Solve for x over the real numbers:
1000 (x + 1)^12 = 2000
Divide both sides by 1000:
(x + 1)^12 = 2
Take the square root of both sides:
(x + 1)^6 = sqrt(2) or (x + 1)^6 = -sqrt(2)
Take the square root of both sides:
(x + 1)^3 = 2^(1/4) or (x + 1)^3 = -2^(1/4) or (x + 1)^6 = -sqrt(2)
Take cube roots of both sides:
x + 1 = 2^(1/12) or (x + 1)^3 = -2^(1/4) or (x + 1)^6 = -sqrt(2)
Subtract 1 from both sides:
x = 2^(1/12) - 1 or (x + 1)^3 = -2^(1/4) or (x + 1)^6 = -sqrt(2)
Take cube roots of both sides:
x = 2^(1/12) - 1 or x + 1 = -2^(1/12) or (x + 1)^6 = -sqrt(2)
Subtract 1 from both sides:
x = 2^(1/12) - 1 or x = -1 - 2^(1/12) or (x + 1)^6 = -sqrt(2)
(x + 1)^6 = -sqrt(2) has no solution since for all x on the real line, (x + 1)^6 = (x + 1)^6 >=0 and -sqrt(2)<0:
Answer: x = 2^(1/12) - 1 or x = -1 - 2^(1/12)
Answer:
Step-by-step explanation:
we know that
the formula to calculate the distance between two points is equal to
we have

substitute the values in the formula
So, if each one costs $54.99
Let "X" be the number of games sold, Let R(X) be the revenue

This function is continuous
Do you have a problem? Parallel lines are like "twins" next to each other, and perpendicular will cross (for example : ∉)