Answer:
(a) There are asymptotes at x=3/2 and x=-1/3
Step-by-step explanation:
The denominator zeros can be found by factoring:
f(x) = (x +1)/((2x -3)(3x +1))
Neither of the denominator factors is cancelled by the numerator factor, so each represents a vertical asyptote, not a function hole.
The asymptotes are at the values of x where the denominator is zero:
2x -3 = 0 ⇒ x = 3/2
3x +1 = 0 ⇒ x = -1/3
Answer:
See below
Step-by-step explanation:
Polynomials are sums of terms in the form of k⋅xⁿ, where k is any number and n is a positive integer. For example, 3x²+2x-5 is a polynomial.
Based on your problem, you only have one given. So, you can't make an equation for this because there are not limits to the equation. The only thing that you know is that the three numbers are consecutive even integers. My way of solution for this is trial-and-error. However, it's really quite easy.
For example: 42 + 46 = 88. I have to increase the numbers more to reach 136. Suppose: 82 + 86 = 168. That exceeded 136. So, it must be between 46 and 82. Suppose again: 66 + 70 = 136. Therefore, the sequence of the consecutive even integers are 66, 68, and 70.
It is <em>1/100 . It is simple </em>
