First, we need to solve for $\displaystyle \dpi{100} \small x$ from the first equation in order to calculate the second quadratic function. To solve for $\displaystyle \dpi{100} \small x$, we need to subtract four on each side of the equation, then we will get

$\displaystyle \dpi{100} \small 5x=15$

The answer for $\displaystyle \dpi{100} \small x$ would be $\displaystyle \dpi{100} \small \frac{15}{5}$, which is $\displaystyle \dpi{100} \small 3$.

So now we can calculate the function by plugging in $\displaystyle \dpi{100} \small x=3$.

$\displaystyle \dpi{100} \small 3^{2}=9$, and $\displaystyle \dpi{100} \small 9\times 4=36$.

$\displaystyle \dpi{100} \small 36-5=31$

The equation that will be used is (rate * number of days = number of fields plowed). From the first part of the question, number of fields plowed $\displaystyle \dpi{100} \small (x)$ is calculated as:

$\displaystyle rate \cdot 5 = x$

To solve for rate both sides are divided by 5.

$\displaystyle rate = \frac{x}{5}$

This rate is used for the second part of the problem. $\displaystyle \frac{x}{5} * days = y$. To solve for days, both sides are divided by $\displaystyle \frac{x}{5}$, which is the same as multiplying by$\displaystyle \frac{5}{x}$, cancelling out the $\displaystyle \frac{x}{5}$ and giving the answer of $\displaystyle days = \frac{5y}{x}$.

Let $\displaystyle \dpi{100} \small x$ = length of the short plank and $\displaystyle \dpi{100} \small x+30$ = length of the long plank.

$\displaystyle \dpi{100} \small x+x+30=70$ is the length of the pre-cut board, or combined length of both planks.

$\displaystyle \dpi{100} \small 2x+30=70$

$\displaystyle \dpi{100} \small 2x=40$

$\displaystyle \dpi{100} \small x=20$

$\displaystyle 2x^{2} - 8x - 24 = 2(x^{2}-4x-12)=0$

Divide both sides by 2: $\displaystyle x^{2}-4x-12=0$. We need to find two numbers that multiply to $\displaystyle \dpi{100} \small -12$ and sum to $\displaystyle \dpi{100} \small -4$. The numbers $\displaystyle \dpi{100} \small -6$ and $\displaystyle \dpi{100} \small 2$ work.

$\displaystyle x^{2}-4x-12= (x-6)(x+2)=0$

$\displaystyle \dpi{100} \small x-6=0$

$\displaystyle \dpi{100} \small x=6$

$\displaystyle \dpi{100} \small or\ x+2=0$

$\displaystyle \dpi{100} \small x=-2$

Multiply both sides of the equation by a: $\displaystyle a-4=2a-7$

Then, solve for $\displaystyle a$.

We need to solve for x in terms of y by isolating x.

$\displaystyle 3x=4y+2$

$\displaystyle x = \frac{4}{3}y + \frac{2}{3}$

We need to plug in the answer choices and see which produce the value 4.

1. $\displaystyle x=3,y=2,z=1:3+2(2)-3(1)=3+4-3=4$, correct

2. $\displaystyle x=1,y=2,z=3:1+2(2)-3(3)=1+4-9=-4$, incorrect

3. $\displaystyle x=8,y=1,z=2:8+2(1)-3(2)=8+2-6=4$, correct

4. $\displaystyle x=2,y=-1,z=5:2+2(-1)-3(5)=2-2-15=-15$, incorrect

Therefore two of the answer choices are correct.

To solve this, we need to plug the answer choices into the equation and see which choice does NOT work.  Let's go through the answer choices.

$\displaystyle \dpi{100} (2,4,6):2(2)-4(4)+6=-6$

This is the correct answer.  If this had been a solution to the equation, the equation would have produced 8.

$\displaystyle \dpi{100} (\frac{11}{2},1,1):2(\frac{11}{2}) -4(1)+1)=8$

$\displaystyle (6,1,0):2(6)-4(1)+0=8$

$\displaystyle (\frac{7}{2},0,1):2(\frac{7}{2})-4(0)+1=8$

$\displaystyle (4+2a-\frac{b}{2}):$ 

Let's let $\displaystyle a=1$ and $\displaystyle b=0$.  Then this ordered pair becomes $\displaystyle (6,1,0)$. 

$\displaystyle 2(6)-4(1)+0=8$.  Any other answer choices of this form will also work.

$\displaystyle x\cdot(x+k)=-7$ is the same as $\displaystyle x^2 + kx + 7 = 0$.

For solutions sets of integers, we are able to divide it as: $\displaystyle (x+\_\_)(x+\_\_) = 0$

When we multiply this, we notice that the constants must return 7.  Since 7 is a prime number, we now know that these constants are 1 and 7.  The signs of these can still switch however.  If both of the numbers are negative, it will also return a product of positive 7.  These are the only ways to make the constant term work.  If they are both positive numbers, then we get $\displaystyle k=8$. If they're both negative numbers then $\displaystyle k= -8$.

For the values of $\displaystyle x$, we simply notice the only way for the overall product to be 0 is for one (or both) of the pieces to be zero.  This is done by having $\displaystyle x$ equal the additive inverse of the constant piece.

Thus we have 2 integer solution sets: $\displaystyle $\left\{1,7\right\}$$ and $\displaystyle $\left\{-1,-7\right\}$$

The following variation equation can be set up: 

$\displaystyle \frac{V_{1} P _{1}}{T_{1}} = \frac{V_{2} P _{2}}{T_{2}}$

But since the initial volume and the current volume are equal, or, equivalently,  $\displaystyle V_{1} = V_{2}$ , 

$\displaystyle \frac{V_{1} P _{1}}{T_{1}} \div V_{1} = \frac{V_{2} P _{2}}{T_{2}}\div V_{2}$

so

$\displaystyle \frac{P _{1}}{T_{1}}= \frac{P _{2}}{T_{2}}$

We substitute $\displaystyle P _{1}= 104, T_{1}=295 ,T_{2}=305$, and solve for $\displaystyle P _{2}$:

$\displaystyle \frac{104}{295}= \frac{P _{2}}{305}$

$\displaystyle \frac{104}{295}\cdot 305 = \frac{P _{2}}{305} \cdot 305$

$\displaystyle P_{2} = \frac{104}{295}\cdot 305 \approx 108 \textrm{ mbar}$