Well as x can never actually be -1 because I'm in the denominator -1 + 1 = 0 and we cannot divide by zero. But we can look at what number it approaches and i assume that is the relative value. sometimes functions will have asymptotes and others will have holes in the graph. this one would have an asymptote going down at a rapid rate. the asymptote would go on forever getting infinitely close to -1 but never touching. So I would say since the asymptote goes down forever that the graph approaches negative infinity
Answer:
B
Step-by-step explanation:
Because the numbers are less than 45 and more than 40
Answer:
x = π/3, x = 5π/3, x = 4π/3
Step-by-step explanation:
Let's split the given equation (2cosx-1)(2sinx+√3 ) = 0 into two parts, and solve each separately. These parts would be 2cos(x) - 1 = 0, and 2sin(x) + √3 = 0.
Remember that the general solutions for cos(x) = 1/2 are x = π/3 + 2πn and x = 5π/3 + 2πn. In this case we are given the interval 0 ≤ x ≤2π, and therefore x = π/3, and x = 5π/3.
Similarly:
The general solutions for sin(x) = - √3/2 are x = 4π/3 + 2πn and x = 5π/3 + 2πn. Therefore x = 4π/3 and x = 5π/3 in this case.
So we have x = π/3, x = 5π/3, and x = 4π/3 as our solutions.
Answer:
<u>This assume the equation is 35y = mx + c, and that r is meant to be x.</u>
Step-by-step explanation:
35y = mx + c
35y = -2(-7) - 3
35y = 14 - 3
35y = 11
y = (11/35)
Please check the formatting of your question.
Answer:
25
Step-by-step explanation:
