What is X+Y?? and ( x+ y)- 7+8

Step-by-step explanation:

What is X+Y?

First note that if x + y = xy then neither x nor y can be equal to 1.

That being said, as a previous user already noted y = x/(x-1) and hence x - y = x - x(x-1) = x(x-2)/(x-1). Assume x - y = D. Then your question becomes what can D be so that the equation x(x-2)/(x-1) = D has a solution different than 1.

Rewrite this equation as x(x-2) = D(x-1) and note that x(x-2) = (x-1)^2 - 1. Replace x - 1 with z. So your question is what can D be so that the equation z^2 - 1 = Dz has a nonzero solution? Given that 0 cannot be a solution anyway of the previous equation. The question is really, for which D does the quadratic equation z^2 - Dz - 1 = 0 have solutions? By elementary 9th grade math, this equation has a solution when its discriminant D^2 + 4 is nonnegative. But D^2 + 4 is strictly positive regardless of the value of D. Hence D can be anything.

How to solve your question

Your question is

(+)−7+8

Simplify

1. Eliminate redundant parentheses

(+)−7+8

+−7+82

Add the numbers

+−7+8

++1

Solution

++1

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

=====================================

part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

=====================================

part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).