Two consecutive numbers can be represented by x and x - 1. the product of two consecutive numbers is equal to 35 more than eleven times the sum of the numbers. a. write a quadratic equation to represent the situation. b. what are the solutions to the equation? c. what are the two consecutive numbers?

a) The numbers are represented by x and x-1, so their product is x(x-1).

Their sum will be (x +(x-1)), and 11 times that is 11(x +(x-1)). The quantity that is 35 more than this value is 35+11(x +(x-1)).

The problem statement tells us these two functions of the numbers are equal, so we can write ...

... x(x-1) = 35 + 11(x +(x-1))

b) When simplified and put in standard form, the equation is ...

... x² -x = 35 +22x -11

... x² -23x -24 = 0 . . . . . subtract the right side from both sides

... (x -24)(x +1) = 0  . . . . factor

Solutions are the values of x that make these factors zero: x=24, x=-1.

c) Since the numbers are x and x-1, the solution x=24 means the numbers are 24 and 24-1 = 23. The solution x=-1 means the numbers are -1 and -1-1 = -2.

The numbers are {23, 24} or {-2, -1}.

Check

For 23, 24

The product is 23·24 = 552

The sum is 23+24 = 47, and 11 times that is 517. 35 more than this is 552, so this answer checks OK.

For -2, -1

The product is (-2)·(-1) = 2.

The sum is -2-1 = -3, and 11 times that is -33. 35 more than this is 35-33 = 2, so this answer checks OK.

h(x) * s(x) = 200(1.05)^(x - 1)

Step-by-step explanation:

Our interest equation is s(x) = (1.05)^(x - 1). This is actually a part of a bigger formula for calculating the amount of money accumulated including interest:

A = P(1 + r)^n, where A is the total, P is the principal amount (initial amount), r is the interest rate, and n is the time

Here, we technically already have the (1 + r)^n part; it's just (1.05)^(x - 1). The principle, though, will actually be the 200 because she starts out at $200.

Thus, to combine these, we simply multiply them together to get:

h(x) * s(x) = 200(1.05)^(x - 1)