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.

well the 1 and -1 are always after the 0 and the 0 is inthe middle. s thats what i would guess. but i dont know what a slope of 2 is sorry

step-by-step explanation:

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}