SAT dread.

1. If n and p are integers greater than 1 and if p is a factor of both n+3 and n+10, what is the value of p?
a. 3 b.7. c.10 d. 13. e 30

2. The company sends a team of 3 plumbers to work on a certain job. The company has 4 plumbers and 4 trainees. 1 team has 1 plumber and 2 trainees, how many different such teams possible?

3. In xy plane, the graph of x=y*y-4 intersects line l at (0,p) and (5,t). What is the greatest possible value of the slope of l?

Thanks in advance!

1. To find the value of p, we need to find a number that is a factor of both n+3 and n+10. One approach is to use the property that if a number is a factor of two numbers, it is also a factor of their difference. So, we can subtract the two given expressions:

(n+10) - (n+3) = n - n + 10 - 3 = 7.

Therefore, p must be a factor of 7. The only integer factors of 7 are 1 and 7. Since the question states that p is greater than 1, the correct answer is b. 7.

2. We need to form teams with 1 plumber and 2 trainees. Since we have 4 plumbers and 4 trainees, we can select 1 plumber in 4 ways and 2 trainees in 4C2 = 6 ways. The total number of teams is the product of these two numbers:

Number of teams = 4 (number of ways to select a plumber) * 6 (number of ways to select 2 trainees)
= 24.

Therefore, there are 24 different such teams possible.

3. Let's consider the intersection points of the graph of x=y^2-4 with line l at (0,p) and (5,t). We can find the slope by using the formula:

slope = (change in y) / (change in x)

The change in y is given by t - p, and the change in x is 5 - 0 = 5.

To maximize the slope, we need to maximize the value of t - p. Since t is the y-coordinate of the point of intersection at x = 5 and p is the y-coordinate of the point of intersection at x = 0, we can maximize the slope by maximizing the value of t.

To find the maximum value of t, we can substitute x = 5 into the equation x = y^2 - 4:

5 = y^2 - 4.

Solving this equation, we find y = ±3. Taking the positive value, y = 3.

Substituting y = 3 into the equation x = y^2 - 4, we find x = 5.

Therefore, the point of intersection at (5,t) is (5,5).

The maximum value of the slope is (t - p) / 5 = (5 - p) / 5. Since p is a factor of y^2 - 4, it must be an integer. The greatest slope occurs when p is minimized, which happens when p = 3.

Substituting p = 3, the maximum slope is (5 - 3) / 5 = 2 / 5.

Therefore, the greatest possible value of the slope of line l is 2/5.