First, we need to find the points of intersection between f(x) and g(x) by setting them equal to one another and solving.

f(x) = g(x)

2x² - 3x + 1 = 13 – x 

Add x to both sides.

2x² – 2x + 1 = 13

Subtract 13 from both sides.

2x² – 2x – 12 = 0.

Divide by two to make the coefficients easier to work with.

x² – x – 6 = 0

Factor.

(x – 3)(x + 2) = 0

Set each of the factors equal to zero and then solve.

x – 3 = 0

x = 3

x + 2 = 0

x = –2

The two functions intersect where x = –2 and where x = 3. 

The question asks us to find the distance between the points of intersection. Therefore, we will need to find the y-coordinates of the points of intersection when x = –2 and when x = 3.

When x = –2, f(–2) = g(–2) = 13 – (–2) = 15.

When x = 3, f(3) = g(3) = 13 – 3 = 10.

Thus, the points of intersection are (–2, 15) and (3, 10).

We can now use the distance formula given below.

The answer is 5√2

We can solve this system of equations by using substitution. Rewriting the first equation, we get x = –5 + 3y. This equation gets substituted into the second equation, then solve for y.  Once we know what y is, we can substitute the value into the first equation to find x. In this case, x = 1 and y = 2.

Define the variables as

x = # of dimes

x + 2 = # of quarters

x + x + 2 = # of nickels

In general, the formula for money problems in V₁N₁ + V₂N₂ + V₃N₃ = $total

0.10x + 0.25(x + 2) + 0.05(2x + 2) = 1.05

Solving the equation we see that there is one dime, three quarters and four nickels.

This problem involves a system of two equations. The first equation is x² – y² = 20, and the second equation is x + y = 10. Let us solve the second equation in terms of y, and then we can substitute this value into the first equation.

x + y = 10

Subtract y from both sides.

x = 10 – y

Substitute 10 - y for x in the first equation.

x² – y² = 20

(10 - y)² – y² = 20

We can use the FOIL method to find (10 – y)². 

(10 – y)² = (10 – y)(10 – y) = 10(10) – 10y – 10y + y² = 100 –20y + y².

Now we can go back to our original equation and replace (10 – y)² with 100 – 20y + y^2.

(100 – 20y + y²) – y² = 20

100 – 20y = 20

Subtract 100 from both sides.

–20y = –80

Divide both sides by –20.

y = 4.

Now that we know that y = 4, we can use either of our original two equations to solve for x. Using the equation x + y = 10 is probably simpler.

x + y = 10

x + 4 = 10

x = 6.

The original question asks for the product of x and y, which would be 4(6), which equals 24.

The answer is 24. 

The correct answer is 5 ¹/₃. The problem is solved by substitution. The first step is to substitute x – 4 into the second equation. Then we have 2x + 4(x – 4) = 16. Next step 2x + 4x – 16 = 16. Then 6x = 32. We then divide 32 by 6 for X and get 5 ¹/₃. 

The question asks us to find the value of –x – 5z, which doesn't include any y terms. Therefore, we need to eliminate y terms from our equations. One way to do this is to solve for y in the second equation and substitute this value into the first one. 

y – 2z = 6

Add 2z to both sides.

y = 6 + 2z

Now, we take 6 + 2z and substitute this in for y in the first equation.

x + 2(6 + 2z) + z = 5

Distribute. 

x + 12 + 4z + z = 5

x + 5z + 12 = 5

Subtract 12 from both sides.

x + 5z = –7

The original question asks for the value of –x – 5z, which is equal to –1(x + 5z). Let's multiply both sides of the equation x + 5z = –7 by negative one.

–1(x + 5z) = –7

–x – 5z = 7

The answer is 7. 

Let t and s represent Tom's and Susan's current ages, respectively.

We are told that six years ago, Tom's age is twice as large as Susan's. We could represent Tom's age six years ago as t – 6, and we could represent Susan's as s – 6. Because t – 6 is twice as large as s – 6, we could write the following equation:

t – 6 = 2(s – 6)

Additionally, we are told that thirteen years ago, Tom was three times as old as Susan. Thirteen years ago, Tom's age would be t – 13, and Susan's would be s – 13. We can then write the following equation: 

t – 13 = 3(s – 13)

We now have two equations and two unknowns. In order to solve this system of equations, we could solve for t in the first equation and substitute this value into the second equation.

t – 6 = 2(s – 6)

Distribute.

t – 6 = 2s – 12

Add six to both sides.

t = 2s – 6

Next, we will substitute 2s - 6 into the second equation.

(2s – 6) – 13 = 3(s – 13)

Distribute.

2s – 6 – 13 = 3s – 39

Combine constants.

2s – 19 = 3s – 39

Subtract 2s from both sides.

–19 = s – 39

Add 39 to both sides.

s = 20

Since t = 2s – 6, t = 2(20) – 6 = 34

This means that Tom is currently 34, and Susan is currently 20. The question asks us how many years older Tom is than Susan, which is 34 – 20 = 14 years.

The answer is 14.

We are asked to find (x – y)². Let's expand (x – y)² using the FOIL method.

(x – y)² = (x – y)(x – y) = x(x) – x(y) – y(x) + y(y) = x² – 2xy + y²

In other words, we need to find the value of x² – 2xy + y². We can use the given information to find the values of x² + y² and –2xy. Then, if we combine the values of x² + y² and –2xy, we will have the value of x² – 2xy +y², which is equal to (x – y)².

The problem states that (x² + y²)^(1/2) = 4. If we were to square both sides of the equation, we can find the value of x² + y². 

((x² + y²)^(1/2))² = 4² = 16

If we use the property of exponents that states that (a^b)^c = a^bc, then ((x² + y²)^(1/2))² becomes (x² + y²)^2(1/2) = x² + y².

Thus, x² + y² = 16.

The second piece of given information states that 4xy = 4. If we divide both sides of the equation by –2, we will have –2xy on the left side.

4xy = 4

Divide both sides by –2.

–2xy = –2

Finally, we will add x² + y²  + –2xy.

x² + y²  + –2xy = 16 + –2 = 14

The answer is 14. 

First, let's expand (x + y)² by using the FOIL method.

(x + y)² = (x + y)(x + y)

According to the FOIL method, we will multiply the first terms of the binomials, then the outer terms, then the inner terms, and then the last terms. We will then add these four products together.

(x + y)(x + y)= x(x) + x(y) + y(x) + y(y) = x² + 2xy + y²

We are told that (x + y)² = 15. Let's replace (x + y)² with x² + 2xy + y².

x² + 2xy + y² = 15

Rearranging, we can write the equation as follows:

x² + y² + 2xy = 15

The second part of the problem tells us that x² + y² = 27. Thus, we can replace x² + y² with 27.

27 + 2xy = 15

Subtract 27 from both sides.

2xy = –12

Divide by two.

xy = –6

The question asks us for x²y². According to one of the properties of exponents, (xy)² = x²y². Thus, if we square both sides of the equation xy = –6, we will obtain the value of x²y².

(xy)² = (–6)²

x²y² = 36

The answer is 36.

First, we need to find points A and B, which we are told form the points of intersection between the graphs y = 9 – x² and y = 3 – x. In order to solve for these two equations, we can set the value of y in the first equation equal to the value of y in the second and then solve for x.

9 – x² = 3 – x

Add x² to both sides.

9 = 3 – x + x²

Subtract 9 from both sides. Then rearrange so that the powers of x are in descending order.

-6 – x + x² = x² – x – 6 = 0

Factor x² – x – 6 by thinking of two numbers that multiply to give –6 and add to give –1. Those two numbers are –3 and 2. 

x² – x – 6 = (x – 3)(x + 2) = 0

Set each factor equal to zero and solve.

x – 3 = 0

x = 3

x + 2 = 0

x = –2

Thus, the points of intersection occur where x = –2 and 3. We can find the y values of the points of intersection by substituting –2 and 3 into either equation. Let's use the equation y = 3 – x.

When x = –2, y = 3 – (–2) = 5. One point of intersection is (–2,5).

When x = 3, y = 3 – 3 = 0. The other point of intersection is (3,0).

Let's assume point A is at (–2,5) and that B is at (3,0). We are told that C is located at (p,0), where p < 0. Let's draw triangle ABC with the information we have so far.

In the figure above, the orange line represents the height from side BC to A.

The area of any triangle is (1/2)bh, where b is the length of the base, and h is the length of the height. We will use BC to represent the base, and the orange line to represent the height.

The length of BC will be equal to 3 – p, since both points lie on the x-axis. The length of the orange line is the distance from CB to point A, which is 5. We can now find a formula for the area and set it equal to 50.

Area of ABC = (1/2)(3 – p)(5) = 50

Multiply both sides by 2.

(3 – p)(5) = 100

Divide by 5.

3 – p = 20

Subtract 3 from both sides. 

–p = 17

Multiply both sides by –1.

p = –17.

The answer is –17.