For vertical asymptotes, find the values which make the function indetermine in this case x=-7,so this is the only vertical asymptote.
For horizontal asymptotes, find the limit when x tends to infinity:
=(5x/x-15/x)/(2x/x+14/x) = 5/2, this is the horizontal asymptote y=5/2
For obliques, you have to meet the degree of the numerator is exactly a greater degree than the denominator, in this case they are the same degree so no oblique asymptote.
Great question! A midpoint is exactly what it sounds like; the point in the middle of two objects (in this case it's the middle of these two coordinates). Another way to think of the middle of something in math is by using the average. To find the midpoint of two points, we just find the average of the x-values and the average of the y-values:
((x1 + x2)/2 , (y1 + y2)/2)
((5 + -13) / 2 , (-4 + 12) / 2)
( -8 / 2 , 8 / 2)
(-4 , 4)
Answer:
Step-by-step explanation:
♨Rage♨
Answer:
k = 32/ cos 64
Step-by-step explanation:
Since this is a right triangle, we can use trig functions
cos theta = adj side / hypotenuse
cos 64 = 32/k
k cos 64 = 32
k = 32/ cos 64
Answer:
b i thingk
Step-by-step explanation:
adsfdsf