Setting x and the number of quarters he has and y as the numbver of nickels. x + y = 14 (total coins), 0.25x + 0.05y = 1.50 (total amount). Substituting x = 14 – y from the first equation into the second, we get y = 10. Therefore Joey has 10 nickels.

Setting t as the time elapsed we need to find when 8t = 12 + 4t (this is the distance traveled by the ball compared to the distance traveled by the player+difference from origin). Solving for t we get a travel time of 3 seconds. If the player runs for 3 seconds at 4m/s, the player travels 12m before receiving the ball.

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.