes5+(-2)	imes6\end{bmatrix}=\begin{bmatrix}-12 + 15&30 - 30\-6 + 6&15 - 12\end{bmatrix}=\begin{bmatrix}3&0\0&3\end{bmatrix}=3I$. So if we divide by 3, we get $I$. So the inverse is $\frac{1}{3}\begin{bmatrix}-2&5\-3&6\end{bmatrix}$, but option b is $\begin{bmatrix}-2&5\-3&6\end{bmatrix}$, which is 3 times the inverse? Wait, no, wait, maybe I messed up the formula. Wait, no, the formula is correct. Wait, maybe the question is wrong, or maybe I made a mistake. Wait, no, let's recalculate the determinant: $6*(-2) - (-5)*3 = -12 + 15 = 3$. Correct. Then inverse is (1/3)*[[ -2, 5], [ -3, 6]]. But option b is [[-2,5],[-3,6]]. So maybe the question intended the determinant to be 1, but no. Wait, maybe the options are wrong, but among the options, option b when multiplied by the original matrix gives 3I, which is a scalar multiple of identity, so maybe the question has a mistake, but option b is the closest. Alternatively, maybe I made a mistake. Wait, let's check option a: $\begin{bmatrix}-6&5\-3&2\end{bmatrix}$. Multiply with original: $\begin{bmatrix}6&-5\3&-2\end{bmatrix}\begin{bmatrix}-6&5\-3&2\end{bmatrix}=\begin{bmatrix}6*(-6)+(-5)*(-3)&6*5+(-5)*2\3*(-6)+(-2)*(-3)&3*5+(-2)*2\end{bmatrix}=\begin{bmatrix}-36 + 15&30 - 10\-18 + 6&15 - 4\end{bmatrix}=\begin{bmatrix}-21&20\-12&11\end{bmatrix}
eq I$. Option c: $\begin{bmatrix}-\frac{2}{3}&\frac{5}{3}\-1&2\end{bmatrix}$. Multiply with original: $\begin{bmatrix}6&-5\3&-2\end{bmatrix}\begin{bmatrix}-\frac{2}{3}&\frac{5}{3}\-1&2\end{bmatrix}=\begin{bmatrix}6*(-\frac{2}{3})+(-5)*(-1)&6*(\frac{5}{3})+(-5)*2\3*(-\frac{2}{3})+(-2)*(-1)&3*(\frac{5}{3})+(-2)*2\end{bmatrix}=\begin{bmatrix}-4 + 5&10 - 10\-2 + 2&5 - 4\end{bmatrix}=\begin{bmatrix}1&0\0&1\end{bmatrix}$. Oh! Wait, I made a mistake earlier. Let's recalculate option c. Wait, option c is $\begin{bmatrix}-\frac{2}{3}&\frac{5}{3}\-1&2\end{bmatrix}$. Multiply with original matrix:

First row first column: $6*(-\frac{2}{3}) + (-5)*(-1) = -4 + 5 = 1$.

First row second column: $6*(\frac{5}{3}) + (-5)*2 = 10 - 10 = 0$.

Second row first column: $3*(-\frac{2}{3}) + (-2)*(-1) = -2 + 2 = 0$.

Second row second column: $3*(\frac{5}{3}) + (-2)*2 = 5 - 4 = 1$.

Oh! So option c is the inverse. Wait, but earlier I thought option b was, but I made a mistake. Wait, let's recalculate the inverse properly. The formula is $\frac{1}{\det}\begin{bmatrix}d&-b\-c&a\end{bmatrix}$. So $\det = 3$, so inverse is $\frac{1}{3}\begin{bmatrix}-2&5\-3&6\end{bmatrix}=\begin{bmatrix}-\frac{2}{3}&\frac{5}{3}\-1&2\end{bmatrix}$, which is option c. Oh! I see, I made a mistake earlier when checking option c. So the correct answer is c? Wait, no, wait, let's check again. Wait, the original matrix is $\begin{bmatrix}6&-5\3&-2\end{bmatrix}$. The inverse should be $\frac{1}{(6)(-2)-(-5)(3)}\begin{bmatrix}-2&5\-3&6\end{bmatrix}=\frac{1}{3}\begin{bmatrix}-2&5\-3&6\end{bmatrix}=\begin{bmatrix}-\frac{2}{3}&\frac{5}{3}\-1&2\end{bmatrix}$, which is option c. So my initial mistake was in calculating option c. So the correct answer is c? Wait, but let's check the options again. Option c is $\begin{bmatrix}-\frac{2}{3}&\frac{5}{3}\-1&2\end{bmatrix}$. Yes, that's the inverse. So I made a mistake earlier. So the correct answer is c? Wait, no, wait, let's multiply option c with the original matrix:

$\begin{bmatrix}6&-5\3&-2\end{bmatrix}\begin{bmatrix}-\frac{2}{3}&\frac{5}{3}\-1&2\end{bmatrix}=\begin{bmatrix}6*(-\frac{2}{3}) + (-5)*(-1)&6*(\frac{5}{3}) + (-5)*2\3*(-\frac{2}{3}) + (-2)*(-1)&3*(\frac{5}{3}) + (-2)*2\end{bmatrix}=\begin{bmatrix}-4 + 5&10 - 10\-2 + 2&5 - 4\end{bmatrix}=\begin{bmatrix}1&0\0&1\end{bmatrix}$. Yes! So option c is the inverse. So my earlier mistake was in checking option b. So the correct answer is c? Wait, but the options are:

a. $\begin{bmatrix}-6&5\-3&2\end{bmatrix}$

b. $\begin{bmatrix}-2&5\-3&6\end{bmatrix}$

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

d. undefined

So the correct answer is c. Wait, but let's recalculate the determinant again: $6*(-2) - (-5)*3 = -12 + 15 = 3
eq 0$, so inverse exists. So the inverse is $\frac{1}{3}\begin{bmatrix}-2&5\-3&6\end{bmatrix}=\begin{bmatrix}-\frac{2}{3}&\frac{5}{3}\-1&2\end{bmatrix}$, which is option c. So the correct answer is c.

### Second Question (Parametric Equations)
# Answer:
b. $x = 2.5\cos45^{\circ}t$ and $y = 2.5\sin45^{\circ}t - 16t^2 + 1$
# Explanation:
## Step1: Horizontal motion (x - component)
The horizontal velocity is $v_0\cos	heta$, where $v_0 = 2.5$ ft/sec and $	heta = 45^{\circ}$. So $x = v_0\cos	heta \cdot t = 2.5\cos45^{\circ}t$.
## Step2: Vertical motion (y - component)
The vertical motion is a projectile motion with initial vertical velocity $v_0\sin	heta$, acceleration due to gravity $-32$ ft/sec² (so the quadratic term is $\frac{1}{2}at^2 = -16t^2$), and initial height $1$ ft. So the vertical position $y = v_0\sin	heta \cdot t + \frac{1}{2}at^2 + h_0 = 2.5\sin45^{\circ}t - 16t^2 + 1$.

For the first question, after correcting the earlier mistake, the correct answer is:

# Answer (First Question):
c. $\begin{bmatrix}-\frac{2}{3}&\frac{5}{3}\-1&2\end{bmatrix}$
# Explanation (First Question):
## Step1: Recall inverse formula for $2	imes2$ matrix $\begin{bmatrix}a&b\c&d\end{bmatrix}$: $\frac{1}{ad - bc}\begin{bmatrix}d&-b\-c&a\end{bmatrix}$, where $\det = ad - bc$.
## Step2: Calculate $\det$ for $\begin{bmatrix}6&-5\3&-2\end{bmatrix}$: $\det = (6)(-2) - (-5)(3) = -12 + 15 = 3$.
## Step3: Apply inverse formula: $\frac{1}{3}\begin{bmatrix}-2&5\-3&6\end{bmatrix} = \begin{bmatrix}-\frac{2}{3}&\frac{5}{3}\-1&2\end{bmatrix}$, which matches option c.

For the second question, the correct answer is:

# Answer (Second Question):
b. $x = 2.5\cos45^{\circ}t$ and $y = 2.5\sin45^{\circ}t - 16t^2 + 1$
# Explanation (Second Question):
## Step1: Horizontal motion: $x$ - component of velocity is $v_0\cos	heta$, so $x = 2.5\cos45^{\circ}t$.
## Step2: Vertical motion: $y$ - component has initial velocity $v_0\sin	heta$, acceleration $-32$ ft/sec² (so $\frac{1}{2}at^2 = -16t^2$), and initial height $1$ ft, so $y = 2.5\sin45^{\circ}t - 16t^2 + 1$.

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QUESTION IMAGE

Question

Explanation:

First Question (Matrix Inverse)

Step1: Recall inverse formula

Step2: Calculate determinant

Step3: Apply inverse formula

Step1: Horizontal motion (x - component)

Step2: Vertical motion (y - component)

Step1: Recall inverse formula for $2\times2$ matrix

Step2: Calculate $\det$ for

Step3: Apply inverse formula: $\frac{1}{3}

Answer: