50PTS!
Describe the vertical asymptotes and holes for the graph of y = x-5/x^2 -1
2 answers:
Answer:
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Step-by-step explanation:
Ok
first of all, for q(x)/p(x)
if the degree of q(x) is less than the degree of p(x),then the horizontal assemtote is 0
then simplify
any factors you factored out is now a hole, remember them
to find the vertical assemtotes of a function, set the SIMPLIFIED denomenator equal to 0 and solve
so
y=(x-5)/(x^2-1)
q(x)<p(x)
horizontal assemtote is y=0
no factors to simplify so no holes
set denomenator to 0 to find vertical assemtote
x^2-1=0
(x-1)(x+1)=0
x-1=0
x=1
x+1=0
x=-1
the horizontal assemtotes are x=1 and -1
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<span>{(1, 16), (2, 64), (3, 144), (4, 256), (5, 400)} Hope this helps </span>
Answer:
The last one (4 sqare root 3)
Step-by-step explanation:
The square root of 16 is 4 so then it would be 4 sqroot 3
Answer:
yes
Step-by-step explanation:
y=-16r^2+63r+4
maximum value of y :
y'=0
-32r +63=0
r= 63/32
y = -16(63/32)²+63(63/32)+4
= -4 + 124 + 4 = 124 ft
124 > 68
the Arrow can pass over the tree
Y – y1 = m(x – x1)
y-(-1)=-5(x-2)
Y+1=-5(x-2)
5,500 millimeters= 0.0055 liters