Hey hey hey I see no picc
Answer:
The Least Common Denominator of 3/4, 4/5 and 2/3
Would be,
4 × 5 × 3 = 60
<em><u>Hence</u></em><em><u>,</u></em>
<em><u>60</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>L.C.D</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>3</u></em><em><u>/</u></em><em><u>4</u></em><em><u>,</u></em><em><u> </u></em><em><u>4</u></em><em><u>/</u></em><em><u>5</u></em><em><u> </u></em><em><u>and</u></em><em><u> </u></em><em><u>2</u></em><em><u>/</u></em><em><u>3</u></em>
-.5 or negative 1/-2
same thing just in a dec and frac
You need to use pemdas so first you do inside the parenthesis 6+21=27 then you add 4 27+4=31 that is your answer
The function
... y = 1/x
has derivative
... y' = -1/x²
which has no zeros. It is undefined at x=0, the only critical point. The derivative is negative for all values of x, so the function is decreasing everywhere in its domain.
Your function
... y = (x+1)/(x-3)
can be written as
... y = 1 +4/(x-3)
which is a version of y = 1/x that has been vertically scaled by a factor of 4, then shifted 1 unit up and 3 units to the right. Shifting the function to the right means x=3 is excluded from the domain (and the interval on which the function is decreasing).
The critical point is x=3.
The function is decreasing on (-∞, 3) ∪ (3, ∞), increasing nowhere.