To calculate the age of the pig in human years, first recall the equation that given in the question.

\(\displaystyle H=2p+4\)

Now, identify the known information given in the question.

Since the pig is four years old, the mathematical statement is

\(\displaystyle p=4\)

Next, substitute in this value into the equation to calculate the human age.

\(\displaystyle H=2p+4\)

\(\displaystyle \\H=2(4)+4 \\H=8+4 \\H=12\)

Therefore, the four year old pig is twelve years old in human years.

To graph the function

\(\displaystyle f(x)=1-2x\),

create an \(\displaystyle x\), \(\displaystyle y\) table to find values and then plot those values on a coordinate grid and connect the points.

\(\displaystyle \begin{tabular}{|c|c|} \hline x &y\\ \hline -1&3\\ 0&1\\ 1&-1\\ 2&-3\\ \hline \end{tabular}\)

\(\displaystyle y = -4x + 5\) is in slope-intercept form.

To get

\(\displaystyle 8x+2 y = 7\)

in slope-intercept form, add \(\displaystyle -8x\) to both sides, then multiply both sides by \(\displaystyle \frac{1}{2}\), as follows:

\(\displaystyle 8x+2 y + (-8x) = 7 + (-8x)\)

\(\displaystyle 2y = -8x + 7\)

\(\displaystyle \frac{1}{2} (2y )= \frac{1}{2} (-8x + 7)\)

\(\displaystyle y = -4x+ \frac{7}{2}\)

The system can thus be stated as

\(\displaystyle y = -4x + 5\)

\(\displaystyle y = -4x+ \frac{7}{2}\)

The slope of each line is x-coefficient \(\displaystyle -4\); the lines have different y-intercepts. This makes them distinct parallel lines.

To solve this problem, first recall the formula to calculate simple interest.

\(\displaystyle I=A_0\times r\times t\)

where

\(\displaystyle \\A_0=\text{initial amount} \\r=\text{rate} \\t=\text{time} \\I=\text{interest accumulated}\)

From the question it's known that,

\(\displaystyle \\A_0=1350\\r=0.13 \\t=4\)

Thus the equation becomes,

\(\displaystyle \\I=1350\times 0.13\times 4 \\I=702\)

Therefore $702 in interested accumulated after four years.

If \(\displaystyle P_{0}\) is deposited in a bank account with interest rate \(\displaystyle r\) ( converted to a decimal) compounded \(\displaystyle n\) times a year for \(\displaystyle t\) years, then the final balance of the account is

\(\displaystyle P = P_{0} \left (1+ \frac{r}{n} \right ) ^{nt}\)

Dividing both sides of the formula by \(\displaystyle P_{0}\), this is

\(\displaystyle \left (1+ \frac{r}{n} \right ) ^{nt} = \frac{P}{P_{0} }\)

The money doubles, so substitute 2 for \(\displaystyle \frac{P}{P_{0} }\). Also, set \(\displaystyle r = 0.07\) and \(\displaystyle n = 6\) (bimonthly), and calculate \(\displaystyle t\) as follows:

\(\displaystyle \left (1+ \frac{0.07}{6} \right ) ^{6t} = 2\)

\(\displaystyle \ln \left [ \left (1+ \frac{0.07}{6} \right ) ^{6t} \right ]= \ln 2\)

\(\displaystyle 6t \ln \left (1+ \frac{0.07}{6} \right ) = \ln 2\)

\(\displaystyle t = \frac{\ln 2}{ 6 \ln \left (1+ \frac{0.07}{6} \right ) }\)

\(\displaystyle \approx \frac{\ln 2}{ 6 \ln \left (1.011666667 \right ) }\)

\(\displaystyle \approx \frac{0.693147181 }{ 6 (0.011599136) }\)

\(\displaystyle \approx 9.960\)

\(\displaystyle 0.960 \times 12 \approx 11.52\), so this is 9 years and 11.52 months; rounding up to the next bimonth, this is 10 years even.

To find the value of \(\displaystyle n\) when,

\(\displaystyle 6\times 7\equiv n\ (\text{mod } 5)\) first multiply six and seven together.

\(\displaystyle 6\times 7=42\)

Now, recall that mod means the remainder after division occurs.

In this case

\(\displaystyle 6\times 7\equiv n\ (\text{mod } 5)\)

       \(\displaystyle 8\)

\(\displaystyle 5)\overline{42\ }\)

 \(\displaystyle -40\)

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       \(\displaystyle 2\)

Therefore, the remainder is two.

\(\displaystyle n=2\)

This question is testing the matrix operation. Remember to use order of operations and perform the algebraic operation that is inside the parentheses first.

\(\displaystyle (\%\ \#\ \%)\ \$\ \%\)

First look at the # table.

\(\displaystyle \begin{tabular}{ c|cc } \# & \%& * \\ \hline \% & *& \% \\ * & \% & * \\ \end{tabular}\)

Multiplying % by % results in *.

Now go to the $ table and multiply * by %.

\(\displaystyle \begin{tabular}{ c|cc } \$ & * & \% \\ \hline * & \%& * \\ \% & * & * \\ \end{tabular}\)

\(\displaystyle *\ \$\ \%=*\)

Therefore,

\(\displaystyle (\%\ \#\ \%)\ \$\ \%=*\)

To find the value of \(\displaystyle n\) when,

\(\displaystyle 3\times 9\equiv n\ (\text{mod } 4)\) first multiply three and nine together.

\(\displaystyle 3\times 9=27\)

Now, recall that mod means the remainder after division occurs.

In this case

\(\displaystyle 3\times 9\equiv n\ (\text{mod } 4)\)

       \(\displaystyle 6\)

\(\displaystyle 4)\overline{27\ }\)

 \(\displaystyle -24\)

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       \(\displaystyle 3\)

Therefore, the remainder is three.

\(\displaystyle n=3\)

The determinant of a two-by-two matrix

\(\displaystyle A= \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}\)

can be found by evaluating the expression

\(\displaystyle \det A = a_{11}a_{22} - a_{12}a_{23}\)

Substitute the corresponding elements to get

\(\displaystyle \det A = 3(4)- (-1)(-2) = 12 - 2 = 10\).

One way to identify whether the matrix is consistent and independent is to form a matrix of its variable coefficients, and calculate its determinant. The matrix is

\(\displaystyle A = \begin{bmatrix} 1 & 8 \\ -4 & 32 \end{bmatrix}\)

The determinant of a two-by-two matrix

\(\displaystyle A= \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}\)

can be found by evaluating the expression

\(\displaystyle \det A = a_{11}a_{22} - a_{12}a_{23}\)

Substitute the corresponding elements to get

\(\displaystyle \det A = 1(32)- 8(-4) = 32 -(-32) = 64\)

Since \(\displaystyle \det A \ne 0\),

it follows that the system is consistent and independent.